Search Colleges, Exams, Schools & more
Login
RD Sharma Class 12 Exercise 5.2 Determinant Solutions Maths - Download PDF Free Online
The RD Sharma Solution books are the best form of tutors for the class 12 students at their home. Mathematics is a subject that gives chances for a lot of doubts to arise. Every mark that a student scores in the public exam is essential; students should not risk losing their marks. RD Sharma class 12th exercise 5.2, Determinants, is a part where students can score easily. Hence utmost care should be taken in solving the problems. With the help of the RD Sharma Class 12th exercise 5.2, students can do their homework effortlessly.
Also Read - RD Sharma Solution for Class 9 to 12 Maths
RD Sharma Class 12 Solutions Chapter 5 Determinants - Other Exercise
- Chapter 5 - Determinants Ex 5.1
- Chapter 5 - Determinants Ex 5.3
- Chapter 5 - Determinants Ex 5.4
- Chapter 5 - Determinants Ex 5.5
- Chapter 5 - Determinants Ex FBQ
- Chapter 5 - Determinants Ex MCQ
- Chapter 5 - Determinants Ex VSA
determinants Exercise 5.2 Question 1
Answer:$\left|\begin{array}{ccc} 1 & 3 & 5 \\ 2 & 6 & 10 \\ 31 & 11 & 38 \end{array}\right|=0$Hint: We will solve the determinant with respect to row one.
Given:$\left|\begin{array}{ccc} 1 & 3 & 5 \\ 2 & 6 & 10 \\ 31 & 11 & 38 \end{array}\right|$
Solution:$\left|\begin{array}{ccc} 1 & 3 & 5 \\ 2 & 6 & 10 \\ 31 & 11 & 38 \end{array}\right|$
$\begin{aligned} &=1\left|\begin{array}{ll} 6 & 10 \\ 11 & 38 \end{array}\right|-3\left|\begin{array}{cc} 2 & 10 \\ 31 & 38 \end{array}\right|+5\left|\begin{array}{cc} 2 & 6 \\ 31 & 11 \end{array}\right| \\ &=1(6 \times 38-10 \times 11)-3(2 \times 38-10 \times 31)+5(2 \times 11-6 \times 31) \\ &=(228-110)-3(76-310)+5(22-186) \\ &=118-3(-234)+5(-164) \\ &=118+702-820 \\ &=820-820 \\ &=0 \end{aligned}$
determinants Exercise 5.2 Question 1 Sub Question 2
Answer:
Answer: $\left|\begin{array}{lll} 67 & 19 & 21 \\ 39 & 13 & 14 \\ 81 & 24 & 26 \end{array}\right|=-43$Hint: First we will use column operation to get a simple determinant.
Given: $\left|\begin{array}{lll} 67 & 19 & 21 \\ 39 & 13 & 14 \\ 81 & 24 & 26 \end{array}\right|$
Solution: $\left|\begin{array}{lll} 67 & 19 & 21 \\ 39 & 13 & 14 \\ 81 & 24 & 26 \end{array}\right|$
$\begin{aligned} &\text { Applying } \mathrm{C}_{1} \rightarrow C_{1}-4 C_{3} \\ &=\left|\begin{array}{ccc} -17 & 19 & 21 \\ -17 & 13 & 14 \\ -23 & 24 & 26 \end{array}\right| \\ &\text { Applying } \mathrm{R}_{1} \rightarrow R_{1}-R_{2} \\ &=\left|\begin{array}{ccc} 0 & 6 & 7 \\ -17 & 13 & 14 \\ -23 & 24 & 26 \end{array}\right| \\ &\text { Applying } \mathrm{R}_{2} \rightarrow R_{2}-2 R_{1} \\ &\left|\begin{array}{ccc} 0 & 6 & 7 \\ -17 & 1 & 0 \\ -23 & 24 & 26 \end{array}\right| \end{aligned}$
on expanding w.r.t row
$\begin{aligned} &=0\left|\begin{array}{cc} 1 & 0 \\ 24 & 26 \end{array}\right|-6\left|\begin{array}{cc} -17 & 0 \\ -23 & 26 \end{array}\right|+7\left|\begin{array}{cc} -17 & 1 \\ -23 & 24 \end{array}\right| \\ &=0-6(-17 \times 26-0 \times-23)+7(-17 \times 24-1 \times-23) \\ &=-6(-442)+7(-408+23) \\ &=2652+7(-385) \\ &=2652-2695 \\ &=-43 \end{aligned}$
determinants Exercise 5.2 Question 1 Sub Question 3
Answer:$\left|\begin{array}{lll} a & h & g \\ h & b & f \\ g & f & c \end{array}\right|$$\left|\begin{array}{lll} a & h & g \\ h & b & f \\ g & f & c \end{array}\right|=a b c-a f^{2}-c h^{2}+2 f g h-b g^{2}$ $=abc-af^{2}-ch^{2}+2fgh-bg^{2}$$\left|\begin{array}{lll} a & h & g \\ h & b & f \\ g & f & c \end{array}\right|=a b c-a f^{2}-c h^{2}+2 f g h-b g^{2}$$\left|\begin{array}{lll} a & h & g \\ h & b & f \\ g & f & c \end{array}\right|=a b c-a f^{2}-c h^{2}+2 f g h-b g^{2}$$\left|\begin{array}{lll} a & h & g \\ h & b & f \\ g & f & c \end{array}\right|=a b c-a f^{2}-c h^{2}+2 f g h-b g^{2}$Hint: We will expand it w.r.t $R_{1}$
Given: $\left|\begin{array}{lll} a & h & g \\ h & b & f \\ g & f & c \end{array}\right|=a b c-a f^{2}-c h^{2}+2 f g h-b g^{2}$$\left|\begin{array}{lll} a & h & g \\ h & b & f \\ g & f & c \end{array}\right|$
Solution: $\left|\begin{array}{lll} a & h & g \\ h & b & f \\ g & f & c \end{array}\right|$
Expanding w.r.t $R_{1}$
$\begin{aligned} &=a\left|\begin{array}{ll} b & f \\ f & c \end{array}\right|-h\left|\begin{array}{ll} h & f \\ g & c \end{array}\right|+g\left|\begin{array}{ll} h & b \\ g & f \end{array}\right| \\ &=a\left(b c-f^{2}\right)-h(h c-g f)+g(h f-b g) \\ &=a b c-a f^{2}-h^{2} c+f g h+f g h-b g^{2} \\ &=a b c-a f^{2}-c h^{2}+2 f g h-b g^{2} \end{aligned}$
determinants Exercise 5.2 Question 1 sub question 4
Answer:$\left|\begin{array}{ccc} 1 & -3 & 2 \\ 4 & -1 & 2 \\ 3 & 5 & 2 \end{array}\right|=40$Hint: First we will use row operation to make some element 0
Given: $\left|\begin{array}{ccc} 1 & -3 & 2 \\ 4 & -1 & 2 \\ 3 & 5 & 2 \end{array}\right|$
Solution:
$\begin{aligned} &\text { Applying } \mathrm{R}_{1} \rightarrow R_{1}-R_{2} \text { and } \mathrm{R}_{2} \rightarrow R_{2}-R_{3} \\ &=\left|\begin{array}{ccc} -3 & -2 & 0 \\ 1 & -6 & 0 \\ 3 & 5 & 2 \end{array}\right| \end{aligned}$
$\begin{aligned} &\text { Now we will expand it w.r.t } R_{1}\\ &=-3\left|\begin{array}{cc} -6 & 0 \\ 5 & 2 \end{array}\right|-(-2)\left|\begin{array}{cc} 1 & 0 \\ 3 & 2 \end{array}\right|+0\left|\begin{array}{cc} 1 & -6 \\ 3 & 5 \end{array}\right|\\ &=-3(-12-5 \times 0)+2(1 \times 2-3 \times 0)+0\\ &=-3(-12)+2(2)\\ &=36+4\\ &=40 \end{aligned}$
determinants Exercise 5.2 Question 1 sub question 5
Answer:$\left|\begin{array}{ccc} 1 & 4 & 9 \\ 4 & 9 & 16 \\ 9 & 16 & 25 \end{array}\right|=-8$Hint: We will expand it w.r.t $R_{1}$
Given:$\left|\begin{array}{ccc} 1 & 4 & 9 \\ 4 & 9 & 16 \\ 9 & 16 & 25 \end{array}\right|$
Solution:$\left|\begin{array}{ccc} 1 & 4 & 9 \\ 4 & 9 & 16 \\ 9 & 16 & 25 \end{array}\right|$
Expanding w.r.t $R_{1}$
$\begin{aligned} &=1\left|\begin{array}{cc} 9 & 16 \\ 16 & 25 \end{array}\right|-4\left|\begin{array}{ll} 4 & 16 \\ 9 & 25 \end{array}\right|+9\left|\begin{array}{cc} 4 & 9 \\ 9 & 16 \end{array}\right| \\ &=1(9 \times 25-16 \times 16)-4(4 \times 25-9 \times 16)+9(4 \times 16-9 \times 9) \\ &=(225-256)-4(100-144)+9(64-81) \\ &=-31+176-153 \\ &=176-184 \\ &=-8 \end{aligned}$
determinants Exercise 5.2 Question 1 sub question 6
Answer:$\left|\begin{array}{ccc} 6 & -3 & 2 \\ 2 & -1 & 2 \\ -10 & 5 & 2 \end{array}\right|=0$Hint: In this question we will convert some elements into 0
Given:$\left|\begin{array}{ccc} 6 & -3 & 2 \\ 2 & -1 & 2 \\ -10 & 5 & 2 \end{array}\right|$
Solution:
$\left|\begin{array}{ccc} 6 & -3 & 2 \\ 2 & -1 & 2 \\ -10 & 5 & 2 \end{array}\right|$
$\begin{aligned} &\text { Applying } \mathrm{R}_{1} \rightarrow R_{1}-R_{2} \text { and } \mathrm{R}_{2} \rightarrow R_{2}-R_{3} \\ &=\left|\begin{array}{ccc} 4 & -2 & 0 \\ 12 & -6 & 0 \\ -10 & 5 & 2 \end{array}\right| \end{aligned}$
On expanding it w.r.t $R_{1}$
$\begin{aligned} &=4\left|\begin{array}{ll} -6 & 0 \\ 5 & 2 \end{array}\right|-(-2)\left|\begin{array}{cc} 12 & 0 \\ -10 & 2 \end{array}\right|+0\left|\begin{array}{cc} 12 & -6 \\ -10 & 5 \end{array}\right| \\ &=4(-6 \times 2-5 \times 0)+2(12 \times 2-(-10) \times 0)+0 \\ &=4(-12)+2(24) \\ &=-48+48 \\ &=0 \end{aligned}$
Hence$\left|\begin{array}{ccc} 6 & -3 & 2 \\ 2 & -1 & 2 \\ -10 & 5 & 2 \end{array}\right|=0$
determinants Exercise 5.2 Question 1 sub question 7
Answer:$\left|\begin{array}{cccc} 1 & 3 & 9 & 27 \\ 3 & 9 & 27 & 1 \\ 9 & 27 & 1 & 3 \\ 27 & 1 & 3 & 9 \end{array}\right|=512000$Hint: First we will make a column with same elements
Given:$\left|\begin{array}{cccc} 1 & 3 & 9 & 27 \\ 3 & 9 & 27 & 1 \\ 9 & 27 & 1 & 3 \\ 27 & 1 & 3 & 9 \end{array}\right|$
Solution:$\left|\begin{array}{cccc} 1 & 3 & 9 & 27 \\ 3 & 9 & 27 & 1 \\ 9 & 27 & 1 & 3 \\ 27 & 1 & 3 & 9 \end{array}\right|$
$\begin{aligned} &\text { Applying } \mathrm{C}_{1} \rightarrow \mathrm{C}_{1}+\mathrm{C}_{2}+\mathrm{C}_{3}+\mathrm{C}_{4}\\ &=\left|\begin{array}{cccc} 40 & 3 & 9 & 27 \\ 40 & 9 & 27 & 1 \\ 40 & 27 & 1 & 3 \\ 40 & 1 & 3 & 9 \end{array}\right|\\ &\text { Applying } \mathrm{R}_{1} \rightarrow R_{1}-R_{2} \text { and } \mathrm{R}_{2} \rightarrow R_{2}-R_{3} \text { and } \mathrm{R}_{3} \rightarrow R_{3}-R_{4}\\ &=\left|\begin{array}{cccc} 0 & -6 & -18 & 26 \\ 0 & -18 & 26 & -2 \\ 0 & 26 & -2 & -6 \\ 40 & 1 & 3 & 9 \end{array}\right| \end{aligned}$
On changing all rows into columns
$=\left|\begin{array}{cccc} 0 & 0 & 0 & 40 \\ -6 & -18 & 26 & 1 \\ -18 & 26 & -2 & 3 \\ 26 & -2 & -6 & 9 \end{array}\right|$
If the rows and columns of a determinant are interchanged, the value of the determinant remains the same.
Expanding the determinant w.r.t $R_{1}$
$\begin{aligned} &=0\left|\begin{array}{ccc} -18 & 26 & 1 \\ 26 & -2 & 3 \\ -2 & -6 & 9 \end{array}\right|-0\left|\begin{array}{ccc} -6 & 26 & 1 \\ -18 & -2 & 3 \\ 26 & -6 & 9 \end{array}\right|+0\left|\begin{array}{ccc} -6 & -18 & 1 \\ -18 & 26 & 3 \\ 26 & -2 & 9 \end{array}\right|-40\left|\begin{array}{ccc} -6 & -18 & 26 \\ -18 & 26 & -2 \\ 26 & -2 & -6 \end{array}\right| \\ &=0-0+0-40\left\{-6\left|\begin{array}{ll} 26 & -2 \\ -2 & -6 \end{array}\right|-(-18)\left|\begin{array}{cc} -18 & -2 \\ 26 & -6 \end{array}\right|+26\left|\begin{array}{cc} -18 & 26 \\ 26 & -2 \end{array}\right|\right\} \\ &=-40[-6\{26(-6)-(-2)(-2)\}+18\{(-18)(-6)-(-2)(26)\}+26\{(-18)(-2)-26 \times 26\}] \\ &=-40[-6(-156-4)+18(108+52)+26(36-676)] \\ &=-40[-6(-160)+18(160)+26(-640)] \\ &=-40(960+2880-16640) \\ &=-40(-12800) \\ &=512000 \end{aligned}$
Hence,$\left|\begin{array}{cccc} 1 & 3 & 9 & 27 \\ 3 & 9 & 27 & 1 \\ 9 & 27 & 1 & 3 \\ 27 & 1 & 3 & 9 \end{array}\right|=512000$
determinants Exercise 5.2 Question 1 Sub Question 8
Answer:$\left|\begin{array}{ccc} 102 & 18 & 36 \\ 1 & 3 & 4 \\ 17 & 3 & 6 \end{array}\right|=0$Hint: We will solve the determinant w.r.t $R_{1}$
Given: $\left|\begin{array}{ccc} 102 & 18 & 36 \\ 1 & 3 & 4 \\ 17 & 3 & 6 \end{array}\right|$
Solution:$\left|\begin{array}{ccc} 102 & 18 & 36 \\ 1 & 3 & 4 \\ 17 & 3 & 6 \end{array}\right|$
$\begin{aligned} &=102\left|\begin{array}{ll} 3 & 4 \\ 3 & 6 \end{array}\right|-18\left|\begin{array}{cc} 1 & 4 \\ 17 & 6 \end{array}\right|+36\left|\begin{array}{cc} 1 & 3 \\ 17 & 3 \end{array}\right| \\ &=102(6 \times 3-3 \times 4)-18(6 \times 1-4 \times 17)+36(1 \times 3-17 \times 3) \\ &=102(18-12)-18(6-68)+36(3-51) \\ &=102(6)-18(-62)+36(-48) \\ &=612+1116-1728 \\ &=1728-1728 \\ &=0 \end{aligned}$
Hence $\left|\begin{array}{ccc} 102 & 18 & 36 \\ 1 & 3 & 4 \\ 17 & 3 & 6 \end{array}\right|=0$
determinants Exercise 5.2 Question 2
Answer:$\left|\begin{array}{ccc} 8 & 2 & 7 \\ 12 & 3 & 5 \\ 16 & 4 & 3 \end{array}\right|=0$Hint: We will try to do any two column or row equal
Given:$\left|\begin{array}{ccc} 8 & 2 & 7 \\ 12 & 3 & 5 \\ 16 & 4 & 3 \end{array}\right|$
Solution:$\left|\begin{array}{ccc} 8 & 2 & 7 \\ 12 & 3 & 5 \\ 16 & 4 & 3 \end{array}\right|$
$\begin{aligned} &\text { On taking common } 4 \text { from } \mathrm{C}_{1}\\ &=4\left|\begin{array}{lll} 2 & 2 & 7 \\ 3 & 3 & 5 \\ 4 & 4 & 3 \end{array}\right| \end{aligned}$
If any two rows or columns of a determinant are identical.
The value of the determinant is zero
$\begin{aligned} &=4 \times 0 \quad\left(\because C_{1}=C_{2}\right) \\ &=0 \end{aligned}$
Hence $\left|\begin{array}{ccc} 8 & 2 & 7 \\ 12 & 3 & 5 \\ 16 & 4 & 3 \end{array}\right|=0$
determinants Exercise 5.2 Question 2 Sub Question 2
Answer:$\left|\begin{array}{ccc} 6 & -3 & 2 \\ 2 & -1 & 2 \\ -10 & 5 & 2 \end{array}\right|=0$Hint: We will try to do any two column or row equal
Given:$\left|\begin{array}{ccc} 6 & -3 & 2 \\ 2 & -1 & 2 \\ -10 & 5 & 2 \end{array}\right|$
Solution:$\left|\begin{array}{ccc} 6 & -3 & 2 \\ 2 & -1 & 2 \\ -10 & 5 & 2 \end{array}\right|$
$\begin{aligned} &\text { On taking }(-2) \text { common from } \mathrm{C}_{1}\\ &=(-2)\left|\begin{array}{ccc} -3 & -3 & 2 \\ -1 & -1 & 2 \\ 5 & 5 & 2 \end{array}\right| \end{aligned}$
If any two rows or columns of a determinant are identical.
The value of the determinant is zero
$\begin{aligned} &=(-2) \times 0 \quad\left(\because C_{1}=C_{2}\right) \\ &=0 \end{aligned}$
Hence $\left|\begin{array}{ccc} 6 & -3 & 2 \\ 2 & -1 & 2 \\ -10 & 5 & 2 \end{array}\right|=0$
determinants Exercise 5.2 Question 2 Sub Question 3
Answer:$\left|\begin{array}{ccc} 2 & 3 & 7 \\ 13 & 17 & 5 \\ 15 & 20 & 12 \end{array}\right|=0$Hint: We will try to do any two column or row equal
Given:$\left|\begin{array}{ccc} 2 & 3 & 7 \\ 13 & 17 & 5 \\ 15 & 20 & 12 \end{array}\right|$
Solution:$\left|\begin{array}{ccc} 2 & 3 & 7 \\ 13 & 17 & 5 \\ 15 & 20 & 12 \end{array}\right|$
$\begin{aligned} &\text { Applying } R_{1} \rightarrow R_{1}+R_{2} \\ &=\left|\begin{array}{ccc} 15 & 20 & 12 \\ 13 & 17 & 5 \\ 15 & 20 & 12 \end{array}\right| \end{aligned}$
If any two rows or columns of a determinant are identical.
The value of the determinant is zero
$=0 \quad\left(\because R_{1}=R_{3}\right)$
Hence,$\left|\begin{array}{ccc} 2 & 3 & 7 \\ 13 & 17 & 5 \\ 15 & 20 & 12 \end{array}\right|=0$
determinants Exercise 5.2 Question 2 Sub Question 5
Answer:$\left|\begin{array}{lll} a+b & 2 a+b & 3 a+b \\ 2 a+b & 3 a+b & 4 a+b \\ 4 a+b & 5 a+b & 6 a+b \end{array}\right|=0$Hint: We will try to do any two column or row equal
Given:$\left|\begin{array}{lll} a+b & 2 a+b & 3 a+b \\ 2 a+b & 3 a+b & 4 a+b \\ 4 a+b & 5 a+b & 6 a+b \end{array}\right|$
Solution:$\left|\begin{array}{lll} a+b & 2 a+b & 3 a+b \\ 2 a+b & 3 a+b & 4 a+b \\ 4 a+b & 5 a+b & 6 a+b \end{array}\right|$
$\begin{aligned} &\text { Applying } \mathrm{C}_{2} \rightarrow C_{2}-C_{1} \text { and } \mathrm{C}_{3} \rightarrow C_{3}-C_{2} \\ &=\left|\begin{array}{lll} a+b & (2 a+b)-(a+b) & (3 a+b)-(2 a+b) \\ 2 a+b & (3 a+b)-(2 a+b) & (4 a+b)-(3 a+b) \\ 4 a+b & (5 a+b)-(4 a+b) & (6 a+b)-(5 a+b) \end{array}\right| \\ &=\left|\begin{array}{lll} a+b & 2 a+b-a-b & 3 a+b-2 a-b \\ 2 a+b & 3 a+b-2 a-b & 4 a+b-3 a-b \\ 4 a+b & 5 a+b-4 a-b & 6 a+b-5 a-b \end{array}\right| \\ &=\left|\begin{array}{lll} a+b & a & a \\ 2 a+b & a & a \\ 4 a+b & a & a \end{array}\right| \end{aligned}$
If any two rows or columns of a determinant are identical.
The value of the determinant is zero
$=0 \quad\left(\because C_{2}=C_{3}\right)$
Hence$\left|\begin{array}{lll} a+b & 2 a+b & 3 a+b \\ 2 a+b & 3 a+b & 4 a+b \\ 4 a+b & 5 a+b & 6 a+b \end{array}\right|=0$
determinants Exercise 5.2 Question 2 Sub Question 6
Answer:$\left|\begin{array}{lll} 1 & a & a^{2}-b c \\ 1 & b & b^{2}-a c \\ 1 & c & c^{2}-a b \end{array}\right|=0$Hint: First we will split the determinant into two determinant
Given: $\left|\begin{array}{lll} 1 & a & a^{2}-b c \\ 1 & b & b^{2}-a c \\ 1 & c & c^{2}-a b \end{array}\right|$
Solution:$\left|\begin{array}{lll} 1 & a & a^{2}-b c \\ 1 & b & b^{2}-a c \\ 1 & c & c^{2}-a b \end{array}\right|$
$\begin{aligned} &=\left|\begin{array}{lll} 1 & a & a^{2} \\ 1 & b & b^{2} \\ 1 & c & c^{2} \end{array}\right|-\left|\begin{array}{ccc} 1 & a & b c \\ 1 & b & a c \\ 1 & c & a b \end{array}\right| \\ &\because \text { Using }\left|\begin{array}{lll} x_{1} & x_{2} & x_{3}+a \\ y_{1} & y_{2} & y_{3}+b \\ z_{1} & z_{2} & z_{3}+c \end{array}\right|=\left|\begin{array}{lll} x_{1} & x_{2} & x_{3} \\ y_{1} & y_{2} & y_{3} \\ z_{1} & z_{2} & z_{3} \end{array}\right|+\left|\begin{array}{lll} x_{1} & x_{2} & a \\ y_{1} & y_{2} & b \\ z_{1} & z_{2} & c \end{array}\right| \\ &==\left|\begin{array}{lll} 1 & a & a^{2} \\ 1 & b & b^{2} \\ 1 & c & c^{2} \end{array}\right|-\left|\begin{array}{lll} 1 & a & b c \\ 1 & b & a c \\ 1 & c & a b \end{array}\right| \end{aligned}$
We will multiply and divide by abc in second determinent
$=\left|\begin{array}{lll} 1 & a & a^{2} \\ 1 & b & b^{2} \\ 1 & c & c^{2} \end{array}\right|-\frac{a b c}{a b c}\left|\begin{array}{lll} 1 & a & b c \\ 1 & b & a c \\ 1 & c & a b \end{array}\right|$
$\begin{aligned} &\text { On multiplying } R_{1} \text { by } a, R_{2} \text { by } b \text { and } R_{3} \text { by } c \text { in determinant }(2)\\ &=\left|\begin{array}{lll} 1 & a & a^{2} \\ 1 & b & b^{2} \\ 1 & c & c^{2} \end{array}\right|-\frac{1}{a b c}\left|\begin{array}{lll} a & a^{2} & a b c \\ b & b^{2} & a b c \\ c & c^{2} & a b c \end{array}\right|\\ &=\left|\begin{array}{lll} 1 & a & a^{2} \\ 1 & b & b^{2} \\ 1 & c & c^{2} \end{array}\right|-\frac{a b c}{a b c}\left|\begin{array}{lll} a & a^{2} & 1 \\ b & b^{2} & 1 \\ c & c^{2} & 1 \end{array}\right|\\ &=\left|\begin{array}{lll} 1 & a & a^{2} \\ 1 & b & b^{2} \\ 1 & c & c^{2} \end{array}\right|-\left|\begin{array}{lll} a & a^{2} & 1 \\ b & b^{2} & 1 \\ c & c^{2} & 1 \end{array}\right| \end{aligned}$
$\begin{aligned} &C_{1} \leftrightarrow C_{2} \text { in determinant }(1)\\ &=-\left|\begin{array}{lll} a & 1 & a^{2} \\ b & 1 & b^{2} \\ c & 1 & c^{2} \end{array}\right|-\left|\begin{array}{lll} a & a^{2} & 1 \\ b & b^{2} & 1 \\ c & c^{2} & 1 \end{array}\right| \text { and } \end{aligned}$
If any two rows or columns of a determinant are interchanged,
then sign of the determinant is changed.
$\begin{aligned} &\text { Again } C_{2} \leftrightarrow C_{3} \text { in determinant }(1)\\ &=(-)(-)\left|\begin{array}{lll} a & a^{2} & 1 \\ b & b^{2} & 1 \\ c & c^{2} & 1 \end{array}\right|-\left|\begin{array}{lll} a & a^{2} & 1 \\ b & b^{2} & 1 \\ c & c^{2} & 1 \end{array}\right|\\ &=\left|\begin{array}{lll} a & a^{2} & 1 \\ b & b^{2} & 1 \\ c & c^{2} & 1 \end{array}\right|-\left|\begin{array}{lll} a & a^{2} & 1 \\ b & b^{2} & 1 \\ c & c^{2} & 1 \end{array}\right|\\ &=0 \end{aligned}$
Hence$\left|\begin{array}{lll} 1 & a & a^{2}-b c \\ 1 & b & b^{2}-a c \\ 1 & c & c^{2}-a b \end{array}\right|=0$
determinants Exercise 5.2 Question 2 Sub Question 7
Answer:$\left|\begin{array}{lll} 49 & 1 & 6 \\ 39 & 7 & 4 \\ 26 & 2 & 3 \end{array}\right|=0$Hint: We will try to do any two column or row equal
Given:$\left|\begin{array}{ccc} 49 & 1 & 6 \\ 39 & 7 & 4 \\ 26 & 2 & 3 \end{array}\right|$
Solution:$\left|\begin{array}{ccc} 49 & 1 & 6 \\ 39 & 7 & 4 \\ 26 & 2 & 3 \end{array}\right|$
$\begin{aligned} &\text { Applying } C_{3} \rightarrow 8 \times C_{3}+C_{2} \\ &=\left|\begin{array}{lll} 49 & 1 & 6 \times 8+1 \\ 39 & 7 & 4 \times 8+7 \\ 26 & 2 & 3 \times 8+2 \end{array}\right| \\ &=\left|\begin{array}{lll} 49 & 1 & 48+1 \\ 39 & 7 & 32+7 \\ 26 & 2 & 24+2 \end{array}\right| \\ &=\left|\begin{array}{lll} 49 & 1 & 49 \\ 39 & 7 & 39 \\ 26 & 2 & 26 \end{array}\right| \end{aligned}$
If any two rows or columns of a determinant are identical.
The value of the determinant is zero
$=0 \quad\left(\because C_{1}=C_{3}\right)$
Hence$\left|\begin{array}{lll} 49 & 1 & 6 \\ 39 & 7 & 4 \\ 26 & 2 & 3 \end{array}\right|=0$
determinants Exercise 5.2 Question 2 Sub Question 8
Answer: $\left|\begin{array}{ccc} 0 & x & y \\ -x & 0 & z \\ -y & -z & 0 \end{array}\right|=0$Hint: We will try to do any two column or row equal
Given: $\left|\begin{array}{ccc} 0 & x & y \\ -x & 0 & z \\ -y & -z & 0 \end{array}\right|$
Solution: $\left|\begin{array}{ccc} 0 & x & y \\ -x & 0 & z \\ -y & -z & 0 \end{array}\right|$
$\begin{aligned} &\text { On multiplying } C_{1}, C_{2} \text { and } C_{3} \text { by } z, y \text { and } x \text { respectively }\\ &=\frac{1}{x y z}\left|\begin{array}{ccc} 0 & x y & y x \\ -x z & 0 & z x \\ -y z & -z y & 0 \end{array}\right| \end{aligned}$
$\begin{aligned} &\text { Take } y, x \text { and } z \text { common from } \mathrm{R}_{1}, R_{2} \text { and } \mathrm{R}_{3} \text { respectively }\\ &=\frac{1}{x y z}(y z x)\left|\begin{array}{ccc} 0 & x & x \\ -z & 0 & z \\ -y & -y & 0 \end{array}\right|\\ &=\left|\begin{array}{ccc} 0 & x & x \\ -z & 0 & z \\ -y & -y & 0 \end{array}\right|\\ &\text { On applying } C_{2} \rightarrow C_{2}-C_{3}\\ &=\left|\begin{array}{ccc} 0 & 0 & x \\ -z & -z & z \\ -y & -y & 0 \end{array}\right| \end{aligned}$
If any two rows or columns of a determinant are identical.
The value of the determinant is zero
$=0 \quad\left(\because C_{1}=C_{2}\right)$
Hence $\left|\begin{array}{ccc} 0 & x & y \\ -x & 0 & z \\ -y & -z & 0 \end{array}\right|=0$
determinants Exercise 5.2 Question 2 Sub Question 9
Answer:$\left|\begin{array}{lll} 1 & 43 & 6 \\ 7 & 35 & 4 \\ 3 & 17 & 2 \end{array}\right|=0$Hint: We will try to do any two column or row equal
Given:$\left|\begin{array}{lll} 1 & 43 & 6 \\ 7 & 35 & 4 \\ 3 & 17 & 2 \end{array}\right|$
Solution:$\left|\begin{array}{lll} 1 & 43 & 6 \\ 7 & 35 & 4 \\ 3 & 17 & 2 \end{array}\right|$
$\begin{aligned} &\text { Applying } C_{3} \rightarrow 7 \times C_{3}+C_{1} \\ &=\left|\begin{array}{lll} 1 & 43 & 6 \times 7+1 \\ 7 & 35 & 4 \times 7+7 \\ 3 & 17 & 2 \times 7+3 \end{array}\right| \\ &=\left|\begin{array}{lll} 1 & 43 & 42+1 \\ 7 & 35 & 28+7 \\ 3 & 17 & 14+3 \end{array}\right| \\ &=\left|\begin{array}{lll} 1 & 43 & 43 \\ 7 & 35 & 35 \\ 3 & 17 & 17 \end{array}\right| \end{aligned}$
If any two rows or columns of a determinant are identical.
The value of the determinant is zero
$=0 \quad\left(\because C_{2}=C_{3}\right)$
Hence $\left|\begin{array}{lll} 1 & 43 & 6 \\ 7 & 35 & 4 \\ 3 & 17 & 2 \end{array}\right|=0$
determinants Exercise 5.2 Question 10
Answer:$\left|\begin{array}{llll} 1^{2} & 2^{2} & 3^{2} & 4^{2} \\ 2^{2} & 3^{2} & 4^{2} & 5^{2} \\ 3^{2} & 4^{2} & 5^{2} & 6^{2} \\ 4^{2} & 5^{2} & 6^{2} & 7^{2} \end{array}\right|=0$Hint: We will try to do any two column or row equal
Given:$\left|\begin{array}{cccc} 1^{2} & 2^{2} & 3^{2} & 4^{2} \\ 2^{2} & 3^{2} & 4^{2} & 5^{2} \\ 3^{2} & 4^{2} & 5^{2} & 6^{2} \\ 4^{2} & 5^{2} & 6^{2} & 7^{2} \end{array}\right|$
Solution:$\left|\begin{array}{cccc} 1^{2} & 2^{2} & 3^{2} & 4^{2} \\ 2^{2} & 3^{2} & 4^{2} & 5^{2} \\ 3^{2} & 4^{2} & 5^{2} & 6^{2} \\ 4^{2} & 5^{2} & 6^{2} & 7^{2} \end{array}\right|$
$\begin{aligned} &\text { On applying } C_{3} \rightarrow C_{3}-C_{2} \text { and } C_{4} \rightarrow C_{4}-C_{1}\\ &=\left|\begin{array}{cccc} 1^{2} & 2^{2} & 3^{2}-2^{2} & 4^{2}-1^{2} \\ 2^{2} & 3^{2} & 4^{2}-3^{2} & 5^{2}-2^{2} \\ 3^{2} & 4^{2} & 5^{2}-4^{2} & 6^{2}-3^{2} \\ 4^{2} & 5^{2} & 6^{2}-5^{2} & 7^{2}-4^{2} \end{array}\right|\\ &=\left|\begin{array}{cccc} 1^{2} & 2^{2} & 9-4 & 16-1 \\ 2^{2} & 3^{2} & 16-9 & 25-4 \\ 3^{2} & 4^{2} & 25-16 & 36-9 \\ 4^{2} & 5^{2} & 36-25 & 49-16 \end{array}\right|\\ &=\left|\begin{array}{cccc} 1^{2} & 2^{2} & 5 & 15 \\ 2^{2} & 3^{2} & 7 & 21 \\ 3^{2} & 4^{2} & 9 & 27 \\ 4^{2} & 5^{2} & 11 & 33 \end{array}\right|\\ &\text { On taking common } 3 \text { from } \mathrm{C}_{4}\\ &=3\left|\begin{array}{cccc} 1^{2} & 2^{2} & 5 & 5 \\ 2^{2} & 3^{2} & 7 & 7 \\ 3^{2} & 4^{2} & 9 & 9 \\ 4^{2} & 5^{2} & 11 & 11 \end{array}\right| \end{aligned}$
If any two rows or columns of a determinant are identical.
The value of the determinant is zero
$\begin{aligned} &=3 \times 0 \quad\left(\because C_{3}=C_{4}\right) \\ &=0 \end{aligned}$
Hence$\left|\begin{array}{cccc} 1^{2} & 2^{2} & 3^{2} & 4^{2} \\ 2^{2} & 3^{2} & 4^{2} & 5^{2} \\ 3^{2} & 4^{2} & 5^{2} & 6^{2} \\ 4^{2} & 5^{2} & 6^{2} & 7^{2} \end{array}\right|=0$
determinants Exercise 5.2 Question 12
Answer:$\left|\begin{array}{lll} \left(2^{x}+2^{-x}\right)^{2} & \left(2^{x}-2^{-x}\right)^{2} & 1 \\ \left(3^{x}+3^{-x}\right)^{2} & \left(3^{x}-3^{-x}\right)^{2} & 1 \\ \left(4^{x}+4^{-x}\right)^{2} & \left(4^{x}-4^{-x}\right)^{2} & 1 \end{array}\right|=0$Hint: We will try to do any two column or row equal
Given:$\left|\begin{array}{lll} \left(2^{x}+2^{-x}\right)^{2} & \left(2^{x}-2^{-x}\right)^{2} & 1 \\ \left(3^{x}+3^{-x}\right)^{2} & \left(3^{x}-3^{-x}\right)^{2} & 1 \\ \left(4^{x}+4^{-x}\right)^{2} & \left(4^{x}-4^{-x}\right)^{2} & 1 \end{array}\right|$
Solution:
$\begin{aligned} &\text { First }(a+b)^{2}-(a-b)^{2} \\ &=\left(a^{2}+2 a b+b^{2}\right)-\left(a^{2}-2 a b+b^{2}\right) \\ &=a^{2}+2 a b+b^{2}-a^{2}+2 a b-b^{2} \\ &=4 a b \end{aligned}$
$\left|\begin{array}{lll} \left(2^{x}+2^{-x}\right)^{2} & \left(2^{x}-2^{-x}\right)^{2} & 1 \\ \left(3^{x}+3^{-x}\right)^{2} & \left(3^{x}-3^{-x}\right)^{2} & 1 \\ \left(4^{x}+4^{-x}\right)^{2} & \left(4^{x}-4^{-x}\right)^{2} & 1 \end{array}\right|$
$\begin{aligned} &\text { On applying } \mathrm{C}_{1} \rightarrow C_{1}-C_{2} \\ &\qquad \begin{array}{lll} \left(2^{x}+2^{-x}\right)^{2}-\left(2^{x}-2^{-x}\right)^{2} & \left(2^{x}-2^{-x}\right)^{2} & 1 \\ \left(3^{x}+3^{-x}\right)^{2}-\left(3^{x}-3^{-x}\right)^{2} & \left(3^{x}-3^{-x}\right)^{2} & 1 \\ \left(4^{x}+4^{-x}\right)^{2}-\left(4^{x}-4^{-x}\right)^{2} & \left(4^{x}-4^{-x}\right)^{2} & 1 \end{array} \mid \\ &=\left|\begin{array}{lll} 4 \times 2^{x} \times 2^{-x} & \left(2^{x}-2^{-x}\right)^{2} & 1 \\ 4 \times 3^{x} \times 3^{-x} & \left(3^{x}-3^{-x}\right)^{2} & 1 \\ 4 \times 4^{x} \times 4^{-x} & \left(4^{x}-4^{-x}\right)^{2} & 1 \end{array}\right| \end{aligned}$
$\begin{aligned} &\left\{\begin{array}{l} (a+b)^{2}-(a-b)^{2}=4 a b \\ a^{m} \times a^{n}=a^{m+n} \end{array}\right\} \\ &=\left|\begin{array}{lll} 4 \times 2^{x+(-x)} & \left(2^{x}-2^{-x}\right)^{2} & 1 \\ 4 \times 3^{x+(-x)} & \left(3^{x}-3^{-x}\right)^{2} & 1 \\ 4 \times 4^{x+(-x)} & \left(4^{x}-4^{-x}\right)^{2} & 1 \end{array}\right| \\ &=\left|\begin{array}{lll} 4 \times 2^{0} & \left(2^{x}-2^{-x}\right)^{2} & 1 \\ 4 \times 3^{0} & \left(3^{x}-3^{-x}\right)^{2} & 1 \\ 4 \times 4^{0} & \left(4^{x}-4^{-x}\right)^{2} & 1 \end{array}\right| \end{aligned}$
$\begin{aligned} &=\left|\begin{array}{lll} 4 & \left(2^{x}-2^{-x}\right)^{2} & 1 \\ 4 & \left(3^{x}-3^{-x}\right)^{2} & 1 \\ 4 & \left(4^{x}-4^{-x}\right)^{2} & 1 \end{array}\right|\\ &\text { Taking } 4 \text { common from } \mathrm{C}_{1}\\ &=4\left|\begin{array}{lll} 1 & \left(2^{x}-2^{-x}\right)^{2} & 1 \\ 1 & \left(3^{x}-3^{-x}\right)^{2} & 1 \\ 1 & \left(4^{x}-4^{-x}\right)^{2} & 1 \end{array}\right| \end{aligned}$
If any two rows or columns of a determinant are identical.
The value of the determinant is zero
$\begin{aligned} &=4 \times 0 \quad\left(\because C_{1}=C_{3}\right) \\ &=0 \end{aligned}$
Hence $\left|\begin{array}{lll} \left(2^{x}+2^{-x}\right)^{2} & \left(2^{x}-2^{-x}\right)^{2} & 1 \\ \left(3^{x}+3^{-x}\right)^{2} & \left(3^{x}-3^{-x}\right)^{2} & 1 \\ \left(4^{x}+4^{-x}\right)^{2} & \left(4^{x}-4^{-x}\right)^{2} & 1 \end{array}\right|=0$
determinants Exercise 5.2 Question 13
Answer:$\left|\begin{array}{lll} \sin \alpha & \cos \alpha & \cos (\alpha+\delta) \\ \sin \beta & \cos \beta & \cos (\beta+\delta) \\ \sin \gamma & \cos \gamma & \cos (\gamma+\delta) \end{array}\right|=0$Hint: We will try to do any two column or row equal
Given:$\left|\begin{array}{lll} \sin \alpha & \cos \alpha & \cos (\alpha+\delta) \\ \sin \beta & \cos \beta & \cos (\beta+\delta) \\ \sin \gamma & \cos \gamma & \cos (\gamma+\delta) \end{array}\right|$
Solution:$\left|\begin{array}{lll} \sin \alpha & \cos \alpha & \cos (\alpha+\delta) \\ \sin \beta & \cos \beta & \cos (\beta+\delta) \\ \sin \gamma & \cos \gamma & \cos (\gamma+\delta) \end{array}\right|$
$\begin{aligned} &\text { On applying } \mathrm{C}_{2} \rightarrow \cos \delta \cdot \mathrm{C}_{2}-\sin \delta . C_{1} \\ &\left|\begin{array}{lll} \sin \alpha & \cos \alpha \cdot \cos \delta-\sin \alpha \cdot \sin \delta & \cos (\alpha+\delta) \\ \sin \beta & \cos \beta \cdot \cos \delta-\sin \beta \cdot \sin \delta & \cos (\beta+\delta) \\ \sin \gamma & \cos \gamma \cdot \cos \delta-\sin \gamma \cdot \sin \delta & \cos (\gamma+\delta) \end{array}\right| \\ &\{\because \cos (\alpha+\delta)=\cos \alpha \cos \delta-\sin \alpha \sin \delta\} \\ &=\left|\begin{array}{lll} \sin \alpha & \cos (\alpha+\delta) & \cos (\alpha+\delta) \\ \sin \beta & \cos (\beta+\delta) & \cos (\beta+\delta) \\ \sin \gamma & \cos (\gamma+\delta) & \cos (\gamma+\delta) \end{array}\right| \end{aligned}$
If any two rows or columns of a determinant are identical.
The value of the determinant is zero
$=0 \quad\left(\because C_{2}=C_{3}\right)$
Hence $\left|\begin{array}{lll} \sin \alpha & \cos \alpha & \cos (\alpha+\delta) \\ \sin \beta & \cos \beta & \cos (\beta+\delta) \\ \sin \gamma & \cos \gamma & \cos (\gamma+\delta) \end{array}\right|=0$
deteminants Exercise 5.2 Question 15
Answer:$\left|\begin{array}{ccc} \cos (x+y) & -\sin (x+y) & \cos 2 y \\ \sin x & \cos x & \sin y \\ -\cos x & \sin x & -\cos y \end{array}\right|=0$Hint: We will try to do any two column or row identical
Given: $\left|\begin{array}{ccc} \cos (x+y) & -\sin (x+y) & \cos 2 y \\ \sin x & \cos x & \sin y \\ -\cos x & \sin x & -\cos y \end{array}\right|$
Solution:$\left|\begin{array}{ccc} \cos (x+y) & -\sin (x+y) & \cos 2 y \\ \sin x & \cos x & \sin y \\ -\cos x & \sin x & -\cos y \end{array}\right|$
$\begin{aligned} &\text { On applying } \mathrm{R}_{3} \rightarrow-\cos y \times R_{3}-\sin y \times R_{2} \\ &=\left|\begin{array}{ccc} \cos (x+y) & -\sin (x+y) & \cos 2 y \\ \sin x & \cos x & \sin y \\ (-\cos y)(-\cos x)-\sin y \sin x & (-\cos y)(\sin x)-\sin y \cos x & (-\cos y)(-\cos y)-\sin y \sin y \end{array}\right| \end{aligned}$
$=\left|\begin{array}{ccc} \cos (x+y) & -\sin (x+y) & \cos 2 y \\ \sin x & \cos x & \sin y \\ \cos x \cos y-\sin y \sin x & -[\sin x \cos y+\sin y \cos x] & \left(\cos ^{2} y-\sin ^{2} y\right) \end{array}\right|$
$\begin{aligned} &\because \cos (x+y)=\cos x \cos y-\sin y \sin x \\ &\sin (x+y)=\sin x \cos y+\sin y \cos x \\ &\cos ^{2} y-\sin ^{2} y=\cos 2 y \end{aligned}$
$=\left|\begin{array}{ccc} \cos (x+y) & -\sin (x+y) & \cos 2 y \\ \sin x & \cos x & \sin y \\ \cos (x+y) & -\sin (x+y) & \cos 2 y \end{array}\right|$
If any two rows or columns of a determinant are identical.
The value of the determinant is zero
$=0 \quad\left(\because R_{1}=R_{3}\right)$
Hence $\left|\begin{array}{ccc} \cos (x+y) & -\sin (x+y) & \cos 2 y \\ \sin x & \cos x & \sin y \\ -\cos x & \sin x & -\cos y \end{array}\right|=0$
deteminants Exercise 5.2 Question 16
Answer:$\left|\begin{array}{ccc} \sqrt{23}+\sqrt{3} & \sqrt{5} & \sqrt{5} \\ \sqrt{15}+\sqrt{46} & 5 & \sqrt{10} \\ 3+\sqrt{115} & \sqrt{15} & 5 \end{array}\right|=0$Hint: We will try to make all the elements of two row or column zero
Given: $\left|\begin{array}{ccc} \sqrt{23}+\sqrt{3} & \sqrt{5} & \sqrt{5} \\ \sqrt{15}+\sqrt{46} & 5 & \sqrt{10} \\ 3+\sqrt{115} & \sqrt{15} & 5 \end{array}\right|$
Solution:$\left|\begin{array}{ccc} \sqrt{23}+\sqrt{3} & \sqrt{5} & \sqrt{5} \\ \sqrt{15}+\sqrt{46} & 5 & \sqrt{10} \\ 3+\sqrt{115} & \sqrt{15} & 5 \end{array}\right|$
$\text { On taking } \sqrt{5} \text { common from } \mathrm{C}_{2} \text { and } \mathrm{C}_{3}$
$=\sqrt{5} \times \sqrt{5}\left|\begin{array}{lll} \sqrt{23}+\sqrt{3} & 1 & 1 \\ \sqrt{15}+\sqrt{46} & \sqrt{5} & \sqrt{2} \\ 3+\sqrt{115} & \sqrt{3} & \sqrt{5} \end{array}\right|$
$\begin{aligned} &\text { On applying } \mathrm{C}_{1} \rightarrow C_{1}-\sqrt{3} C_{2}-\sqrt{23} C_{3} \\ &=5\left|\begin{array}{ccc} \sqrt{23}+\sqrt{3}-\sqrt{3}-\sqrt{23} & 1 & 1 \\ \sqrt{15}+\sqrt{46}-\sqrt{15}-\sqrt{46} & \sqrt{5} & \sqrt{2} \\ 3+\sqrt{115}-3-\sqrt{115} & \sqrt{3} & \sqrt{5} \end{array}\right| \\ &=5\left|\begin{array}{ccc} 0 & 1 & 1 \\ 0 & \sqrt{5} & \sqrt{2} \\ 0 & \sqrt{3} & \sqrt{5} \end{array}\right| \end{aligned}$
If all the elements of a row or column are zero
The value of the determinant is zero
$=5\times 0$
$=0$
Hence $\left|\begin{array}{ccc} \sqrt{23}+\sqrt{3} & \sqrt{5} & \sqrt{5} \\ \sqrt{15}+\sqrt{46} & 5 & \sqrt{10} \\ 3+\sqrt{115} & \sqrt{15} & 5 \end{array}\right|=0$
deteminants Exercise 5 .2 Question 17
Answer: $\left|\begin{array}{lll} \sin ^{2} A & \cot A & 1 \\ \sin ^{2} B & \cot B & 1 \\ \sin ^{2} C & \cot C & 1 \end{array}\right|=0$where A,B,C are the angles of ΔABCHint: We will try to convert some elements of determinant zero or 1
Given:$\left|\begin{array}{ccc} \sin ^{2} A & \cot A & 1 \\ \sin ^{2} B & \cot B & 1 \\ \sin ^{2} C & \cot C & 1 \end{array}\right|$
Solution:$\left|\begin{array}{ccc} \sin ^{2} A & \cot A & 1 \\ \sin ^{2} B & \cot B & 1 \\ \sin ^{2} C & \cot C & 1 \end{array}\right|$
$\begin{aligned} &\text { On applying } \mathrm{R}_{1} \rightarrow R_{1}-R_{2} \text { and } \mathrm{R}_{2} \rightarrow R_{2}-R_{3}\\ &=\left|\begin{array}{ccc} \sin ^{2} A-\sin ^{2} B & \cot A-\cot B & 0 \\ \sin ^{2} B-\sin ^{2} C & \cot B-\cot C & 0 \\ \sin ^{2} C & \cot C & 1 \end{array}\right|\\ &\because \sin ^{2} A-\sin ^{2} B=\sin (A+B) \cdot \sin (A-B) \backslash\\ &\because A+B+C=180^{\circ}\\ &A-B=180-C\\ &\sin (180-C)=\sin C\\ &\sin (180-A)=\sin A \end{aligned}$
$\begin{aligned} &\text { Now, }\\ &\cot A-\cot B\\ &\frac{\cos A}{\sin A}-\frac{\cos B}{\sin B}\\ &=\frac{\sin B \cos A-\cos B \sin A}{\sin A \sin B} \Rightarrow \frac{\sin (B-A)}{\sin A \sin B}\\ &=\frac{\sin \{-(A-B)\}}{\sin A \sin B}=\frac{-\sin (A-B)}{\sin A \sin B} \end{aligned}$
$\begin{aligned} &\text { Now from (1) }\\ &=\left|\begin{array}{ccc} \sin (A+B) \sin (A-B) & \frac{\sin (B-A)}{\sin A \sin B} & 0 \\ \sin (B+C) \sin (B-C) & \frac{\sin (C-B)}{\sin B \sin C} & 0 \\ \sin ^{2} C & \cot C & 1 \end{array}\right| \end{aligned}$
$\begin{aligned} &=\left|\begin{array}{ccc} \sin (180-C) \sin (A-B) & \frac{-\sin (A-B)}{\sin A \sin B} & 0 \\ \sin (180-A) \sin (B-C) & \frac{-\sin (B-C)}{\sin B \sin C} & 0 \\ \sin ^{2} C & \cot C & 1 \end{array}\right| \\ &=\left|\begin{array}{ccc} \sin (A) \sin (B-C) & \frac{-\sin (B-C)}{\sin B \sin C} & 0 \\ \sin ^{2} C & \cot C & 1 \end{array}\right| \end{aligned}$
$\begin{aligned} &\text { On taking } \sin (A-B) \text { common from } \mathrm{R}_{1} \text { and } \sin (B-C) \text { from } \mathrm{R}_{2}\\ &=\sin (A-B) \cdot \sin (B-C)\left|\begin{array}{lcc} \sin C & \frac{-1}{\sin A \sin B} & 0 \\ \sin A & \frac{-1}{\sin B \sin C} & 0 \\ \sin ^{2} C & \cot C & 1 \end{array}\right| \end{aligned}$
$\begin{aligned} &\text { On expanding w.r.t } \mathrm{C}_{3}\\ &=\sin (A-B) \cdot \sin (B-C)\left[0\left|\begin{array}{lc} \sin A & \frac{-1}{\sin B \sin C} \\ \sin ^{2} C & \cot C \end{array}\right|-0\left|\begin{array}{lc} \sin C & \frac{-1}{\sin A \sin B} \\ \sin ^{2} C & \cot C \end{array}\right|+1\left|\begin{array}{ll} \sin C & \frac{-1}{\sin A \sin B} \\ \sin A & \frac{-1}{\sin B \sin C} \end{array}\right|\right]\\ &=\sin (A-B) \cdot \sin (B-C)\left[0-0+1\left(\frac{-\sin C}{\sin B \sin C}+\frac{\sin A}{\sin A \sin B}\right)\right]\\ &=\sin (A-B) \cdot \sin (B-C)\left[\frac{-1}{\sin B}+\frac{1}{\sin B}\right]\\ &=\sin (A-B) \cdot \sin (B-C) \times 0\\ &=0 \end{aligned}$
Hence $\left|\begin{array}{lll} \sin ^{2} A & \cot A & 1 \\ \sin ^{2} B & \cot B & 1 \\ \sin ^{2} C & \cot C & 1 \end{array}\right|=0$
deteminants Exercise MCQs Question 2 sub question 3
Answer:-$\left|\begin{array}{lll} a & b+c & a^{2} \\ b & c+a & b^{2} \\ c & a+b & c^{2} \end{array}\right|=-(a+b+c)(a-b)(b-c)(c-a)$Hint: We will try to make some elements of the determinant into zero
Given:-$\left|\begin{array}{lll} a & b+c & a^{2} \\ b & c+a & b^{2} \\ c & a+b & c^{2} \end{array}\right|$
Solution :-$\left|\begin{array}{lll} a & b+c & a^{2} \\ b & c+a & b^{2} \\ c & a+b & c^{2} \end{array}\right|$
$\begin{aligned} &\text { On applying } \mathrm{C}_{1} \rightarrow C_{1}+C_{2}\\ &=\left|\begin{array}{lll} a+b+c & b+c & a^{2} \\ b+c+a & c+a & b^{2} \\ c+a+b & a+b & c^{2} \end{array}\right|\\ &\text { On taking common }(a+b+c) \text { from } \mathrm{C}_{1}\\ &=a+b+c\left|\begin{array}{lll} 1 & b+c & a^{2} \\ 1 & c+a & b^{2} \\ 1 & a+b & c^{2} \end{array}\right| \end{aligned}$
$\begin{aligned} &\text { On applying } \mathrm{C}_{1} \rightarrow C_{1}-C_{2} \text { and } \mathrm{C}_{2} \rightarrow C_{2}-C_{3}\\ &=a+b+c\left|\begin{array}{ccc} 0 & (b+c)-(c+a) & a^{2}-b^{2} \\ 0 & (c+a)-(a+b) & b^{2}-c^{2} \\ 1 & a+b & c^{2} \end{array}\right| \end{aligned}$
$\begin{aligned} &=a+b+c\left|\begin{array}{ccc} 0 & b+c-c-a & (a+b)(a-b) \\ 0 & c+a-a-b & (b+c)(b-c) \\ 1 & a+b & c^{2} \end{array}\right| \quad\left(\because a^{2}-b^{2}=(a+b)(a-b)\right) \\ &=a+b+c\left|\begin{array}{ccc} 0 & b-a & (a+b)(a-b) \\ 0 & c-b & (b+c)(b-c) \\ 1 & a+b & c^{2} \end{array}\right| \\ &=a+b+c\left|\begin{array}{ccc} 0 & -(a-b) & (a+b)(a-b) \\ 0 & -(b-c) & (b+c)(b-c) \\ 1 & a+b & c^{2} \end{array}\right| \end{aligned}$
$\begin{aligned} &\text { On taking common }(a-b) \text { from } \mathrm{R}_{1} \text { and }(b-c) \text { from } \mathrm{R}_{2}\\ &=(a+b+c)(a-b)(b-c)\left|\begin{array}{ccc} 0 & -1 & (a+b) \\ 0 & -1 & (b+c) \\ 1 & a+b & c^{2} \end{array}\right| \end{aligned}$
$\begin{aligned} &\text { On expanding w.r.t } \mathrm{R}_{1}\\ &=(a+b+c)(a-b)(b-c)\left[0\left|\begin{array}{cc} -1 & b+c \\ a+b & c^{2} \end{array}\right|-(-1)\left|\begin{array}{cc} 0 & b+c \\ 1 & c^{2} \end{array}\right|+(a+b)\left|\begin{array}{cc} 0 & -1 \\ 1 & a+b \end{array}\right|\right]\\ &=(a+b+c)(a-b)(b-c)\left[0+1\left\{0 \times c^{2}-1(b+c)\right\}+(a+b)\{0 \cdot(a+b)-(1)(-1)\}\right]\\ &=(a+b+c)(a-b)(b-c)\{1(-b-c)+(a+b)(1)\}\\ &=(a+b+c)(a-b)(b-c)(-b-c+a+b)\\ &=(a+b+c)(a-b)(b-c)(a-c)\\ &=-(a+b+c)(a-b)(b-c)(c-a) \end{aligned}$
Hence$\left|\begin{array}{lll} a & b+c & a^{2} \\ b & c+a & b^{2} \\ c & a+b & c^{2} \end{array}\right|=-(a+b+c)(a-b)(b-c)(c-a)$
determinants Exercise 5.2 Question 2 sub question 4
Answer:-$\left|\begin{array}{lll} 1 & a & b c \\ 1 & b & c a \\ 1 & c & a b \end{array}\right|=(a-b)(b-c)(c-a)$Hint: We will try to make some elements of the determinant into zero
Given:$\left|\begin{array}{lll} 1 & a & b c \\ 1 & b & c a \\ 1 & c & a b \end{array}\right|$
Solution :$\left|\begin{array}{lll} 1 & a & b c \\ 1 & b & c a \\ 1 & c & a b \end{array}\right|$
$\begin{aligned} &\text { On applying } \mathrm{R}_{1} \rightarrow R_{1}-R_{2} \text { and } \mathrm{R}_{2} \rightarrow R_{2}-R_{3}\\ &=\left|\begin{array}{ccc} 0 & a-b & b c-c a \\ 0 & b-c & c a-a b \\ 1 & c & a b \end{array}\right|\\ &=\left|\begin{array}{ccc} 0 & a-b & -c(a-b) \\ 0 & b-c & -a(b-c) \\ 1 & c & a b \end{array}\right| \end{aligned}$
$\begin{aligned} &\text { On taking common }(a-b) \text { from } \mathrm{R}_{1} \text { and }(b-c) \text { from } \mathrm{R}_{2}\\ &=(a-b)(b-c)\left|\begin{array}{ccc} 0 & 1 & -c \\ 0 & 1 & -a \\ 1 & c & a b \end{array}\right| \end{aligned}$
$\begin{aligned} &\text { On expanding w.r.t } \mathrm{R}_{1}\\ &=(a-b)(b-c)\left[0\left|\begin{array}{ll} 1 & -a \\ c & a b \end{array}\right|-1\left|\begin{array}{cc} 0 & -a \\ 1 & a b \end{array}\right|+(-c)\left|\begin{array}{ll} 0 & 1 \\ 1 & c \end{array}\right|\right]\\ &=(a-b)(b-c)[0-1\{0 \times a b-(-a) \times 1\}-c\{0 \times c-1 \times 1\}]\\ &=(a-b)(b-c)[-1(+a)-c(-1)]\\ &=(a-b)(b-c)(-a+c)\\ &=(a-b)(b-c)(c-a) \end{aligned}$
Hence $\left|\begin{array}{lll} 1 & a & b c \\ 1 & b & c a \\ 1 & c & a b \end{array}\right|=(a-b)(b-c)(c-a)$
determinants Exercise 5.2 Question 2 sub question 5
Answer:$\left|\begin{array}{ccc} x+\lambda & x & x \\ x & x+\lambda & x \\ x & x & x+\lambda \end{array}\right|=\lambda^{2}(3 x+\lambda)$Hint: We will try to make some elements of the determinant into zero
Given:$\left|\begin{array}{ccc} x+\lambda & x & x \\ x & x+\lambda & x \\ x & x & x+\lambda \end{array}\right|$
Solution:$\left|\begin{array}{ccc} x+\lambda & x & x \\ x & x+\lambda & x \\ x & x & x+\lambda \end{array}\right|$
$\begin{aligned} &\text { On applying } \mathrm{R}_{1} \rightarrow R_{1}+R_{2}+R_{3} \\ &=\left|\begin{array}{ccc} 3 x+\lambda & 3 x+\lambda & 3 x+\lambda \\ x & x+\lambda & x \\ x & x & x+\lambda \end{array}\right| \end{aligned}$
$\begin{aligned} &\text { On taking common }(3 x+\lambda) \text { from } \mathrm{R}_{1}\\ &=(3 x+\lambda)\left|\begin{array}{ccc} 1 & 1 & 1 \\ x & x+\lambda & x \\ x & x & x+\lambda \end{array}\right| \end{aligned}$
$\begin{aligned} &\text { On applying } \mathrm{C}_{1} \rightarrow C_{1}-C_{2} \text { and } \mathrm{C}_{2} \rightarrow C_{2}-C_{3}\\ &=(3 x+\lambda)\left|\begin{array}{ccc} 0 & 0 & 1 \\ x-(x+\lambda) & (x+\lambda)-x & x \\ x-x & x-(x+\lambda) & x+\lambda \end{array}\right|\\ &=(3 x+\lambda)\left|\begin{array}{ccc} 0 & 0 & 1 \\ x-x-\lambda & x+\lambda-x & x \\ 0 & x-x-\lambda & x+\lambda \end{array}\right|\\ &=(3 x+\lambda)\left|\begin{array}{ccc} 0 & 0 & 1 \\ -\lambda & \lambda & x \\ 0 & -\lambda & x+\lambda \end{array}\right| \end{aligned}$
$\begin{aligned} &\text { On expanding w.r.t } \mathrm{R}_{1}\\ &=(3 x+\lambda)\left[0\left|\begin{array}{cc} \lambda & x \\ -\lambda & x+\lambda \end{array}\right|-0\left|\begin{array}{cc} -\lambda & x \\ 0 & x+\lambda \end{array}\right|+1\left|\begin{array}{cc} -\lambda & \lambda \\ 0 & -\lambda \end{array}\right|\right]\\ &=(3 x+\lambda)[0-0+1[(-\lambda)(-\lambda)-0 \times \lambda]]\\ &=(3 x+\lambda)\left\{1\left(\lambda^{2}-0\right)\right\}\\ &=\lambda^{2}(3 x+\lambda) \end{aligned}$
Hence$\left|\begin{array}{ccc} x+\lambda & x & x \\ x & x+\lambda & x \\ x & x & x+\lambda \end{array}\right|=\lambda^{2}(3 x+\lambda)$
determinants Exercise 5.2 Question 2 sub question 6
Answer:$\left|\begin{array}{lll} a & b & c \\ c & a & b \\ b & c & a \end{array}\right|=(a+b+c)\left(a^{2}+b^{2}+c^{2}-a b-b c-c a\right)$Hint :We will try to make some elements of the determinant into zero
Given :$\left|\begin{array}{lll} a & b & c \\ c & a & b \\ b & c & a \end{array}\right|$
Solution: $\left|\begin{array}{lll} a & b & c \\ c & a & b \\ b & c & a \end{array}\right|$
$\begin{aligned} &\text { On applying } \mathrm{R}_{1} \rightarrow R_{1}+R_{2}+R_{3} \\ &=\left|\begin{array}{ccc} a+b+c & a+b+c & a+b+c \\ c & a & b \\ b & c & a \end{array}\right| \end{aligned}$
$\begin{aligned} &\text { On taking common }(a+b+c) \text { from } \mathrm{R}_{1}\\ &=(a+b+c)\left|\begin{array}{lll} 1 & 1 & 1 \\ c & a & b \\ b & c & a \end{array}\right|\\ &\text { On applying } \mathrm{C}_{1} \rightarrow C_{1}-C_{2} \text { and } \mathrm{C}_{2} \rightarrow C_{2}-C_{3}\\ &=(a+b+c)\left|\begin{array}{ccc} 0 & 0 & 1 \\ c-a & a-b & b \\ b-c & c-a & a \end{array}\right| \end{aligned}$
$\begin{aligned} &\text { On expanding w.r.t } \mathrm{R}_{1}\\ &=(a+b+c)\left[0\left|\begin{array}{ll} a-b & b \\ c-a & a \end{array}\right|-0\left|\begin{array}{cc} b-a & b \\ 0 & a \end{array}\right|+1\left|\begin{array}{cc} c-a & a-b \\ b-c & c-a \end{array}\right|\right.\\ &=(a+b+c)[0-0+1\{(c-a)(c-a)-(b-c)(a-b)\}]\\ &=(a+b+c)\left\{\left(c^{2}-a c-a c+a^{2}\right)-\left(b a-b^{2}+a c-b c\right)\right\}\\ &=(a+b+c)\left(a^{2}+b^{2}+c^{2}-a b-b c-c a\right) \end{aligned}$
Hence $\left|\begin{array}{lll} a & b & c \\ c & a & b \\ b & c & a \end{array}\right|=(a+b+c)\left(a^{2}+b^{2}+c^{2}-a b-b c-c a\right)$
determinants Exercise 5.2 Question 2 sub question 7
Answer:$\left|\begin{array}{lll} x & 1 & 1 \\ 1 & x & 1 \\ 1 & 1 & x \end{array}\right|=(x+2)(x-1)^{2}$Hint: We will try to make some elements of the determinant into zero
Given:$\left|\begin{array}{lll} x & 1 & 1 \\ 1 & x & 1 \\ 1 & 1 & x \end{array}\right|$
Solution:$\left|\begin{array}{lll} x & 1 & 1 \\ 1 & x & 1 \\ 1 & 1 & x \end{array}\right|$
$\begin{aligned} &\text { On applying } \mathrm{R}_{1} \rightarrow R_{1}+R_{2}+R_{3} \\ &=\left|\begin{array}{ccc} x+2 & x+2 & x+2 \\ 1 & x & 1 \\ 1 & 1 & x \end{array}\right| \end{aligned}$
$\begin{aligned} &\text { On taking common }(x+2) \text { from } \mathrm{R}_{1}\\ &=(x+2)\left|\begin{array}{lll} 1 & 1 & 1 \\ 1 & x & 1 \\ 1 & 1 & x \end{array}\right|\\ &\text { On applying } \mathrm{C}_{1} \rightarrow C_{1}-C_{2} \text { and } \mathrm{C}_{2} \rightarrow C_{2}-C_{3}\\ &=(x+2)\left|\begin{array}{ccc} 0 & 0 & 1 \\ 1-x & x-1 & 1 \\ 1 & 1-x & x \end{array}\right| \end{aligned}$
$\begin{aligned} &\text { On expanding w.r.t } \mathrm{R}_{1}\\ &=(x+2)\left[0\left|\begin{array}{ll} x-1 & 1 \\ 1-x & x \end{array}\right|-0\left|\begin{array}{cc} 1-x & 1 \\ 0 & x \end{array}\right|+1\left|\begin{array}{cc} 1-x & x-1 \\ 0 & 1-x \end{array}\right|\right]\\ &=(x+2)[0-0+1\{(1-x)(1-x)-0(x-1)\}]\\ &=(x+2)\left[1\left\{-(x-1)^{2}-0\right\}\right]\\ &=(x+2)(x-1)^{2} \end{aligned}$
Hence $\left|\begin{array}{lll} x & 1 & 1 \\ 1 & x & 1 \\ 1 & 1 & x \end{array}\right|=(x+2)(x-1)^{2}$
determinants Exercise 5.2 Question 2 sub question 8
Answer:$\left|\begin{array}{ccc} 0 & x y^{2} & x z^{2} \\ x^{2} y & 0 & y z^{2} \\ x^{2} z & z y^{2} & 0 \end{array}\right|=2 x^{3} y^{3} z^{3}$Hint:We will take common from $C_{1},C_{2} \: \: and\: \: \: C_{3}$
Given:$\left|\begin{array}{ccc} 0 & x y^{2} & x z^{2} \\ x^{2} y & 0 & y z^{2} \\ x^{2} z & z y^{2} & 0 \end{array}\right|$
Solution: $\left|\begin{array}{ccc} 0 & x y^{2} & x z^{2} \\ x^{2} y & 0 & y z^{2} \\ x^{2} z & z y^{2} & 0 \end{array}\right|$
$\begin{aligned} &\text { On taking common } x^{2} \text { from } \mathrm{C}_{1}, \mathrm{y}^{2} \text { from } \mathrm{C}_{2} \text { and } \mathrm{z}^{2} \text { from } \mathrm{C}_{3}\\ &=x^{2} y^{2} z^{2}\left|\begin{array}{ccc} 0 & x & x \\ y & 0 & y \\ z & z & 0 \end{array}\right| \end{aligned}$
$\begin{aligned} &\text { On expanding w.r.t } \mathrm{R}_{1}\\ &=x^{2} y^{2} z^{2}\left[0\left|\begin{array}{ll} 0 & y \\ z & 0 \end{array}\right|-x\left|\begin{array}{ll} y & y \\ z & 0 \end{array}\right|+x\left|\begin{array}{ll} y & 0 \\ z & z \end{array}\right|\right]\\ &=x^{2} y^{2} z^{2}[0-x(0 x y-y x z)+x(y x z-0 x z)]\\ &=x^{2} y^{2} z^{2}[-x(0-y z)+x(y z-0)]\\ &=x^{2} y^{2} z^{2}(x y z+x y z)\\ &=x^{2} y^{2} z^{2} \times 2 x y z\\ &=2 x^{3} y^{3} z^{3} \end{aligned}$
Hence$\left|\begin{array}{ccc} 0 & x y^{2} & x z^{2} \\ x^{2} y & 0 & y z^{2} \\ x^{2} z & z y^{2} & 0 \end{array}\right|=2 x^{3} y^{3} z^{3}$
determinants Exercise 5.2 Question 2 sub question 9
Answer:$\left|\begin{array}{ccc} a+x & y & z \\ x & a+y & z \\ x & y & a+z \end{array}\right|=a^{2}(a+x+y+z)$Hint:We will try to make some elements of the determinant into zero
Given :$\left|\begin{array}{ccc} a+x & y & z \\ x & a+y & z \\ x & y & a+z \end{array}\right|$
Solution:$\left|\begin{array}{ccc} a+x & y & z \\ x & a+y & z \\ x & y & a+z \end{array}\right|$
$\begin{aligned} &\text { On applying } \mathrm{C}_{1} \rightarrow C_{1}+C_{2}+C_{3} \\ &=\left|\begin{array}{ccc} a+x+y+z & y & z \\ a+x+y+z & a+y & z \\ a+x+y+z & y & a+z \end{array}\right| \end{aligned}$
$\begin{aligned} &\text { On taking common }(a+x+y+z) \text { from } \mathrm{C}_{1}\\ &=(a+x+y+z)\left|\begin{array}{ccc} 1 & y & z \\ 1 & a+y & z \\ 1 & y & a+z \end{array}\right| \end{aligned}$
$\begin{aligned} &\text { On applying } \mathrm{R}_{1} \rightarrow R_{1}-R_{2} \text { and } \mathrm{R}_{2} \rightarrow R_{2}-R_{3} \\ &=(a+x+y+z)\left|\begin{array}{ccc} 0 & y-(a+y) & z-z \\ 0 & a+y-y & z-(a+z) \\ 1 & y & a+z \end{array}\right| \\ &=(a+x+y+z)\left|\begin{array}{ccc} 0 & y-a-y & 0 \\ 0 & a & z-a-z \\ 1 & y & a+z \end{array}\right| \\ &=(a+x+y+z)\left|\begin{array}{ccc} 0 & -a & 0 \\ 0 & a & -a \\ 1 & y & a+z \end{array}\right| \end{aligned}$
$\begin{aligned} &\text { On expanding w.r.t } \mathrm{R}_{1}\\ &=(a+x+y+z)\left[0\left|\begin{array}{cc} a & -a \\ y & a+z \end{array}\right|-(-a)\left|\begin{array}{cc} 0 & -a \\ 1 & a+z \end{array}\right|+0\left|\begin{array}{ll} 0 & a \\ 1 & y \end{array}\right|\right]\\ &=(a+x+y+z)[0+a\{0 \times(a+z)-(-a) 1\}+0]\\ &=(a+x+y+z)\{a(0+a)\}\\ &=(a+x+y+z) a^{2}\\ &=a^{2}(a+x+y+z) \end{aligned}$
Hence $\left|\begin{array}{ccc} a+x & y & z \\ x & a+y & z \\ x & y & a+z \end{array}\right|=a^{2}(a+x+y+z)$
deteminants Exercise 5.2 Question 10
Answer:-$\Delta=\left|\begin{array}{ccc} 1 & x & x^{2} \\ 1 & y & y^{2} \\ 1 & z & z^{2} \end{array}\right|+\Delta_{1}=\left|\begin{array}{ccc} 1 & 1 & 1 \\ y z & z x & x y \\ x & y & z \end{array}\right|=0$Hint: First we solve $\Delta$ and after we solve for $\Delta _{1}$
We will try to make some elements of the determinant zero
Given:$\Delta=\left|\begin{array}{lll} 1 & x & x^{2} \\ 1 & y & y^{2} \\ 1 & z & z^{2} \end{array}\right| \quad, \quad \Delta_{1}=\left|\begin{array}{ccc} 1 & 1 & 1 \\ y z & z x & x y \\ x & y & z \end{array}\right|$
Solution:Let us solve for $\Delta=\left|\begin{array}{lll} 1 & x & x^{2} \\ 1 & y & y^{2} \\ 1 & z & z^{2} \end{array}\right|$
$\Delta=\left|\begin{array}{lll} 1 & x & x^{2} \\ 1 & y & y^{2} \\ 1 & z & z^{2} \end{array}\right|$
$\begin{aligned} &\text { On applying } R_{1} \rightarrow R_{1}-R_{2} \text { and } R_{2} \rightarrow R_{2}-R_{3}\\ &\Delta=\left|\begin{array}{ccc} 0 & x-y & x^{2}-y^{2} \\ 0 & y-z & y^{2}-z^{2} \\ 1 & z & z^{2} \end{array}\right|\\ &\Delta=\left|\begin{array}{ccc} 0 & x-y & (x+y)(x-y) \\ 0 & y-z & (y+z)(y-z) \\ 1 & z & z^{2} \end{array}\right| \quad \mathrm{a}^{2}-b^{2}=(a+b)(a-b) \end{aligned}$
$\begin{aligned} &\text { On taking common }(x-y) \text { from } \mathrm{R}_{1} \text { and }(y-z) \text { from } \mathrm{R}_{2}\\ &\Delta=(x-y)(y-z)\left|\begin{array}{ccc} 0 & 1 & (x+y) \\ 0 & 1 & (y+z) \\ 1 & z & z^{2} \end{array}\right| \end{aligned}$
$\Delta=(x-y)(y-z)\left|\begin{array}{ccc} 0 & 0 & 1 \\ 1 & 1 & z \\ (x+y) & (y+z) & z^{2} \end{array}\right|$
If we convert all columns into rows
then the value of determinant does not change
$\Delta=(x-y)(y-z)\left|\begin{array}{ccc} 0 & 0 & 1 \\ 1 & 1 & z \\ (x+y) & (y+z) & z^{2} \end{array}\right|$
$\begin{aligned} &\text { On expanding w.r.t } \mathrm{R}_{1}\\ &\Delta=(x-y)(y-z)\left[0\left|\begin{array}{cc} 1 & z \\ y+z & z^{2} \end{array}\right|-0\left|\begin{array}{cc} 1 & z \\ x+y & z^{2} \end{array}\right|+1\left|\begin{array}{cc} 1 & 1 \\ x+y & y+z \end{array}\right|\right]\\ &\Delta=(x-y)(y-z)[0-0+1\{1(y+z)-1(x+y)\}]\\ &\Delta=(x-y)(y-z)(y+z-x-y)\\ &\Delta=(x-y)(y-z)(z-x) \quad \ldots \ldots \ldots(1) \end{aligned}$
Now Lets us Solve for $\Delta_{1}=\left|\begin{array}{ccc} 1 & 1 & 1 \\ y z & z x & x y \\ x & y & z \end{array}\right|$
$\Delta_{1}=\left|\begin{array}{ccc} 1 & 1 & 1 \\ y z & z x & x y \\ x & y & z \end{array}\right|$
$\begin{aligned} &\text { On applying } \mathrm{C}_{1} \rightarrow C_{1}-C_{2} \text { and } \mathrm{C}_{2} \rightarrow C_{2}-C_{3}\\ &\Delta_{1}=\left|\begin{array}{ccc} 0 & 0 & 1 \\ y z-z x & z x-x y & x y \\ x-y & y-z & z \end{array}\right|\\ &\Delta_{1}=\left|\begin{array}{ccc} 0 & 0 & 1 \\ -z(x-y) & -x(y-z) & x y \\ x-y & y-z & z \end{array}\right| \end{aligned}$
$\begin{aligned} &\text { On taking common }(x-y) \text { from } \mathrm{C}_{1} \text { and }(y-z) \text { from } \mathrm{C}_{2}\\ &\Delta_{1}=(x-y)(y-z)\left|\begin{array}{ccc} 0 & 0 & 1 \\ -z & -x & x y \\ 1 & 1 & z \end{array}\right| \end{aligned}$
On expanding w.r.t $R_{1}$
$\begin{aligned} &\Delta_{1}=(x-y)(y-z)\left[0\left|\begin{array}{cc} -x & x y \\ 1 & z \end{array}\right|-0\left|\begin{array}{cc} -z & x y \\ 1 & z \end{array}\right|+1 \mid \begin{array}{cc} -z & -x \\ 1 & 1 \end{array}\right] \\ &\Delta_{1}=(x-y)(y-z)[0-0+1\{(-z)(1)-(-x)(1)\}] \\ &\Delta_{1}=(x-y)(y-z)(-z+x) \\ &\Delta_{1}=-(x-y)(y-z)(z-x) \end{aligned}$
Now L.H.S
$\Delta+\Delta_{1}=(x-y)(y-z)(z-x)+\{-(x-y)(y-z)(z-x)\}$
From equation (1) and (2)
$\begin{aligned} &\Delta+\Delta_{1}=0 \\ &=R \cdot H \cdot S \end{aligned}$
Hence it is proved that $\Delta+\Delta_{1}=0$
determinants Exercise 5.2 Question 11
Answer:$\left|\begin{array}{ccc} a & b & c \\ a-b & b-c & c-a \\ b+c & c+a & a+b \end{array}\right|=a^{3}+b^{3}+c^{3}-3 a b c$Hint: We will try to make some elements of the determinant zero
Given:$\left|\begin{array}{ccc} a & b & c \\ a-b & b-c & c-a \\ b+c & c+a & a+b \end{array}\right|$
Solution: L.H.S$\left|\begin{array}{ccc} a & b & c \\ a-b & b-c & c-a \\ b+c & c+a & a+b \end{array}\right|$
$\begin{aligned} &\text { On applying } \mathrm{C}_{1} \rightarrow C_{1}+C_{2}+C_{3} \\ &=\left|\begin{array}{ccc} a+b+c & b & c \\ a-b+b-c+c-a & b-c & c-a \\ b+c+c+a+a+b & c+a & a+b \end{array}\right| \\ &=\left|\begin{array}{ccc} a+b+c & b & c \\ 0 & b-c & c-a \\ 2 a+2 b+2 c & c+a & a+b \end{array}\right| \\ &=\left|\begin{array}{ccc} a+b+c & b & c \\ 0 & b-c & c-a \\ 2(a+b+c) & c+a & a+b \end{array}\right| \end{aligned}$
$\begin{aligned} &\text { On taking common }(a+b+c) \text { from } \mathrm{C}_{1}\\ &=(a+b+c)\left|\begin{array}{ccc} 1 & b & c \\ 0 & b-c & c-a \\ 2 & c+a & a+b \end{array}\right| \end{aligned}$
$\begin{aligned} &\text { On expanding wr.t } \mathrm{R}_{1}\\ &=(a+b+c)\left[1\left|\begin{array}{ll} b-c & c-a \\ c+a & a+b \end{array}\right|-b\left|\begin{array}{ll} 0 & c-a \\ 2 & a+b \end{array}\right|+c \mid \begin{array}{ll} 0 & b-c \\ 2 & c+a \end{array}\right]\\ &=(a+b+c)[1\{(b-c)(a+b)-(c-a)(c+a)\}-b\{0 \times(a+b)-2(c-a)\}+c\{0 \times(c+a)-2(b-c)\}]\\ &=(a+b+c)\left[1\left\{\left(a b-c a+b^{2}-b c\right)-\left(c^{2}-a^{2}\right)\right\}-b\{0-2 c+2 a\}+c\{0-2 b+2 c\}\right]\\ &=(a+b+c)\left[a b-c a+b^{2}-b c-c^{2}+a^{2}+2 b c-2 a b-2 b c+2 c^{2}\right] \end{aligned}$
$\begin{aligned} &=(a+b+c)\left(a^{2}+b^{2}+c^{2}-a b-b c-c a\right) \\ &=a^{3}+b^{3}+c^{3}-3 a b c \quad \because x^{3}+y^{3}+z^{3}-3 x y z=(x+y+z)\left(x^{2}+y^{2}+z^{2}-x y-y z-z x\right) \end{aligned}$
$=RHS$
Hence it is proved that
$\left|\begin{array}{ccc} a & b & c \\ a-b & b-c & c-a \\ b+c & c+a & a+b \end{array}\right|=a^{3}+b^{3}+c^{3}-3 a b c$
determinants Exercise 5.2 Question 12
Answer:$\left|\begin{array}{lll} b+c & a-b & a \\ c+a & b-c & b \\ a+b & c-a & c \end{array}\right|=3 a b c-a^{3}-b^{3}-c^{3}$Hint:We will try to make some elements of the determinant zero
Given:$\left|\begin{array}{lll} b+c & a-b & a \\ c+a & b-c & b \\ a+b & c-a & c \end{array}\right|$
Solution:L.H.S$\left|\begin{array}{lll} b+c & a-b & a \\ c+a & b-c & b \\ a+b & c-a & c \end{array}\right|$
$\begin{aligned} &\text { On applying } \mathrm{R}_{1} \rightarrow R_{1}+R_{2}+R_{3} \\ &=\left|\begin{array}{ccc} b+c+c+a+a+b & (a-b+b-c+c-a) & a+b+c \\ c+a & b-c & b \\ a+b & c-a & c \end{array}\right| \\ &=\left|\begin{array}{ccc} 2 a+2 b+2 c & 0 & a+b+c \\ c+a & b-c & b \\ a+b & c-a & c \end{array}\right| & \\ &=\left|\begin{array}{ccc} 2(a+b+c) & 0 & a+b+c \\ c+a & b-c & b \\ a+b & c-a & c \end{array}\right| \end{aligned}$
$\begin{aligned} &\text { On taking common }(a+b+c) \text { from } \mathrm{R}_{1}\\ &=(a+b+c)\left|\begin{array}{ccc} 2 & 0 & 1 \\ c+a & b-c & b \\ a+b & c-a & c \end{array}\right| \end{aligned}$
$\begin{aligned} &\text { On expanding w.r.t } \mathrm{R}_{1}\\ &=(a+b+c)\left[2\left|\begin{array}{ll} b-c & b \\ c-a & c \end{array}\right|-0\left|\begin{array}{ll} c+a & b \\ a+b & c \end{array}\right|+1\left|\begin{array}{ll} c+a & b-c \\ a+b & c-a \end{array}\right|\right.\\ &=(a+b+c)[1\{(b-c)(c)-(b)(c-a)\}-0+1\{(c+a)(c-a)-(b-c)(a+b)\}]\\ &=(a+b+c)\left[2\left(b c-c^{2}-b c+a b\right)+1\left\{c^{2}-a^{2}-\left(a b-a c+b^{2}-b c\right)\right\}\right]\\ &=(a+b+c)\left\{2\left(-c^{2}+a b\right)+1\left(c^{2}-a^{2}-a b+a c-b^{2}+b c\right)\right\}\\ &=(a+b+c)\left(-2 c^{2}+2 a b+c^{2}-a^{2}-a b+a c-b^{2}+b c\right)\\ &=(a+b+c)\left(c^{2}-a^{2}-b^{2}-a b+a c+b c\right)\\ &=-(a+b+c)\left(a^{2}+b^{2}+c^{2}-a b-b c-c a\right)\\ &=-\left(a^{3}+b^{3}+c^{3}-3 a b c\right)\\ &=-a^{3}-b^{3}-c^{3}+3 a b c \quad \because x^{3}+y^{3}+z^{3}-3 x y z=(x+y+z)\left(x^{2}+y^{2}+z^{2}-x y-y z-z x\right)\\ &=3 a b c-a^{3}-b^{3}-c^{3} \end{aligned}$
=RHS
Hence it is proved that
$\left|\begin{array}{lll} b+c & a-b & a \\ c+a & b-c & b \\ a+b & c-a & c \end{array}\right|=3 a b c-a^{3}-b^{3}-c^{3}$
deteminants Exercise 5.2 Question 13
Answer:$\left|\begin{array}{lll} a+b & b+c & c+a \\ b+c & c+a & a+b \\ c+a & a+b & b+c \end{array}\right|=2\left|\begin{array}{lll} a & b & c \\ b & c & a \\ c & a & b \end{array}\right|$Hint: We will make $R_{1}$ of L.H.S according to R.H.S
Given:$\left|\begin{array}{lll} a+b & b+c & c+a \\ b+c & c+a & a+b \\ c+a & a+b & b+c \end{array}\right|=2\left|\begin{array}{lll} a & b & c \\ b & c & a \\ c & a & b \end{array}\right|$
Solution:L.H.S $\left|\begin{array}{lll} a+b & b+c & c+a \\ b+c & c+a & a+b \\ c+a & a+b & b+c \end{array}\right|=2\left|\begin{array}{lll} a & b & c \\ b & c & a \\ c & a & b \end{array}\right|$
$\begin{aligned} &\text { On applying } \mathrm{R}_{1} \rightarrow R_{1}-R_{2}+R_{3} \\ &=\left|\begin{array}{ccc} (a+b)-(b+c)+(c+a) & (b+c)-(c+a)+(a+b) & (c+a)-(a+b)+(b+c) \\ b+c & c+a & a+b \\ c+a & a+b & b+c \end{array}\right| \\ &=\left|\begin{array}{ccc} a+b-b-c+c+a & b+c-c-a+a+b & c+a-a-b+b+c \\ b+c & c+a & a+b \\ c+a & a+b & b+c \end{array}\right| \\ &=\left|\begin{array}{ccc} 2 a & 2 b & 2 c \\ b+c & c+a & a+b \\ c+a & a+b & b+c \end{array}\right| \end{aligned}$
$\begin{aligned} &\text { On taking } 2 \text { common from } \mathrm{R}_{1}\\ &=2\left|\begin{array}{ccc} a & b & c \\ b+c & c+a & a+b \\ c+a & a+b & b+c \end{array}\right| \end{aligned}$
$\begin{aligned} &\text { On applying } \mathrm{R}_{3} \rightarrow R_{3}-R_{1} \\ &=2\left|\begin{array}{ccc} a & b & c \\ b+c & c+a & a+b \\ c+a-a & a+b-b & b+c-c \end{array}\right| \\ &=2\left|\begin{array}{ccc} a & b & c \\ b+c & c+a & a+b \\ c & a & b \end{array}\right| \end{aligned}$
$\begin{aligned} &\text { On applying } \mathrm{R}_{2} \rightarrow R_{2}-R_{3} \\ &=2\left|\begin{array}{ccc} a & b & c \\ b+c-c & c+a-a & a+b-b \\ c & a & b \end{array}\right| \\ &=2\left|\begin{array}{lll} a & b & c \\ b & c & a \\ c & a & b \end{array}\right| \\ &=R \cdot H \cdot S \end{aligned}$
Hence it is proved that
$\left|\begin{array}{lll} a+b & b+c & c+a \\ b+c & c+a & a+b \\ c+a & a+b & b+c \end{array}\right|=2\left|\begin{array}{lll} a & b & c \\ b & c & a \\ c & a & b \end{array}\right|$
deteminants Exercise 5.2 Question 14
Answer:$\left|\begin{array}{ccc} a+b+2 c & a & b \\ c & b+c+2 a & b \\ c & a & c+a+2 b \end{array}\right|=2(a+b+c)^{3}$Hint We will try to make some elements of the determinant zero
Given:$\left|\begin{array}{ccc} a+b+2 c & a & b \\ c & b+c+2 a & b \\ c & a & c+a+2 b \end{array}\right|=2(a+b+c)^{3}$
Solution:$\text { L.H.S }\left|\begin{array}{ccc} a+b+2 c & a & b \\ c & b+c+2 a & b \\ c & a & c+a+2 b \end{array}\right|$
$\begin{aligned} &\text { On applying } \mathrm{C}_{1} \rightarrow C_{1}+C_{2}+C_{3} \\ &=\left|\begin{array}{ccc} 2 a+2 b+2 c & a & b \\ 2 a+2 b+2 c & b+c+2 a & b \\ 2 a+2 b+2 c & a & c+a+2 b \end{array}\right| \\ &=\left|\begin{array}{ccc} 2(a+b+c) & a & b \\ 2(a+b+c) & b+c+2 a & b \\ 2(a+b+c) & a & c+a+2 b \end{array}\right| \end{aligned}$
$\begin{aligned} &\text { On taking common } 2(a+b+c) \text { from } \mathrm{C}_{1}\\ &=2(a+b+c)\left|\begin{array}{ccc} 1 & a & b \\ 1 & b+c+2 a & b \\ 1 & a & c+a+2 b \end{array}\right| \end{aligned}$
$\begin{aligned} &\text { On applying } \mathrm{R}_{1} \rightarrow R_{1}-R_{2}, \mathrm{R}_{2} \rightarrow R_{2}-R_{3} \\ &=2(a+b+c)\left|\begin{array}{ccc} 0 & a-(b+c+2 a) & b-b \\ 0 & b+c+2 a-a & b-(c+a+2 b) \\ 1 & a & c+a+2 b \end{array}\right| \\ &=2(a+b+c)\left|\begin{array}{ccc} 0 & a-b-c-2 a & 0 \\ 0 & b+c+a & b-c-a-2 b \\ 1 & a & c+a+2 b \end{array}\right| \\ &=2(a+b+c)\left|\begin{array}{ccc} 0 & -a-b-c & 0 \\ 0 & a+b+c & -a-b-c \\ 1 & a & c+a+2 b \end{array}\right| \\ &=2(a+b+c)\left|\begin{array}{ccc} 0 & -(a+b+c) & 0 \\ 0 & a+b+c & -(a+b+c) \\ 1 & a & c+a+2 b \end{array}\right| \end{aligned}$
$\begin{aligned} &\text { On taking common }(a+b+c) \text { from } \mathrm{R}_{2}\\ &=2(a+b+c)(a+b+c)\left|\begin{array}{ccc} 0 & -(a+b+c) & 0 \\ 0 & 1 & -1 \\ 1 & a & c+a+2 b \end{array}\right|\\ &=2(a+b+c)^{2}\left|\begin{array}{ccc} 0 & -(a+b+c) & 0 \\ 0 & 1 & -1 \\ 1 & a & c+a+2 b \end{array}\right| \end{aligned}$
$\begin{aligned} &\text { On expanding w.r.t } \mathrm{R}_{1}\\ &=2(a+b+c)^{2}\left[0\left|\begin{array}{cc} 1 & -1 \\ a & c+a+2 b \end{array}\right|-\{-(a+b+c)\}\left|\begin{array}{cc} 0 & -1 \\ 1 & c+a+2 b \end{array}\right|+0 \mid \begin{array}{ll} 0 & 1 \\ 1 & a \end{array}\right]\\ &=2(a+b+c)^{2}[0+(a+b+c)\{0-(-1) 1\}+0]\\ &=2(a+b+c)^{2}[(a+b+c)(1)]\\ &=2(a+b+c)^{2}(a+b+c)\\ &=2(a+b+c)^{3}\\ &=R \cdot H \cdot S \end{aligned}$
Hence it is proved that
$\left|\begin{array}{ccc} a+b+2 c & a & b \\ c & b+c+2 a & b \\ c & a & c+a+2 b \end{array}\right|=2(a+b+c)^{3}$
determinants Exercise 5.2 Question 15
Answer:$\left|\begin{array}{ccc} a-b-c & 2 a & 2 a \\ 2 b & b-c-a & 2 b \\ 2 c & 2 c & c-a-b \end{array}\right|=(a+b+c)^{3}$Hint:We will try to make some elements of the determinant zero
Given:$\left|\begin{array}{ccc} a-b-c & 2 a & 2 a \\ 2 b & b-c-a & 2 b \\ 2 c & 2 c & c-a-b \end{array}\right|=(a+b+c)^{3}$
Solution:
$\begin{aligned} &\text { L.H.S }\left|\begin{array}{ccc} a-b-c & 2 a & 2 a \\ 2 b & b-c-a & 2 b \\ 2 c & 2 c & c-a-b \end{array}\right| \\ &\text { On applying } \mathrm{R}_{1} \rightarrow R_{1}+R_{2}+R_{3} \\ &=\left|\begin{array}{ccc} a-b-c+2 b+2 c & 2 a+b-c-a+2 c & 2 a+2 b+c-a-b \\ 2 b & b-c-a & 2 b \\ 2 c & 2 c & c-a-b \end{array}\right| \end{aligned}$
$=\left|\begin{array}{ccc} a+b+c & a+b+c & a+b+c \\ 2 b & b-c-a & 2 b \\ 2 c & 2 c & c-a-b \end{array}\right|$
$\begin{aligned} &\text { On taking common }(a+b+c) \text { from } \mathrm{C}_{1}\\ &=(a+b+c)\left|\begin{array}{ccc} 1 & 1 & 1 \\ 2 b & b-c-a & 2 b \\ 2 c & 2 c & c-a-b \end{array}\right| \end{aligned}$
$\begin{aligned} &\text { On applying } \mathrm{C}_{1} \rightarrow C_{1}-C_{2}, \mathrm{C}_{2} \rightarrow C_{2}-C_{3} \\ &=(a+b+c)\left|\begin{array}{ccc} 0 & 0 & 1 \\ 2 b-(b-c-a) & b-c-a-2 b & 2 b \\ 2 c-2 c & 2 c-(c-a-b) & c-a-b \end{array}\right| \end{aligned}$
$\begin{aligned} &=(a+b+c)\left|\begin{array}{ccc} 0 & 0 & 1 \\ 2 b-b+c+a & -b-c-a & 2 b \\ 2 c-2 c & 2 c-c+a+b & c-a-b \end{array}\right| \\ &=(a+b+c)\left|\begin{array}{ccc} 0 & 0 & 1 \\ a+b+c & -(a+b+c) & 2 b \\ 0 & a+b+c & c-a-b \end{array}\right| \end{aligned}$
$\begin{aligned} &\text { On taking common }(a+b+c) \text { from } \mathrm{C}_{2}\\ &=(a+b+c)(a+b+c)\left|\begin{array}{ccc} 0 & 0 & 1 \\ a+b+c & -1 & 2 b \\ 0 & 1 & c-a-b \end{array}\right| \end{aligned}$
$\begin{aligned} &\text { On expanding w.r.t } \mathrm{R}_{1}\\ &=(a+b+c)^{2}\left[0\left|\begin{array}{cc} -1 & 2 b \\ 1 & c-a-b \end{array}\right|-0\left|\begin{array}{cc} a+b+c & 2 b \\ 0 & c-a-b \end{array}\right|+1\left|\begin{array}{cc} a+b+c & -1 \\ 0 & 1 \end{array}\right|\right]\\ &=(a+b+c)^{2}[0-0+1\{(a+b+c) \times 1-(-1) \times 0\}]\\ &=(a+b+c)^{2}[1(a+b+c)]\\ &=(a+b+c)^{2}(a+b+c)\\ &=(a+b+c)^{3}\\ &=R \cdot H \cdot S \end{aligned}$
Hence it is proved that
$\left|\begin{array}{ccc} a-b-c & 2 a & 2 a \\ 2 b & b-c-a & 2 b \\ 2 c & 2 c & c-a-b \end{array}\right|=(a+b+c)^{3}$
determinants Exercise 5.2 Question 16
Answer:
Answer:$\left|\begin{array}{lll} 1 & b+c & b^{2}+c^{2} \\ 1 & c+a & c^{2}+a^{2} \\ 1 & a+b & a^{2}+b^{2} \end{array}\right|=(a-b)(b-c)(c-a)$Hint : We will try to make some elements of the determinant zero
Given:$\left|\begin{array}{lll} 1 & b+c & b^{2}+c^{2} \\ 1 & c+a & c^{2}+a^{2} \\ 1 & a+b & a^{2}+b^{2} \end{array}\right|=(a-b)(b-c)(c-a)$
Solution:
$\text { L.H.S }\left|\begin{array}{lll} 1 & b+c & b^{2}+c^{2} \\ 1 & c+a & c^{2}+a^{2} \\ 1 & a+b & a^{2}+b^{2} \end{array}\right|$
If the rows and columns of a determinant are interchanged, the value of the determinant remains the same.
$=\left|\begin{array}{ccc} 1 & 1 & 1 \\ b+c & c+a & a+b \\ b^{2}+c^{2} & c^{2}+a^{2} & a^{2}+b^{2} \end{array}\right|$
$\begin{aligned} &\text { On applying } \mathrm{C}_{1} \rightarrow \mathrm{C}_{1}-\mathrm{C}_{2} \text { and } \mathrm{C}_{2} \rightarrow \mathrm{C}_{2}-\mathrm{C}_{3}\\ &=\left|\begin{array}{ccc} 0 & 0 & 1 \\ b+c-(c-a) & c+a-(a+b) & a+b \\ b^{2}+c^{2}-\left(c^{2}+a^{2}\right) & c^{2}+a^{2}-\left(a^{2}+b^{2}\right) & a^{2}+b^{2} \end{array}\right|\\ &=\left|\begin{array}{ccc} 0 & 0 & 1 \\ b+c-c-a & c+a-a-b & a+b \\ b^{2}+c^{2}-c^{2}-a^{2} & c^{2}+a^{2}-a^{2}-b^{2} & a^{2}+b^{2} \end{array}\right| \end{aligned}$
$\begin{aligned} &=\left|\begin{array}{ccc} 0 & 0 & 1 \\ b-a & c-b & a+b \\ b^{2}-a^{2} & c^{2}-b^{2} & a^{2}+b^{2} \end{array}\right| \\ &=\left|\begin{array}{ccc} 0 & 0 & 1 \\ b-a & c-b & a+b \\ (b-a)(b+a) & (c-b)(c+b) & a^{2}+b^{2} \end{array}\right| \quad \because x^{2}-y^{2}=(x-y)(x+y) \end{aligned}$
$\begin{aligned} &\text { On taking common }(b-a) \text { from } \mathrm{C}_{1} \text { and }(c-b) \text { from } \mathrm{C}_{2}\\ &=(b-a)(c-b)\left|\begin{array}{ccc} 0 & 0 & 1 \\ 1 & 1 & a+b \\ (b+a) & (c+b) & a^{2}+b^{2} \end{array}\right| \end{aligned}$
$\begin{aligned} &\text { On expanding w.r.t } R_{1}\\ &=(b-a)(c-b)\left[0\left|\begin{array}{cc} 1 & a+b \\ c+b & a^{2}+b^{2} \end{array}\right|-0\left|\begin{array}{cc} 1 & a+b \\ b+a & a^{2}+b^{2} \end{array}\right|+1\left|\begin{array}{cc} 1 & 1 \\ b+a & c+b \end{array}\right|\right]\\ &=(-)(a-b)(-)(b-c)[0-0+1\{(c+b) 1-1(b+a)\}]\\ &=(a-b)(b-c)[1\{(c+b-b-a)\}]\\ &=(a-b)(b-c)(c-a)\\ &=R H . S \end{aligned}$
Hence it is proved that
$\left|\begin{array}{lll} 1 & b+c & b^{2}+c^{2} \\ 1 & c+a & c^{2}+a^{2} \\ 1 & a+b & a^{2}+b^{2} \end{array}\right|=(a-b)(b-c)(c-a)$
determinants Exercise 5.2 Question 17
Answer:$\left|\begin{array}{ccc} a & a+b & a+2 b \\ a+2 b & a & a+b \\ a+b & a+2 b & a \end{array}\right|=9(a+b) b^{2}$Hint We will try to make some elements of the determinant zero
Given :$\left|\begin{array}{ccc} a & a+b & a+2 b \\ a+2 b & a & a+b \\ a+b & a+2 b & a \end{array}\right|=9(a+b) b^{2}$
Solution:
$\text { L.H.S }\left|\begin{array}{ccc} a & a+b & a+2 b \\ a+2 b & a & a+b \\ a+b & a+2 b & a \end{array}\right|$
$\begin{aligned} &\text { On applying } \mathrm{R}_{1} \rightarrow \mathrm{R}_{1}+\mathrm{R}_{2}+\mathrm{R}_{3} \\ &=\left|\begin{array}{ccc} a+a+2 b+a+b & a+b+a+a+2 b & a+2 b+a+b+a \\ a+2 b & a & a+b \\ a+b & a+2 b & a \end{array}\right| \\ &=\left|\begin{array}{ccc} 3 a+3 b & 3 a+3 b & 3 a+3 b \\ a+2 b & a & a+b \\ a+b & a+2 b & a \end{array}\right| \end{aligned}$
$\begin{aligned} &\text { Taking common }(3 a+3 b) \text { from } R_{1}\\ &=(3 a+3 b)\left|\begin{array}{ccc} 1 & 1 & 1 \\ a+2 b & a & a+b \\ a+b & a+2 b & a \end{array}\right|\\ &\text { On applying } \mathrm{R}_{2} \rightarrow \mathrm{R}_{2}-\mathrm{R}_{3}\\ &=3(a+b)\left|\begin{array}{ccc} 1 & 1 & 1 \\ a+2 b-(a+b) & a-(a+2 b) & a+b-a \\ a+b & a+2 b & a \end{array}\right|\\ &=3(a+b)\left|\begin{array}{ccc} 1 & 1 & 1 \\ a+2 b-a-b & a-a-2 b & b \\ a+b & a+2 b & a \end{array}\right|\\ &=3(a+b)\left|\begin{array}{ccc} 1 & 1 & 1 \\ b & -2 b & b \\ a+b & a+2 b & a \end{array}\right| \end{aligned}$
$\begin{aligned} &\text { Taking common (b) from } R_{2}\\ &=3(a+b) b\left|\begin{array}{ccc} 1 & 1 & 1 \\ 1 & -2 & 1 \\ a+b & a+2 b & a \end{array}\right|\\ &\text { On applying } \mathrm{C}_{1} \rightarrow C_{1}-\mathrm{C}_{2} \text { and } \mathrm{C}_{2} \rightarrow C_{2}-\mathrm{C}_{3}\\ &=3(a+b) b\left|\begin{array}{ccc} 0 & 0 & 1 \\ 1-(-2) & -2-1 & 1 \\ a+b-(a+2 b) & a+2 b-a & a \end{array}\right|\\ &=3(a+b) b\left|\begin{array}{ccc} 0 & 0 & 1 \\ 1+2 & -2-1 & 1 \\ a+b-a-2 b & a+2 b-a & a \end{array}\right|\\ &=3(a+b) b\left|\begin{array}{ccc} 0 & 0 & 1 \\ 3 & -3 & 1 \\ -b & 2 b & a \end{array}\right| \end{aligned}$
$\begin{aligned} &\text { On expanding w.r.t } R_{1}\\ &=3(a+b) b\left[0\left|\begin{array}{cc} -3 & 1 \\ 2 b & a \end{array}\right|-0\left|\begin{array}{ll} 3 & 1 \\ -b & a \end{array}\right|+1\left|\begin{array}{cc} 3 & -3 \\ -b & 2 b \end{array}\right|\right]\\ &=3(a+b) b[0-0+1\{3 \times 2 b-(-3)(-b)\}]\\ &=3(a+b) b[1(6 b-3 b)]\\ &=3(a+b) b \times 3 b\\ &=9(a+b) b^{2} \end{aligned}$
Hence it is proved that
$\left|\begin{array}{ccc} a & a+b & a+2 b \\ a+2 b & a & a+b \\ a+b & a+2 b & a \end{array}\right|=9(a+b) b^{2}$
deteminants Exercise 5.2 Question 18
Answer:$\left|\begin{array}{lll} 1 & a & b c \\ 1 & b & c a \\ 1 & c & a b \end{array}\right|=\left|\begin{array}{lll} 1 & a & a^{2} \\ 1 & b & b^{2} \\ 1 & c & c^{2} \end{array}\right|$Hint We will make column (3) of L.H.S abc
Given:$\left|\begin{array}{lll} 1 & a & b c \\ 1 & b & c a \\ 1 & c & a b \end{array}\right|=\left|\begin{array}{lll} 1 & a & a^{2} \\ 1 & b & b^{2} \\ 1 & c & c^{2} \end{array}\right|$
Solution:
$\text { L.H.S }\left|\begin{array}{lll} 1 & a & b c \\ 1 & b & c a \\ 1 & c & a b \end{array}\right|$
$\text { On multiplying by 'a' in } R_{1}=\text { by } b \text { ' in } R_{2} \text { and by 'c' in } R_{m}$
$\begin{aligned} &=\frac{1}{a b c}\left|\begin{array}{lll} a & a^{2} & a b c \\ b & b^{2} & a b c \\ c & c^{2} & a b c \end{array}\right|\\ &\text { Taking common }(a b c) \text { from } \mathrm{C}_{3}\\ &=\frac{a b c}{a b c}\left|\begin{array}{lll} a & a^{2} & 1 \\ b & b^{2} & 1 \\ c & c^{2} & 1 \end{array}\right|\\ &=\left|\begin{array}{lll} a & a^{2} & 1 \\ b & b^{2} & 1 \\ c & c^{2} & 1 \end{array}\right| \end{aligned}$
If any two rows or columns of a determinant are interchanged, then sign of the determinant is changed.
$\begin{aligned} &C_{2} \leftrightarrow C_{3} \\ &=(-)\left|\begin{array}{lll} a & 1 & a^{2} \\ b & 1 & b^{2} \\ c & 1 & c^{2} \end{array}\right| \\ &C_{1} \leftrightarrow C_{2} \\ &=(-)(-)\left|\begin{array}{lll} 1 & a & a^{2} \\ 1 & b & b^{2} \\ 1 & c & c^{2} \end{array}\right| \\ &=\left|\begin{array}{lll} 1 & a & a^{2} \\ 1 & b & b^{2} \\ 1 & c & c^{2} \end{array}\right| \\ &=R \cdot H \cdot S \end{aligned}$
Hence it is proved that
$\left|\begin{array}{lll} 1 & a & b c \\ 1 & b & c a \\ 1 & c & a b \end{array}\right|=\left|\begin{array}{lll} 1 & a & a^{2} \\ 1 & b & b^{2} \\ 1 & c & c^{2} \end{array}\right|$
deteminants Exercise 5.2 Question 19.
Answer:$\left|\begin{array}{ccc} z & x & y \\ z^{2} & x^{2} & y^{2} \\ z^{4} & x^{4} & y^{4} \end{array}\right|=\left|\begin{array}{ccc} x & y & z \\ x^{2} & y^{2} & z^{2} \\ x^{4} & y^{4} & z^{4} \end{array}\right|=\left|\begin{array}{ccc} x^{2} & y^{2} & z^{2} \\ x^{4} & y^{4} & z^{4} \\ x & y & z \end{array}\right|$$=x y z(x-y)(y-z)(z-x)(x+y+z)$
Hint:We will prove all the determinant equal by using interchange column or row property
Given:$\begin{aligned} &\left|\begin{array}{lll} z & x & y \\ z^{2} & x^{2} & y^{2} \\ z^{4} & x^{4} & y^{4} \end{array}\right|=\left|\begin{array}{ccc} x & y & z \\ x^{2} & y^{2} & z^{2} \\ x^{4} & y^{4} & z^{4} \end{array}\right|=\left|\begin{array}{ccc} x^{2} & y^{2} & z^{2} \\ x^{4} & y^{4} & z^{4} \\ x & y & z \end{array}\right| \\ &=x y z(x-y)(y-z)(z-x)(x+y+z) \end{aligned}$
Solution:
Let us first solve for
$\left|\begin{array}{ccc} z & x & y \\ z^{2} & x^{2} & y^{2} \\ z^{4} & x^{4} & y^{4} \end{array}\right|$
If any two rows or columns of a determinant are interchanged, then sign of the determinant is changed.
$C_{1} \leftrightarrow C_{2}$
$=(-)\left|\begin{array}{ccc}x & z & y \\ x^{2} & z^{2} & y^{2} \\ x^{4} & z^{4} & y^{4}\end{array}\right|$
$C_{1} \leftrightarrow C_{2}$
$=(-)(-)\left|\begin{array}{ccc}x & y & z \\ x^{2} & y^{2} & z^{2} \\ x^{4} & y^{4} & z^{4}\end{array}\right|$
$=\left|\begin{array}{ccc}x & y & z \\ x^{2} & y^{2} & z^{2} \\ x^{4} & y^{4} & z^{4}\end{array}\right|$
$R_{1} \leftrightarrow R_{2} =(-)\left|\begin{array}{ccc}x^{2} & y^{2} & z^{2} \\ x & y & z \\ x^{4} & y^{4} & z^{4}\end{array}\right| R_{2} \leftrightarrow R_{3} =(-)(-)\left|\begin{array}{ccc}x^{2} & y^{2} & z^{2} \\ x^{4} & y^{4} & z^{4} \\ x & y & z\end{array}\right| =\left|\begin{array}{ccc}x^{2} & y^{2} & z^{2} \\ x^{4} & y^{4} & z^{4} \\ x & y & z\end{array}\right|$.....2
$From equation (1) and (2) we get \left|\begin{array}{ccc}z & x & y \\ z^{2} & x^{2} & y^{2} \\ z^{4} & x^{4} & y^{4}\end{array}\right|=\left|\begin{array}{ccc}x & y & z \\ x^{2} & y^{2} & z^{2} \\ x^{4} & y^{4} & z^{4}\end{array}\right|=\left|\begin{array}{ccc}x^{2} & y^{2} & z^{2} \\ x^{4} & y^{4} & z^{4} \\ x & y & z\end{array}\right| \quad \ldots$(3)
$Now consider- \left|\begin{array}{ccc}x^{2} & y^{2} & z^{2} \\ x^{4} & y^{4} & z^{4} \\ x & y & z\end{array}\right| Taking common x from C_{1}, y from C_{2} and z from C_{3} =x y z\left|\begin{array}{ccc}x & y & z \\ x^{3} & y^{3} & z^{3} \\ 1 & 1 & 1\end{array}\right| On applying C_{1} \rightarrow C_{1}-C_{2} and C_{2} \rightarrow C_{2}-C_{3} =x y z\left|\begin{array}{ccc}x-y & y-z & z \\ x^{3}-y^{3} & y^{3}-z^{3} & z^{3} \\ 0 & 0 & 1\end{array}\right| =x y z\left|\begin{array}{ccc}x-y & y-z & z \\ (x-y)\left(x^{2}+x y+y^{2}\right) & (y-z)\left(y^{2}+y z+z^{2}\right) & z^{3} \\ 0 & 0 & 1\end{array}\right|$
$\because x^{3}-y^{3}=(x-y)\left(x^{2}+x y+y^{2}\right)$
$\begin{aligned} &\text { On taking common }(x-y) \text { common from } \mathrm{C}_{1} \text { and }(y-z) \text { from } \mathrm{C}_{2}\\ &=x y z(x-y)(y-z)\left|\begin{array}{ccc} 1 & 1 & z \\ \left(x^{2}+x y+y^{2}\right) & \left(y^{2}+y z+z^{2}\right) & z^{3} \\ 0 & 0 & 1 \end{array}\right| \end{aligned}$
$\begin{aligned} &\text { On expanding w.r.t } \mathrm{R}_{3}\\ &=x y z(x-y)(y-z)\left[0\left|\begin{array}{cc} 1 & z \\ y^{2}+y z+z^{2} & z^{3} \end{array}\right|-0\left|\begin{array}{cc} 1 & z \\ x^{2}+x y+y^{2} & z^{3} \end{array}\right|+1\left|\begin{array}{cc} 1 & 1 \\ x^{2}+x y+y^{2} & y^{2}+y z+z^{2} \end{array}\right|\right]\\ &=x y z(x-y)(y-z)\left[0-0+1\left\{\left(y^{2}+y z+z^{2}\right)-\left(x^{2}+x y+y^{2}\right)\right\}\right]\\ &=x y z(x-y)(y-z)\left(y^{2}+y z+z^{2}-x^{2}-x y-y^{2}\right)\\ &=x y z(x-y)(y-z)\left(z^{2}-x^{2}+y z-x y\right)\\ &=x y z(x-y)(y-z)\{(z+x)(z-x)+y(z-x)\}\\ &=x y z(x-y)(y-z)(z-x)(z+x+y)\\ &=x y z(x-y)(y-z)(z-x)(x+y+z) \end{aligned}$
From equation (1), (2),(3) and (4)
Hence it is proved that
$\left|\begin{array}{ccc} z & x & y \\ z^{2} & x^{2} & y^{2} \\ z^{4} & x^{4} & y^{4} \end{array}\right|=\left|\begin{array}{ccc} x & y & z \\ x^{2} & y^{2} & z^{2} \\ x^{4} & y^{4} & z^{4} \end{array}\right|=\left|\begin{array}{ccc} x^{2} & y^{2} & z^{2} \\ x^{4} & y^{4} & z^{4} \\ x & y & z \end{array}\right|$
$=x y z(x-y)(y-z)(z-x)(x+y+z)$
determinants Exercise 5.2 Question 20
Answer:$\left|\begin{array}{lll} (b+c)^{2} & a^{2} & b c \\ (c+a)^{2} & b^{2} & c a \\ (a+b)^{2} & c^{2} & a b \end{array}\right|=(a-b)(b-c)(c-a)(a+b+c)\left(a^{2}+b^{2}+c^{2}\right)$Hint We will try to make $\left ( a-b \right )$and $\left ( b-c \right )$any two elements of determinant
Given:$\left|\begin{array}{lll} (b+c)^{2} & a^{2} & b c \\ (c+a)^{2} & b^{2} & c a \\ (a+b)^{2} & c^{2} & a b \end{array}\right|=(a-b)(b-c)(c-a)\left(a^{2}+b^{2}+c^{2}\right)$
Solution:$\text { L.H.S }\left|\begin{array}{lll} (b+c)^{2} & a^{2} & b c \\ (c+a)^{2} & b^{2} & c a \\ (a+b)^{2} & c^{2} & a b \end{array}\right|$
$\begin{aligned} &\text { On applying } \mathrm{R}_{1} \rightarrow R_{1}-R_{2} \text { and } \mathrm{R}_{2} \rightarrow R_{2}-R_{3}\\ &=\left|\begin{array}{ccc} (b+c)^{2}-(c+a)^{2} & a^{2}-b^{2} & b c-c a \\ (c+a)^{2}-(a+b)^{2} & b^{2}-c^{2} & c a-a b \\ (a+b)^{2} & c^{2} & a b \end{array}\right|\\ &x^{2}-y^{2}=(x+y)(x-y) \end{aligned}$
$\begin{aligned} &=\left|\begin{array}{ccc} (b+c+c+a)(b+c-c-a) & (a+b)(a-b) & -c(a-b) \\ (c+a+a+b)(c+a-a-b) & (b+c)(b-c) & -a(b-c) \\ (a+b)^{2} & c^{2} & a b \end{array}\right| \\ &=\left|\begin{array}{ccc} (a+b+2 c)(b-a) & (a+b)(a-b) & -c(a-b) \\ (2 a+b+c)(c-b) & (b+c)(b-c) & -a(b-c) \\ (a+b)^{2} & c^{2} & a b \end{array}\right| \\ &=\left|\begin{array}{ccc} -(a+b+2 c)(a-b) & (a+b)(a-b) & -c(a-b) \\ -(2 a+b+c)(b-c) & (b+c)(b-c) & -a(b-c) \\ (a+b)^{2} & c^{2} & a b \end{array}\right| \end{aligned}$
$\begin{aligned} &\text { Taking common }(a-b) \text { from } \mathrm{R}_{1},(b-c) \text { from } \mathrm{R}_{2}\\ &=(a-b)(b-c)\left|\begin{array}{ccc} -(a+b+2 c) & (a+b) & -c \\ -(2 a+b+c) & (b+c) & -a \\ (a+b)^{2} & c^{2} & a b \end{array}\right|\\ &\text { On applying } \mathrm{R}_{1} \rightarrow R_{1}-R_{2}\\ &=(a-b)(b-c)\left|\begin{array}{ccc} -a-b-2 c+2 a+b+c & a+b-b-c & -c+a \\ -(2 a+b+c) & (b+c) & -a \\ (a+b)^{2} & c^{2} & a b \end{array}\right| \end{aligned}$
$\begin{aligned} &=(a-b)(b-c)\left|\begin{array}{ccc} a-c & a-c & a-c \\ -(2 a+b+c) & (b+c) & -a \\ (a+b)^{2} & c^{2} & a b \end{array}\right|\\ &\text { Taking common }(a-c) \text { from } \mathrm{R}_{1}\\ &=(a-b)(b-c)(a-c)\left|\begin{array}{ccc} 1 & 1 & 1 \\ -(2 a+b+c) & (b+c) & -a \\ (a+b)^{2} & c^{2} & a b \end{array}\right| \end{aligned}$
$\begin{aligned} &\text { On applying } C_{1} \rightarrow C_{1}-C_{2} \text { and } C_{2} \rightarrow C_{2}-C_{3}\\ &=(a-b)(b-c)(a-c)\left|\begin{array}{ccc} 0 & 0 & 1 \\ -2 a-b-c-b-c & b+c+a & -a \\ (a+b)^{2}-c^{2} & c^{2}-a b & a b \end{array}\right|\\ &=(a-b)(b-c)(a-c)\left|\begin{array}{ccc} 0 & 0 & 1 \\ -2 a-2 b-2 c & a+b+c & -a \\ (a+b+c)(a+b-c) & c^{2}-a b & a b \end{array}\right|\\ &=(a-b)(b-c)(a-c)\left|\begin{array}{ccc} 0 & 0 & 1 \\ -2(a+b+c) & a+b+c & -a \\ (a+b+c)(a+b-c) & c^{2}-a b & a b \end{array}\right| \end{aligned}$
Taking common(a+b+c)from$C_{1}$
$\begin{aligned} &=(a+b+c)(a-b)(b-c)(a-c)\left|\begin{array}{ccc} 0 & 0 & 1 \\ -2 & a+b+c & -a \\ (a+b-c) & c^{2}-a b & a b \end{array}\right| \\ &\quad=-(a+b+c)(a-b)(b-c)(c-a)\left|\begin{array}{ccc} 0 & 0 & 1 \\ (a+b-c) & c^{2}-a b & a b \end{array}\right| \end{aligned}$
$\begin{aligned} &\text { On expanding w.r.t } \mathrm{R}_{1}\\ &=-(a+b+c)(a-b)(b-c)(c-a)\left[0\left|\begin{array}{cc} a+b+c & -a \\ c^{2}-a b & a b \end{array}\right|-0\left|\begin{array}{cc} -2 & -a \\ a+b-c & a b \end{array}\right|\right.\\ &\left.+1\left|\begin{array}{cc} -2 & a+b+c \\ a+b-c & c^{2}-a b \end{array}\right|\right]\\ &=-(a+b+c)(a-b)(b-c)(c-a)\left[0-0+1\left\{-2\left(c^{2}-a b\right)-(a+b-c)(a+b+c)\right\}\right]\\ &=-(a+b+c)(a-b)(b-c)(c-a)\left[-2 c^{2}+2 a b-\left\{(a+b)^{2}-c^{2}\right\}\right]\\ &=-(a+b+c)(a-b)(b-c)(c-a)\left[-2 c^{2}+2 a b-\left\{a^{2}+b^{2}+2 a b-c^{2}\right\}\right]\\ &\because(a+b)^{2}=a^{2}+b^{2}+2 a b\\ &=-(a+b+c)(a-b)(b-c)(c-a)\left(-2 c^{2}+2 a b-a^{2}-b^{2}-2 a b-c^{2}\right)\\ &=-(a+b+c)(a-b)(b-c)(c-a)\left(-a^{2}-b^{2}-c^{2}\right) \end{aligned}$
$\begin{aligned} &=(-)(-)(a-b)(b-c)(c-a)(a+b+c)\left(a^{2}+b^{2}+c^{2}\right) \\ &=(a-b)(b-c)(c-a)(a+b+c)\left(a^{2}+b^{2}+c^{2}\right) \end{aligned}$
Hence it is proved that
$\left|\begin{array}{lll} (b+c)^{2} & a^{2} & b c \\ (c+a)^{2} & b^{2} & c a \\ (a+b)^{2} & c^{2} & a b \end{array}\right|=(a-b)(b-c)(c-a)(a+b+c)\left(a^{2}+b^{2}+c^{2}\right)$
deteminants Exercise 5.2 Question 21
Answer:$\left|\begin{array}{lll} (a+1)(a+2) & a+2 & 1 \\ (a+2)(a+3) & a+3 & 1 \\ (a+3)(a+4) & a+4 & 1 \end{array}\right|=-2$
Hint: $\text { First we will make two elements of } \mathrm{C}_{3} \text { zero. then we will expand it }$
Given:$\left|\begin{array}{lll} (a+1)(a+2) & a+2 & 1 \\ (a+2)(a+3) & a+3 & 1 \\ (a+3)(a+4) & a+4 & 1 \end{array}\right|=-2$
Solution:
$\text { L.H.S }\left|\begin{array}{lll} (a+1)(a+2) & a+2 & 1 \\ (a+2)(a+3) & a+3 & 1 \\ (a+3)(a+4) & a+4 & 1 \end{array}\right|$
$\begin{aligned} &\text { On applying } \mathrm{R}_{1} \rightarrow R_{1}-R_{2} \text { and } \mathrm{R}_{2} \rightarrow R_{2}-R_{3} \\ &=\left|\begin{array}{ccc} (a+1)(a+2)-(a+2)(a+3) & (a+2)-(a+3) & 0 \\ (a+2)(a+3)-(a+3)(a+4) & (a+3)-(a+4) & 0 \\ (a+3)(a+4) & a+4 & 1 \end{array}\right| \\ &=\left|\begin{array}{ccc} (a+2)(a+1-a-3) & a+2-a-3 & 0 \\ (a+3)(a+2-a-4) & a+3-a-4 & 0 \\ (a+3)(a+4) & a+4 & 1 \end{array}\right| \\ &=\left|\begin{array}{ccc} (a+2)(-2) & -1 & 0 \\ (a+3)(-2) & -1 & 0 \\ (a+3)(a+4) & a+4 & 1 \end{array}\right| \end{aligned}$
$\begin{aligned} &\text { On expanding w.r.t } \mathrm{C}_{3}\\ &=0\left|\begin{array}{cc} (a+3)(-2) & -1 \\ (a+3)(a+4) & a+4 \end{array}\right|-0\left|\begin{array}{cc} (a+2)(-2) & -1 \\ (a+3)(a+4) & a+4 \end{array}\right|+1 \mid\left(\begin{array}{ll} a+2)(-2) & -1 \\ (a+3)(-2) & -1 \end{array} \mid\right.\\ &=0-0+1\{(a+2)(-2)(-1)-(-1)(-2)(a+3)\}\\ &=1\{2(a+2)-2(a+3)\}\\ &=2 a+4-2 a-6\\ &=-2\\ &=R \cdot H \cdot S \end{aligned}$
Hence it is proved that
$\left|\begin{array}{lll} (a+1)(a+2) & a+2 & 1 \\ (a+2)(a+3) & a+3 & 1 \\ (a+3)(a+4) & a+4 & 1 \end{array}\right|=-2$
deteminants Exercise 5.2 Question 22
Answer:
Answer:$\left|\begin{array}{lll} a^{2} & a^{2}-(b-c)^{2} & b c \\ b^{2} & b^{2}-(c-a)^{2} & c a \\ c^{2} & c^{2}-(a-b)^{2} & a b \end{array}\right|=(a-b)(b-c)(c-a)(a+b+c)\left(a^{2}+b^{2}+c^{2}\right)$Hint $\text { We will try to convert some elements of the determinant }(a-b),(b-c) \text { and }(c-a) \text { to get the final answer }$
Given:$\left|\begin{array}{lll} a^{2} & a^{2}-(b-c)^{2} & b c \\ b^{2} & b^{2}-(c-a)^{2} & c a \\ c^{2} & c^{2}-(a-b)^{2} & a b \end{array}\right|=(a-b)(b-c)(c-a)(a+b+c)\left(a^{2}+b^{2}+c^{2}\right)$
Solution:
$\text { L.H.S }\left|\begin{array}{lll} a^{2} & a^{2}-(b-c)^{2} & b c \\ b^{2} & b^{2}-(c-a)^{2} & c a \\ c^{2} & c^{2}-(a-b)^{2} & a b \end{array}\right|$
$\begin{aligned} &\text { On applying } C_{1} \rightarrow C_{2}-2 C_{1}-2 C_{3}\\ &=\left|\begin{array}{lll} a^{2} & a^{2}-(b-c)^{2}-2 a^{2}-2 b c & b c \\ b^{2} & b^{2}-(c-a)^{2}-2 b^{2}-2 c a & c a \\ c^{2} & c^{2}-(a-b)^{2}-2 c^{2}-2 a b & a b \end{array}\right|\\ &=\left|\begin{array}{lll} a^{2} & -a^{2}-(b-c)^{2}-2 b c & b c \\ b^{2} & -b^{2}-(c-a)^{2}-2 c a & c a \\ c^{2} & -c^{2}-(a-b)^{2}-2 a b & a b \end{array}\right|\\ &=\left|\begin{array}{lll} a^{2} & -\left\{a^{2}+(b-c)^{2}+2 b c\right\} & b c \\ b^{2} & -\left\{b^{2}+(c-a)^{2}+2 c a\right\} & c a \\ c^{2} & -\left\{c^{2}+(a-b)^{2}+2 a b\right\} & a b \end{array}\right|\\ &=\left|\begin{array}{lll} a^{2} & -\left(a^{2}+b^{2}+c^{2}-2 b c+2 b c\right) & b c \\ b^{2} & -\left(b^{2}+c^{2}+a^{2}-2 c a+2 c a\right) & c a \\ c^{2} & -\left(c^{2}+a^{2}+b^{2}-2 a b+2 a b\right) & a b \end{array}\right| \end{aligned}$
$\begin{aligned} &=\left|\begin{array}{lll} a^{2} & -\left(a^{2}+b^{2}+c^{2}\right) & b c \\ b^{2} & -\left(a^{2}+b^{2}+c^{2}\right) & c a \\ c^{2} & -\left(a^{2}+b^{2}+c^{2}\right) & a b \end{array}\right|\\ &\text { Taking common }-\left(a^{2}+b^{2}+c^{2}\right) \text { from } \mathrm{C}_{2}\\ &=-\left(a^{2}+b^{2}+c^{2}\right)\left|\begin{array}{ccc} a^{2} & 1 & b c \\ b^{2} & 1 & c a \\ c^{2} & 1 & a b \end{array}\right| \end{aligned}$
$\begin{aligned} &\text { On applying } \mathrm{R}_{2} \rightarrow R_{2}-R_{1} \text { and } \mathrm{R}_{3} \rightarrow R_{3}-R_{1}\\ &=-\left(a^{2}+b^{2}+c^{2}\right)\left|\begin{array}{ccc} a^{2} & 1 & b c \\ b^{2}-a^{2} & 0 & c(a-b) \\ c^{2}-a^{2} & 0 & b(a-c) \end{array}\right|\\ &=-\left(a^{2}+b^{2}+c^{2}\right)\left|\begin{array}{ccc} a^{2} & 1 & b c \\ -(a+b)(a-b) & 0 & c(a-b) \\ (c+a)(c-a) & 0 & -b(c-a) \end{array}\right|\\ &\text { On taking common }(a-b) \text { from } \mathrm{R}_{2} \text { and }(c-a) \text { from } \mathrm{R}_{3}\\ &=-\left(a^{2}+b^{2}+c^{2}\right)(a-b)(c-a)\left|\begin{array}{ccc} a^{2} & 1 & b c \\ -(a+b) & 0 & c \\ (c+a) & 0 & -b \end{array}\right| \end{aligned}$
$\begin{aligned} &\text { On expanding w.r.t } \mathrm{R}_{1} \\ &\left.=-\left(a^{2}+b^{2}+c^{2}\right)(a-b)(c-a)\left[a^{2}\left|\begin{array}{cc} 0 & c \\ 0 & -b \end{array}\right|-1\left|\begin{array}{cc} -(a+b) & c \\ c+a & -b \end{array}\right|+b c \mid \begin{array}{cc} -(a+b) & 0 \\ c+a & 0 \end{array}\right]\right] \\ &=-\left(a^{2}+b^{2}+c^{2}\right)(a-b)(c-a)\left[a^{2}(0-0)-1\{(-b)(-(a+b)-c(c+a))+b c(0-0)\}\right] \\ &=-\left(a^{2}+b^{2}+c^{2}\right)(a-b)(c-a)\left[0-1\left\{b(a+b)-c^{2}-a c\right\}\right] \\ &=-\left(a^{2}+b^{2}+c^{2}\right)(a-b)(c-a)\left\{-1\left(a b+b^{2}-c^{2}-a c\right)\right\} \\ &=-\left(a^{2}+b^{2}+c^{2}\right)(a-b)(c-a)\left(-a b-b^{2}+c^{2}+a c\right) \end{aligned}$
$\begin{aligned} &=-\left(a^{2}+b^{2}+c^{2}\right)(a-b)(c-a)\left(a c-a b-b^{2}+c^{2}\right) \\ &=-\left(a^{2}+b^{2}+c^{2}\right)(a-b)(c-a)\left\{-a(b-c)-\left(b^{2}-c^{2}\right)\right\} \\ &=-\left(a^{2}+b^{2}+c^{2}\right)(a-b)(c-a)\{-a(b-c)-(b+c)(b-c)\} \\ &=-\left(a^{2}+b^{2}+c^{2}\right)(a-b)(c-a)(b-c)(-a-b-c) \\ &=(-)(-)\left(a^{2}+b^{2}+c^{2}\right)(a-b)(c-a)(b-c)(a+b+c) \\ &=(a-b)(b-c)(c-a)(a+b+c)\left(a^{2}+b^{2}+c^{2}\right) \\ &=R H . S \end{aligned}$
Hence it is proved that
$\left|\begin{array}{lll} a^{2} & a^{2}-(b-c)^{2} & b c \\ b^{2} & b^{2}-(c-a)^{2} & c a \\ c^{2} & c^{2}-(a-b)^{2} & a b \end{array}\right|=(a-b)(b-c)(c-a)(a+b+c)\left(a^{2}+b^{2}+c^{2}\right)$
determinants Exercise 5.2 Question 23
Answer:$\left|\begin{array}{lll} 1 & a^{2}+b c & a^{3} \\ 1 & b^{2}+c a & b^{3} \\ 1 & c^{2}+a b & c^{3} \end{array}\right|=-(a-b)(b-c)(c-a)\left(a^{2}+b^{2}+c^{2}\right)$Hint We will make some elements of the determinant zero
Given:$\left|\begin{array}{lll} 1 & a^{2}+b c & a^{3} \\ 1 & b^{2}+c a & b^{3} \\ 1 & c^{2}+a b & c^{3} \end{array}\right|=-(a-b)(b-c)(c-a)\left(a^{2}+b^{2}+c^{2}\right)$
Solution:
$\text { L.H.S }\left|\begin{array}{ccc} 1 & a^{2}+b c & a^{3} \\ 1 & b^{2}+c a & b^{3} \\ 1 & c^{2}+a b & c^{3} \end{array}\right|$
$\begin{aligned} &\text { On applying } \mathrm{R}_{1} \rightarrow R_{1}-R_{2} \text { and } \mathrm{R}_{2} \rightarrow R_{2}-R_{3}\\ &=\left|\begin{array}{ccc} 0 & a^{2}+b c-\left(b^{2}+c a\right) & a^{3}-b^{3} \\ 0 & b^{2}+c a-\left(c^{2}+a b\right) & b^{3}-c^{3} \\ 1 & c^{2}+a b & c^{3} \end{array}\right|\\ &=\left|\begin{array}{ccc} 0 & a^{2}+b c-b^{2}-c a & (a-b)\left(a^{2}+a b+b^{2}\right) \\ 0 & b^{2}+c a-c^{2}-a b & (b-c)\left(b^{2}+b c+c^{2}\right) \\ 1 & c^{2}+a b & c^{3} \end{array}\right| \quad \ldots . .(1) \end{aligned}$
$\begin{aligned} &\text { Consider } a^{2}+b c-b^{2}-c a \\ &=a^{2}-b^{2}+b c-c a \\ &=(a-b)(a+b)-(a-b) \\ &=(a-b)(a+b-c) \end{aligned}$
$\begin{aligned} &\text { Similarly }\\ &b^{2}+c a-c^{2}-a b=(b-c)(b+c-a)\\ &\because a^{3}-b^{3}=(a-b)\left(a^{2}+a b+b^{2}\right) \end{aligned}$
$\begin{aligned} &\text { Now from }(1)\\ &=\left|\begin{array}{ccc} 0 & (a-b)(a+b-c) & (a-b)\left(a^{2}+a b+b^{2}\right) \\ 0 & (b-c)(b+c-a) & (b-c)\left(b^{2}+b c+c^{2}\right) \\ 1 & c^{2}+a b & c^{3} \end{array}\right| \end{aligned}$
$\begin{aligned} &\text { On taking common }(a-b) \text { from } \mathrm{R}_{1} \text { and }(b-c) \text { from } \mathrm{R}_{2}\\ &=(a-b)(b-c)\left|\begin{array}{ccc} 0 & (a+b-c) & \left(a^{2}+a b+b^{2}\right) \\ 0 & (b+c-a) & \left(b^{2}+b c+c^{2}\right) \\ 1 & c^{2}+a b & c^{3} \end{array}\right| \end{aligned}$
$\begin{aligned} &\text { On applying } \mathrm{R}_{1} \rightarrow R_{1}-R_{2}\\ &=(a-b)(b-c)\left|\begin{array}{ccc} 0 & (a+b-c)-(b+c-a) & \left(a^{2}+a b+b^{2}\right)-\left(b^{2}+b c+c^{2}\right) \\ 0 & (b+c-a) & \left(b^{2}+b c+c^{2}\right) \\ 1 & c^{2}+a b & c^{3} \end{array}\right|\\ &=(a-b)(b-c)\left|\begin{array}{ccc} 0 & a+b-c-b-c+a & a^{2}+a b+b^{2}-b^{2}-b c-c^{2} \\ 0 & (b+c-a) & \left(b^{2}+b c+c^{2}\right) \\ 1 & c^{2}+a b & c^{3} \end{array}\right|\\ &=(a-b)(b-c)\left|\begin{array}{ccc} 0 & 2 a-2 c & a^{2}+a b-b c-c^{2} \\ 0 & (b+c-a) & \left(b^{2}+b c+c^{2}\right) \\ 1 & c^{2}+a b & c^{3} \end{array}\right| \quad \ldots . . \end{aligned}$(2)
$\begin{aligned} &\text { Consider } a^{2}+a b-b c-c^{2} \\ &=a^{2}-c^{2}+a b-b c \\ &=(a+c)(a-c)+b(a-c) \\ &=(a-c)(a+c+b) \end{aligned}$
From (2)
$\begin{aligned} &=(a-b)(b-c)\left|\begin{array}{ccc} 0 & 2(a-c) & (a-c)(a+c+b) \\ 0 & b+c-a & b^{2}+b c+c^{2} \\ 1 & c^{2}+a b & c^{3} \end{array}\right|\\ &\text { On taking common }(a-c) \text { from } \mathrm{R}_{1}\\ &=(a-b)(b-c)(a-c)\left|\begin{array}{ccc} 0 & 2 & (a+c+b) \\ 0 & b+c-a & b^{2}+b c+c^{2} \\ 1 & c^{2}+a b & c^{3} \end{array}\right| \end{aligned}$
$\begin{aligned} &\text { On expanding w.r.t } \mathrm{C}_{1}\\ &=-(a-b)(b-c)(c-a)\left[0\left|\begin{array}{ccc} b+c-a & b^{2}+b c+c^{2} \\ c^{2}+a b & c^{3} \end{array}\right|-0\left|\begin{array}{cc} 2 & a+b+c \\ c^{2}+a b & c^{3} \end{array}\right|+1 \mid \begin{array}{cc} 2 & a+b+c \\ b+c+a & b^{2}+b c+c^{2} \end{array}\right]\\ &=-(a-b)(b-c)(c-a)\left[0-0+\left\{2\left(b^{2}+b c+c^{2}\right)-(b+c-a)(b+c-a)\right\}\right]\\ &=-(a-b)(b-c)(c-a)\left[2 b^{2}+2 b c+2 c^{2}-\left\{(b+c)^{2}-a^{2}\right\}\right]\\ &=-(a-b)(b-c)(c-a)\left\{2 b^{2}+2 b c+2 c^{2}-\left(b^{2}+2 b c+c^{2}-a^{2}\right)\right\}\\ &=-(a-b)(b-c)(c-a)\left(2 b^{2}+2 b c+2 c^{2}-b^{2}-2 b c-c^{2}+a^{2}\right)\\ &=-(a-b)(b-c)(c-a)\left(a^{2}+b^{2}+c^{2}\right) \end{aligned}$
Hence it is proved that
$\left|\begin{array}{lll} 1 & a^{2}+b c & a^{3} \\ 1 & b^{2}+c a & b^{3} \\ 1 & c^{2}+a b & c^{3} \end{array}\right|=-(a-b)(b-c)(c-a)\left(a^{2}+b^{2}+c^{2}\right)$
determinants Exercise 5.2 Question 24
Answer:$\left|\begin{array}{ccc} a^{2} & b c & a c+c^{2} \\ a^{2}+a b & b^{2} & a c \\ a b & b^{2}+b c & c^{2} \end{array}\right|=4 a^{2} b^{2} c^{2}$Hint We will convert some elements of the determinant into zero
Given: $\left|\begin{array}{ccc} a^{2} & b c & a c+c^{2} \\ a^{2}+a b & b^{2} & a c \\ a b & b^{2}+b c & c^{2} \end{array}\right|=4 a^{2} b^{2} c^{2}$
Solution:
$\begin{aligned} &\text { L.H.S }\left|\begin{array}{ccc} a^{2} & b c & a c+c^{2} \\ a^{2}+a b & b^{2} & a c \\ a b & b^{2}+b c & c^{2} \end{array}\right| \\ &=\left|\begin{array}{ccc} a^{2} & b c & c(a+c) \\ a(a+b) & b^{2} & a c \\ a b & b(b+c) & c^{2} \end{array}\right| \end{aligned}$
$\begin{aligned} &\text { Taking out 'a' common from } \mathrm{C}_{1}, \text { and } \mathrm{b} \text { from } \mathrm{C}_{2} \text { and } \mathrm{c} \text { from } \mathrm{C}_{3}\\ &=a b c\left|\begin{array}{ccc} a & c & a+c \\ a+b & b & a \\ b & b+c & c \end{array}\right| \end{aligned}$
$\begin{aligned} &\text { On applying } \mathrm{C}_{1} \rightarrow \mathrm{C}_{1}+\mathrm{C}_{2}+\mathrm{C}_{3} \\ &=a b c\left|\begin{array}{ccc} a+c+a+c & c & a+c \\ a+b+b+a & b & a \\ b+b+c+c & b+c & c \end{array}\right| \\ &=a b c\left|\begin{array}{ccc} 2 a+2 c & c & a+c \\ 2 a+2 b & b & a \\ 2 b+2 c & b+c & c \end{array}\right| \\ &=a b c\left|\begin{array}{ccc} 2(a+c) & c & a+c \\ 2(a+b) & b & a \\ 2(b+c) & b+c & c \end{array}\right| \end{aligned}$
$\begin{aligned} &\text { On taking } 2 \text { common from } \mathrm{C}_{1}\\ &=2 a b c\left|\begin{array}{ccc} (a+c) & c & a+c \\ (a+b) & b & a \\ (b+c) & b+c & c \end{array}\right| \end{aligned}$
$\begin{aligned} &\text { On applying } \mathrm{C}_{1} \rightarrow C_{2}-\mathrm{C}_{1}, C_{3} \rightarrow C_{3}-C_{1} \\ &=2 a b c\left|\begin{array}{ccc} (a+c) & c-a-c & (a+c)-(a+c) \mid \\ (a+b) & b-a-b & a-a-b \\ (b+c) & (b+c)-(b+c) & c-b-c \end{array}\right| \\ &=2 a b c\left|\begin{array}{ccc} (a+c) & -a & 0 \\ (a+b) & -a & -b \\ (b+c) & 0 & -b \end{array}\right| \end{aligned}$
$\text { On taking }(-a) \text { common from } \mathrm{C}_{2} \text { and }(-b) \text { from } \mathrm{C}_{3}$
$=2 a b c(-a)(-b)\left|\begin{array}{lll} (a+c) & 1 & 0 \\ (a+b) & 1 & 1 \\ (b+c) & 0 & 1 \end{array}\right|$
$\begin{aligned} &\text { On expanding w.r.t } \mathrm{R}_{1}\\ &=2 a^{2} b^{2} c\left[(a+c)\left|\begin{array}{ll} 1 & 1 \\ 0 & 1 \end{array}\right|-1\left|\begin{array}{ll} a+b & 1 \\ b+c & 1 \end{array}\right|+0\left|\begin{array}{ll} a+b & 1 \\ b+c & 0 \end{array}\right|\right]\\ &=2 a^{2} b^{2} c[(a+c)(1-0)-1(a+b-b-c)+0]\\ &=2 a^{2} b^{2} c[a+c-1(a-c)]\\ &=2 a^{2} b^{2} c(a+c-a+c)\\ &=2 a^{2} b^{2} c \times 2 c\\ &=4 a^{2} b^{2} c^{2}\\ &=R \cdot H \cdot S \end{aligned}$
Hence it is proved that
$\left|\begin{array}{ccc} a^{2} & b c & a c+c^{2} \\ a^{2}+a b & b^{2} & a c \\ a b & b^{2}+b c & c^{2} \end{array}\right|=4 a^{2} b^{2} c^{2}$
determinants Exercise 5.2 Question 25
Answer:
Answer:$\left|\begin{array}{ccc} x+4 & x & x \\ x & x+4 & x \\ x & x & x+4 \end{array}\right|=16(3 x+4)$Hint First we will try to convert some elements into zero
Given:$\left|\begin{array}{ccc} x+4 & x & x \\ x & x+4 & x \\ x & x & x+4 \end{array}\right|=16(3 x+4)$
Solution:
$\text { L.H.S }\left|\begin{array}{ccc} x+4 & x & x \\ x & x+4 & x \\ x & x & x+4 \end{array}\right|$
$\begin{aligned} &\text { On applying } \mathrm{R}_{1} \rightarrow R_{1}+R_{2}+R_{3}\\ &=\left|\begin{array}{ccc} 3 x+4 & 3 x+4 & 3 x+4 \\ x & x+4 & x \\ x & x & x+4 \end{array}\right|\\ &\text { On taking } 3 x+4 \text { common from } \mathrm{R}_{1}\\ &=(3 x+4)\left|\begin{array}{ccc} 1 & 1 & 1 \\ x & x+4 & x \\ x & x & x+4 \end{array}\right| \end{aligned}$
$\begin{aligned} &\text { On applying } \mathrm{C}_{1} \rightarrow \mathrm{C}_{1}-\mathrm{C}_{2}, \mathrm{C}_{2} \rightarrow \mathrm{C}_{2}-C_{3} \\ &=(3 x+4)\left|\begin{array}{ccc} 0 & 0 & 1 \\ x-x-4 & x+4-x & x \\ x-x & x-x-4 & x+4 \end{array}\right| \\ &=(3 x+4)\left|\begin{array}{ccc} 0 & 0 & 1 \\ -4 & 4 & x \\ 0 & -4 & x+4 \end{array}\right| \end{aligned}$
$\begin{aligned} &\text { On taking }(-4) \text { common from } \mathrm{C}_{1} \text { and }(-4) \text { from } \mathrm{C}_{2}\\ &=(3 x+4)(-4)(-4)\left|\begin{array}{ccc} 0 & 0 & 1 \\ 1 & -1 & x \\ 0 & 1 & x+4 \end{array}\right|\\ &=16(3 x+4)\left|\begin{array}{ccc} 0 & 0 & 1 \\ 1 & -1 & x \\ 0 & 1 & x+4 \end{array}\right| \end{aligned}$
$\begin{aligned} &\text { On expanding w.r.t } \mathrm{R}_{1}\\ &=16(3 x+4)\left[0\left|\begin{array}{cc} -1 & x \\ 1 & x+4 \end{array}\right|-0\left|\begin{array}{cc} 1 & x \\ 0 & x+4 \end{array}\right|+1\left|\begin{array}{cc} 1 & -1 \\ 0 & 1 \end{array}\right|\right]\\ &=16(3 x+4)[0-0+1(1-0)]\\ &=16(3 x+4) \times 1\\ &=16(3 x+4)\\ &=R \cdot H \cdot S \end{aligned}$
Hence it is proved that
$\left|\begin{array}{ccc} x+4 & x & x \\ x & x+4 & x \\ x & x & x+4 \end{array}\right|=16(3 x+4)$
determinants Exercise 5.2 Question 26
Answer:
Answer:$\left|\begin{array}{ccc} 1 & 1+p & 1+p+q \\ 2 & 3+2 p & 4+3 p+2 q \\ 3 & 6+3 p & 10+6 p+3 q \end{array}\right|=1$Hint we will convert element of $C_{1}$ into zero
Given:$\left|\begin{array}{ccc} 1 & 1+p & 1+p+q \\ 2 & 3+2 p & 4+3 p+2 q \\ 3 & 6+3 p & 10+6 p+3 q \end{array}\right|=1$
Solution:
$\begin{aligned} &\text { L.H.S }\left|\begin{array}{ccc} 1 & 1+p & 1+p+q \\ 2 & 3+2 p & 4+3 p+2 q \\ 3 & 6+3 p & 10+6 p+3 q \end{array}\right|\\ &\text { On applying } \mathrm{R}_{2} \rightarrow R_{2}-2 R_{1} \text { and } \mathrm{R}_{3} \rightarrow R_{3}-3 R_{1}\\ &=\left|\begin{array}{ccc} 1 & 1+p & 1+p+q \\ 0 & 1 & 2+p \\ 0 & 3 & 7+3 p \end{array}\right| \end{aligned}$
$\begin{aligned} &\text { On expanding w.r.t } \mathrm{C}_{1}\\ &=1\left|\begin{array}{cc} 1 & 2+p \\ 3 & 7+3 p \end{array}\right|-0\left|\begin{array}{cc} 1+p & 1+p+q \\ 3 & 7+3 p \end{array}\right|+0\left|\begin{array}{cc} 1+p & 1+p+q \\ 1 & 2+p \end{array}\right|\\ &=1\{1(7+3 p)-3(2+p)\}-0+0\\ &=7+3 p-6-3 p\\ &=1\\ &=R \cdot H \cdot S \end{aligned}$
Hence it is proved that
$\left|\begin{array}{ccc} 1 & 1+p & 1+p+q \\ 2 & 3+2 p & 4+3 p+2 q \\ 3 & 6+3 p & 10+6 p+3 q \end{array}\right|=1$
determinants Exercise 5.2 Question 27
Answer:$\left|\begin{array}{ccc} a & b-c & c-b \\ a-c & b & c-a \\ a-b & b-a & c \end{array}\right|=(a+b-c)(b+c-a)(c+a-b)$Hint First we will make elements of $C_{1}(a+b-c) \text { or } 0$
Given:$\left|\begin{array}{ccc} a & b-c & c-b \\ a-c & b & c-a \\ a-b & b-a & c \end{array}\right|=(a+b-c)(b+c-a)(c+a-b)$
Solution:
$\begin{aligned} &\text { L.H.S }\left|\begin{array}{ccc} a & b-c & c-b \\ a-c & b & c-a \\ a-b & b-a & c \end{array}\right|\\ &\text { On applying } \mathrm{C}_{1} \rightarrow C_{1}+C_{2} \text { and } \mathrm{C}_{2} \rightarrow C_{2}+C_{3}\\ &=\left|\begin{array}{ccc} a+b-c & 0 & c-b \\ a+b-c & b+c-a & c-a \\ 0 & b+c-a & c \end{array}\right| \end{aligned}$
$\text { On taking }(a+b-c) \text { common from } \mathrm{C}_{1} \text { and }(b+c-a) \text { common from } \mathrm{C}_{2}$
$=(a+b-c)(b+c-a)\left|\begin{array}{ccc} 1 & 0 & c-b \\ 1 & 1 & c-a \\ 0 & 1 & c \end{array}\right|$
$\begin{aligned} &\text { On expanding w.r.t } \mathrm{R}_{1}\\ &=(a+b-c)(b+c-a)\left[1\left|\begin{array}{cc} 1 & c-a \\ 1 & c \end{array}\right|-0\left|\begin{array}{cc} 1 & c-a \\ 0 & c \end{array}\right|+(c-b)\left|\begin{array}{ll} 1 & 1 \\ 0 & 1 \end{array}\right|\right]\\ &=(a+b-c)(b+c-a)[1\{c-(c-a)\}-0+(c-b)(1-0)]\\ &=(a+b-c)(b+c-a)(c-c+a+c-b)\\ &=(a+b-c)(b+c-a)(c+a-b)\\ &=R \cdot H \cdot S \end{aligned}$
Hence it is proved that
$\left|\begin{array}{ccc} a & b-c & c-b \\ a-c & b & c-a \\ a-b & b-a & c \end{array}\right|=(a+b-c)(b+c-a)(c+a-b)$
deteminants Exercise 5.2 Question 28
Answer:
Answer:$\left|\begin{array}{ccc} a^{2} & 2 a b & b^{2} \\ b^{2} & a^{2} & 2 a b \\ 2 a b & b^{2} & a^{2} \end{array}\right|=\left(a^{3}+b^{3}\right)^{2}$Hint First we will try to make some elements of $R_{1} \text { or } C_{1} \quad(a+b)^{2}=a^{2}+2 a b+b^{2}$
Because from R.H.S $\begin{aligned} &\left(a^{3}+b^{3}\right)^{2}=\left[(a+b)\left(a^{2}-a b+b^{2}\right)\right]^{2} \\ &=(a+b)^{2}\left(a^{2}-a b+b^{2}\right)^{2} \end{aligned}$
Given: $\left|\begin{array}{ccc} a^{2} & 2 a b & b^{2} \\ b^{2} & a^{2} & 2 a b \\ 2 a b & b^{2} & a^{2} \end{array}\right|=\left(a^{3}+b^{3}\right)^{2}$
Solution:
$\text { L.H.S }\left|\begin{array}{ccc} a^{2} & 2 a b & b^{2} \\ b^{2} & a^{2} & 2 a b \\ 2 a b & b^{2} & a^{2} \end{array}\right|$
$\begin{aligned} &\text { On applying } \mathrm{R}_{1} \rightarrow R_{1}+R_{2}+R_{3} \\ &=\left|\begin{array}{ccc} a^{2}+b^{2}+2 a b & 2 a b+a^{2}+b^{2} & b^{2}+2 a b+a^{2} \\ b^{2} & a^{2} & 2 a b \\ 2 a b & b^{2} & a^{2} \end{array}\right| \\ &\because(a+b)^{2}=a^{2}+2 a b+b^{2} \\ &=\left|\begin{array}{ccc} (a+b)^{2} & (a+b)^{2} & (a+b)^{2} \\ b^{2} & a^{2} & 2 a b \\ 2 a b & b^{2} & a^{2} \end{array}\right| \end{aligned}$
$\begin{aligned} &\text { On taking }(a+b)^{2} \text { common from } \mathrm{R}_{1}\\ &=(a+b)^{2}\left|\begin{array}{ccc} 1 & 1 & 1 \\ b^{2} & a^{2} & 2 a b \\ 2 a b & b^{2} & a^{2} \end{array}\right| \end{aligned}$
$\begin{aligned} &\text { On expanding w.r.t } \mathrm{R}_{1}\\ &=(a+b)^{2}\left[0\left|\begin{array}{cc} a^{2}-2 a b & 2 a b \\ b^{2}-a^{2} & a^{2} \end{array}\right|-0\left|\begin{array}{cc} b^{2}-a^{2} & 2 a b \\ 2 a b-b^{2} & a^{2} \end{array}\right|+1\left|\begin{array}{cc} b^{2}-a^{2} & a^{2}-2 a b \\ 2 a b-b^{2} & b^{2}-a^{2} \end{array}\right|\right]\\ &=(a+b)^{2}\left[0-0+1\left\{\left(b^{2}-a^{2}\right)\left(b^{2}-a^{2}\right)\right\}-\left(a^{2}-2 a b\right)\left(2 a b-b^{2}\right)\right]\\ &=(a+b)^{2}\left(b^{4}+a^{4}-2 a^{2} b^{2}-2 a^{3} b+a^{2} b^{2}+4 a^{2} b^{2}-2 a b^{3}\right) \end{aligned}$
$\begin{aligned} &=(a+b)^{2}\left[\left(a^{2}\right)^{2}+\left(b^{2}\right)^{2}+2 a^{2} b^{2}-2 a^{3} b-2 a b^{3}+a^{2} b^{2}\right] \\ &=(a+b)^{2}\left[\left(a^{2}\right)^{2}+(a b)^{2}+\left(b^{2}\right)^{2}-2 a^{3} b-2 a b^{3}+2 a^{2} b^{2}\right] \\ &=(a+b)^{2}\left(a^{2}-a b+b^{2}\right)^{2} \\ &=\left(a^{3}+b^{3}\right)^{2} \\ &=R \cdot H . S \end{aligned}$
Hence it is proved that
$\left|\begin{array}{ccc} a^{2} & 2 a b & b^{2} \\ b^{2} & a^{2} & 2 a b \\ 2 a b & b^{2} & a^{2} \end{array}\right|=\left(a^{3}+b^{3}\right)^{2}$
determinants Exercise 5.2 Question 29
Answer:
Answer:$\left|\begin{array}{ccc} a^{2}+1 & a b & a c \\ a b & b^{2}+1 & b c \\ c a & c b & c^{2}+1 \end{array}\right|=1+a^{2}+b^{2}+c^{2}$Hint We will try to convert some elements of determinant into zero
Given:$\left|\begin{array}{ccc} a^{2}+1 & a b & a c \\ a b & b^{2}+1 & b c \\ c a & c b & c^{2}+1 \end{array}\right|=1+a^{2}+b^{2}+c^{2}$
Solution:
$\text { L.H.S }\left|\begin{array}{ccc} a^{2}+1 & a b & a c \\ a b & b^{2}+1 & b c \\ c a & c b & c^{2}+1 \end{array}\right|$
$\begin{aligned} &=\left|\begin{array}{ccc} 1+a^{2} & a b & a c \\ a b & 1+b^{2} & b c \\ c a & c b & 1+c^{2} \end{array}\right|\\ &\text { Multiply } C_{1} \text { by } a, C_{2} \text { by } b \text { and } C_{3} \text { by } c\\ &=\frac{1}{a b c}\left|\begin{array}{ccc} a\left(1+a^{2}\right) & a b^{2} & a c^{2} \\ a^{2} b & b\left(1+b^{2}\right) & b c^{2} \\ c a^{2} & c b^{2} & c\left(1+c^{2}\right) \end{array}\right| \end{aligned}$
$\begin{aligned} &\text { On taking } a \text { common from } \mathrm{R}_{1}, \mathrm{~b} \text { from } \mathrm{R}_{2} \text { and } \mathrm{c} \text { from } \mathrm{R}_{3}\\ &=\frac{a b c}{a b c}\left|\begin{array}{ccc} 1+a^{2} & b^{2} & c^{2} \\ a^{2} & 1+b^{2} & c^{2} \\ a^{2} & b^{2} & 1+c^{2} \end{array}\right|\\ &=\left|\begin{array}{ccc} 1+a^{2} & b^{2} & c^{2} \\ a^{2} & 1+b^{2} & c^{2} \\ a^{2} & b^{2} & 1+c^{2} \end{array}\right|\\ &\text { On applying } \mathrm{C}_{1} \rightarrow C_{1}+C_{2}+C_{3}\\ &=\left|\begin{array}{ccc} 1+a^{2}+b^{2}+c^{2} & b^{2} & c^{2} \\ 1+a^{2}+b^{2}+c^{2} & 1+b^{2} & c^{2} \\ 1+a^{2}+b^{2}+c^{2} & b^{2} & 1+c^{2} \end{array}\right| \end{aligned}$
$\begin{aligned} &\text { On taking }\left(1+a^{2}+b^{2}+c^{2}\right) \text { common from } \mathrm{C}_{1}\\ &=\left(1+a^{2}+b^{2}+c^{2}\right)\left|\begin{array}{ccc} 1 & b^{2} & c^{2} \\ 1 & 1+b^{2} & c^{2} \\ 1 & b^{2} & 1+c^{2} \end{array}\right|\\ &\text { On applying } \mathrm{R}_{1} \rightarrow R_{1}-R_{2}, R_{2} \rightarrow R_{2}-R_{3}\\ &=\left(1+a^{2}+b^{2}+c^{2}\right)\left|\begin{array}{ccc} 0 & b^{2}-1-b^{2} & c^{2} \\ 0 & 1+b^{2}-b^{2} & c^{2}-1-c^{2} \\ 1 & b^{2} & 1+c^{2} \end{array}\right|\\ &=\left(1+a^{2}+b^{2}+c^{2}\right)\left|\begin{array}{ccc} 0 & -1 & 0 \\ 0 & 1 & -1 \\ 1 & b^{2} & 1+c^{2} \end{array}\right| \end{aligned}$
$\begin{aligned} &\text { On expanding w.r.t } \mathrm{R}_{1}\\ &=\left(1+a^{2}+b^{2}+c^{2}\right)\left[0\left|\begin{array}{cc} 1 & -1 \\ b^{2} & 1+c^{2} \end{array}\right|-(-1)\left|\begin{array}{cc} 0 & -1 \\ 1 & 1+c^{2} \end{array}\right|+0\left|\begin{array}{cc} 0 & 1 \\ 1 & b^{2} \end{array}\right|\right]\\ &=\left(1+a^{2}+b^{2}+c^{2}\right)[0+1\{0-(-1) 1\}+0]\\ &=\left(1+a^{2}+b^{2}+c^{2}\right)(1 \times 1)\\ &=\left(1+a^{2}+b^{2}+c^{2}\right)\\ &=R \cdot H \cdot S \end{aligned}$
Hence it is proved that
$\left|\begin{array}{ccc} a^{2}+1 & a b & a c \\ a b & b^{2}+1 & b c \\ c a & c b & c^{2}+1 \end{array}\right|=1+a^{2}+b^{2}+c^{2}$
determinants Exercise 5.2 Question 30
Answer: $\left|\begin{array}{ccc} 1 & a & a^{2} \\ a^{2} & 1 & a \\ a & a^{2} & 1 \end{array}\right|=\left(a^{3}-1\right)^{2}$Hint We will try to convert some elements of determinant into zero
Given:$\left|\begin{array}{ccc} 1 & a & a^{2} \\ a^{2} & 1 & a \\ a & a^{2} & 1 \end{array}\right|=\left(a^{3}-1\right)^{2}$
Solution:
$\text { L.H.S }\left|\begin{array}{ccc} 1 & a & a^{2} \\ a^{2} & 1 & a \\ a & a^{2} & 1 \end{array}\right|$
$\begin{aligned} &\text { On applying } \mathrm{R}_{1} \rightarrow R_{1}+R_{2}+R_{3} \\ &=\left|\begin{array}{ccc} 1+a^{2}+a & a+1+a^{2} & a^{2}+a+1 \\ a^{2} & 1 & a \\ a & a^{2} & 1 \end{array}\right| \end{aligned}$
$\begin{aligned} &\text { On taking }\left(a^{2}+a+1\right) \text { common from } \mathrm{R}_{1}\\ &=\left(a^{2}+a+1\right)\left|\begin{array}{ccc} 1 & 1 & 1 \\ a^{2} & 1 & a \\ a & a^{2} & 1 \end{array}\right|\\ &\text { On applying } \mathrm{C}_{1} \rightarrow C_{1}-C_{2}, C_{2} \rightarrow C_{2}-C_{3}\\ &=\left(a^{2}+a+1\right)\left|\begin{array}{ccc} 0 & 0 & 1 \\ a^{2}-1 & 1-a & a \\ a-a^{2} & a^{2}-1 & 1 \end{array}\right|\\ &=\left(a^{2}+a+1\right)\left|\begin{array}{ccc} 0 & 0 & 1 \\ (a+1)(a-1) & -(a-1) & a \\ -a(a-1) & (a+1)(a-1) & 1 \end{array}\right| \quad \because a^{2}-b^{2}=(a+b)(a-b) \end{aligned}$
$\begin{aligned} &\text { On taking common }(a-1) \text { from } \mathrm{C}_{1} \text { and }(a-1) \text { from } \mathrm{C}_{2}\\ &=\left(a^{2}+a+1\right)(a-1)(a-1)\left|\begin{array}{ccc} 0 & 0 & 1 \\ (a+1) & -1 & a \\ -a & (a+1) & 1 \end{array}\right| \end{aligned}$
$\begin{aligned} &\text { On expanding w.r.t } \mathrm{R}_{1}\\ &=\left(a^{2}+a+1\right)(a-1)^{2}\left[0\left|\begin{array}{cc} -1 & a \\ a+1 & 1 \end{array}\right|-0\left|\begin{array}{cc} a+1 & a \\ -a & 1 \end{array}\right|+1\left|\begin{array}{cc} a+1 & -1 \\ -a & a+1 \end{array}\right|\right]\\ &=(a-1)^{2}\left(a^{2}+a+1\right)\left[0-0+1\left\{(a+1)^{2}-(-a)(-1)\right\}\right]\\ &=(a-1)^{2}\left(a^{2}+a+1\right)\left(a^{2}+2 a+1-a\right)\\ &=(a-1)^{2}\left(a^{2}+a+1\right)\left(a^{2}+a+1\right)\\ &=\left\{(a-1)\left(a^{2}+a+1\right)\right\}^{2}\\ &=\left(a^{3}-1\right)^{2}\\ &=R \cdot H \cdot S \end{aligned}$
Hence it is proved that
$\left|\begin{array}{ccc} 1 & a & a^{2} \\ a^{2} & 1 & a \\ a & a^{2} & 1 \end{array}\right|=\left(a^{3}-1\right)^{2}$
determinants Exercise 5.2 Question 31
Answer: $2(a+b)(b+c)(c+a)$Hint Use determinant formula
Given:$\left|\begin{array}{ccc} a+b+c & -c & -b \\ -c & a+b+c & -a \\ -b & -a & a+b+c \end{array}\right|=2(a+b)(b+c)(c+a)$
Solution:
$\begin{aligned} &\text { L.H.S }\left|\begin{array}{ccc} a+b+c & -c & -b \\ -c & a+b+c & -a \\ -b & -a & a+b+c \end{array}\right| \\ &\text { Use } \mathrm{C}_{1} \rightarrow C_{1}+\mathrm{C}_{2} \text { and } \mathrm{C}_{2} \rightarrow C_{2}+\mathrm{C}_{3} \\ &|\Delta|=\left|\begin{array}{ccc} a+b & -b-c & -b \\ a+b & b+c & -a \\ -a-b & b+c & a+b+c \end{array}\right| \end{aligned}$
$\begin{aligned} &\text { Common }(a+b)(b+c) \text { from } \mathrm{C}_{1} \& C_{2} \\ &|\Delta|=(a+b)(b+c)\left|\begin{array}{ccc} 1 & -1 & -b \\ 1 & 1 & -a \\ -1 & 1 & a+b+c \end{array}\right| \\ &\text { Use } \mathrm{R}_{1} \rightarrow R_{1}+\mathrm{R}_{2} \text { and } \mathrm{R}_{2} \rightarrow R_{2}+\mathrm{R}_{3} \\ &|\Delta|=(a+b)(b+c)\left|\begin{array}{ccc} 2 & 0 & -b+a \\ 0 & 2 & b+c \\ -1 & 1 & a+b+c \end{array}\right| \end{aligned}$
$\begin{aligned} &\text { Expanding w.r.t } \mathrm{C}_{1}\\ &|\Delta|=(a+b)(b+c)[2(2 a+2 b+2 c-b-c)+(-1)(2 b+2 a)]\\ &|\Delta|=(a+b)(b+c)[4 a+4 b+4 c-2 b-2 c-2 b-2 a]\\ &|\Delta|=2(a+b)(b+c)(c+a) \end{aligned}$
determinants Exercise 5.2 Question 32
Answer:$4\: abc$Hint: Use determinant formula
Given: $\left|\begin{array}{ccc} b+c & a & a \\ b & a+c & b \\ c & c & a+b \end{array}\right|=4 a b c$
Solution:
$\text { L.H.S }\left|\begin{array}{ccc} b+c & a & a \\ b & a+c & b \\ c & c & a+b \end{array}\right|$
$\begin{aligned} &\text { Use } R_{1} \rightarrow R_{1}+R_{2}+R_{3}\\ &|\Delta|=\left|\begin{array}{ccc} 2(b+c) & 2(c+a) & 2(a+b) \\ b & a+c & b \\ c & c & a+b \end{array}\right|\\ &\text { Now } R_{2} \rightarrow R_{2}-R_{1} \text { and } \mathrm{R}_{3} \rightarrow R_{3}-R_{1} \text { and } 2 \text { common }\\ &|\Delta|=2\left|\begin{array}{ccc} b+c & c+a & a+b \\ -c & 0 & -a \\ -b & -a & 0 \end{array}\right| \end{aligned}$
$\begin{aligned} &\text { Use } R_{1} \rightarrow R_{1}+R_{2}+R_{3}\\ &|\Delta|=2\left|\begin{array}{ccc} 0 & c & b \\ -c & 0 & -a \\ -b & -a & 0 \end{array}\right|\\ &\text { Expanding w.r.t } \mathrm{C}_{1}\\ &|\Delta|=2[-c(0-a b)+b(a c-0)]\\ &|\Delta|=2(a b c+a b c)\\ &|\Delta|=2(2 a b c)=4 a b c\\ &=R \cdot H . S \end{aligned}$
determinants Exercise 5.2 Question 33
Answer: $4 a^{2} b^{2} c^{2}$Hint: Use determinant formula
Given: $\left|\begin{array}{ccc} b^{2}+c^{2} & a b & a c \\ b a & c^{2}+a^{2} & b c \\ c a & c b & a^{2}+b^{2} \end{array}\right|=4 a^{2} b^{2} c^{2}$
Solution:
$\begin{aligned} &\text { L.H.S }\left|\begin{array}{ccc} b^{2}+c^{2} & a b & a c \\ b a & c^{2}+a^{2} & b c \\ c a & c b & a^{2}+b^{2} \end{array}\right| \\ &\text { Use } R_{1} \rightarrow a R_{1}, R_{2} \rightarrow b R_{2}, R_{3} \rightarrow c R_{3} \\ &=\frac{1}{a b c}\left|\begin{array}{ccc} a\left(b^{2}+c^{2}\right) & a^{2} b & a^{2} c \\ b^{2} a & b\left(c^{2}+a^{2}\right) & b^{2} c \\ c^{2} a & c^{2} b & c\left(a^{2}+b^{2}\right) \end{array}\right| \end{aligned}$
$\begin{aligned} &\text { Take common a from } \mathrm{R}_{1}, b \text { from } \mathrm{R}_{2} \text { and } \mathrm{c} \text { from } \mathrm{R}_{3}\\ &=\frac{a b c}{a b c}\left|\begin{array}{ccc} \left(b^{2}+c^{2}\right) & a^{2} & a^{2} \\ b^{2} & \left(c^{2}+a^{2}\right) & b^{2} \\ c^{2} & c^{2} & \left(a^{2}+b^{2}\right) \end{array}\right| \end{aligned}$
$\begin{aligned} &\text { Use } R_{1} \rightarrow R_{1}-R_{2}-R_{3} \\ &=\left|\begin{array}{ccc} 0 & -2 c^{2} & -2 b^{2} \\ b^{2} & \left(c^{2}+a^{2}\right) & b^{2} \\ c^{2} & c^{2} & \left(a^{2}+b^{2}\right) \end{array}\right| \end{aligned}$
$\begin{aligned} &\text { Expanding w.r.t } C_{1}\\ &=-b^{2}\left|\begin{array}{cc} -2 c^{2} & -2 b^{2} \\ c^{2} & a^{2}+b^{2} \end{array}\right|+c^{2}\left|\begin{array}{cc} -2 c^{2} & -2 b^{2} \\ c^{2}+a^{2} & b^{2} \end{array}\right|\\ &=-b^{2}\left(-2 a^{2} c^{2}-2 b^{2} c^{2}+2 b^{2} c^{2}\right)+c^{2}\left(-2 c^{2} b^{2}+2 b^{2} c^{2}+2 a^{2} b^{2}\right)\\ &=2 a^{2} b^{2} c^{2}+2 a^{2} b^{2} c^{2}=4 a^{2} b^{2} c^{2}\\ &=R . H . S \end{aligned}$
deteminants Exercise 5.2 Question 34
Answer:$2 a^{3} b^{3} c^{3}$Hint Use determinant formula
Given: $\left|\begin{array}{ccc} 0 & a b^{2} & a c^{2} \\ a^{2} b & 0 & b c^{2} \\ a^{2} c & b^{2} c & 0 \end{array}\right|=2 a^{3} b^{3} c^{3}$
Solution:
$\text { L.H.S }\left|\begin{array}{ccc} 0 & a b^{2} & a c^{2} \\ a^{2} b & 0 & b c^{2} \\ a^{2} c & b^{2} c & 0 \end{array}\right|$
$\begin{aligned} &\text { Use } R_{1} \rightarrow R_{1}+R_{2}+R_{3}\\ &|\Delta|=\left|\begin{array}{ccc} a^{2}(b+c) & b^{2}(a+c) & c^{2}(a+b) \\ a^{2} b & 0 & b c^{2} \\ a^{2} c & b^{2} c & 0 \end{array}\right|\\ &\text { Taking out } \mathrm{a}^{2} \text { from } \mathrm{C}_{1}, b^{2} \text { from } \mathrm{C}_{2} \text { and } \mathrm{c}^{2} \text { from } \mathrm{C}_{3}\\ &=a^{2} b^{2} c^{2}\left|\begin{array}{ccc} b+c & a+c & a+b \\ b & 0 & b \\ c & c & 0 \end{array}\right| \end{aligned}$
$\begin{aligned} &\text { Use } R_{1} \rightarrow R_{1}-\left(R_{2}+R_{3}\right)\\ &=a^{2} b^{2} c^{2}\left|\begin{array}{ccc} b+c-(b+c) & a+c-c & a+b-b \\ b & 0 & b \\ c & c & 0 \end{array}\right|\\ &=a^{2} b^{2} c^{2}\left|\begin{array}{lll} 0 & a & a \\ b & 0 & b \\ c & c & 0 \end{array}\right|\\ &\text { Expanding w.r.t } \mathrm{R}_{1}\\ &=a^{2} b^{2} c^{2}(-a(0-b c)+a(b c-0))\\ &=a^{2} b^{2} c^{2} \cdot 2 a b c\\ &=2 a^{3} b^{3} c^{3}\\ &=R \cdot H . S \end{aligned}$
determinants Exercise 5.2 Question 35
Answer:$4 a b c$Hint Use determinant formula
Given :$\left|\begin{array}{ccc} \frac{a^{2}+b^{2}}{c} & c & c \\ a & \frac{b^{2}+c^{2}}{a} & a \\ b & b & \frac{c^{2}+a^{2}}{b} \end{array}\right|=4 a b c$
Solution:
$\text { L.H.S }\left|\begin{array}{ccc} \frac{a^{2}+b^{2}}{c} & c & c \\ a & \frac{b^{2}+c^{2}}{a} & a \\ b & b & \frac{c^{2}+a^{2}}{b} \end{array}\right|$
$\begin{aligned} &R_{1} \rightarrow c R_{1}, R_{2} \rightarrow c R_{2}, R_{3} \rightarrow b R_{3} \\ &=\frac{1}{a b c}\left|\begin{array}{ccc} a^{2}+b^{2} & c^{2} & c^{2} \\ a^{2} & b^{2}+c^{2} & a^{2} \\ b^{2} & b^{2} & c^{2}+a^{2} \end{array}\right| \end{aligned}$
$\begin{aligned} &\text { Use } R_{1} \rightarrow R_{1}-R_{2}-R_{3} \\ &=\frac{1}{a b c}\left|\begin{array}{ccc} 0 & -2 b^{2} & -2 a^{2} \\ a^{2} & b^{2}+c^{2} & a^{2} \\ b^{2} & b^{2} & c^{2}+a^{2} \end{array}\right| \\ &\text { Now } \mathrm{C}_{2} \rightarrow C_{2}-C_{1} \text { and } \mathrm{C}_{3} \rightarrow C_{3}-C_{1} \\ &=\frac{1}{a b c} \mid \begin{array}{ccc} 0 & -2 b^{2} & -2 a^{2} \\ a^{2} & b^{2}+c^{2}-a^{2} & a^{2} \\ b^{2} & b^{2} & c^{2}+a^{2}-b^{2} \end{array} \end{aligned}$
$\begin{aligned} &\text { Expanding w.r.t } \mathrm{C}_{1}\\ &=\frac{1}{a b c}\left[c-a^{2}\left(-2 b^{2}\left(c^{2}+a^{2}-b^{2}\right)-a\right)+b^{2}\left(0-\left(2 a^{2}\right)\left(b^{2}+c^{2}-a^{2}\right)\right)\right]\\ &=\frac{1}{a b c}\left[2 a^{2} b^{2}\left(c^{2}+a^{2}-b^{2}\right)+b^{2}\left(\left(2 a^{2}\right)\left(b^{2}+c^{2}-a^{2}\right)\right)\right]\\ &=\frac{2 a^{2} b^{2}}{a b c}\left(c^{2}+a^{2}-b^{2}+b^{2}+c^{2}-a^{2}\right)\\ &=\frac{2 a b}{c}\left(2 c^{2}\right)=4 a b c\\ &=R \cdot H \cdot S \end{aligned}$
determinants Exercise 5.2 Question 36
Answer:$(a b+b c+c a)^{3}$Hint Use determinant formula
Given:$\left|\begin{array}{ccc} -b c & b^{2}+b c & c^{2}+a b \\ a^{2}+a c & -a c & c^{2}+a c \\ a^{2}+a b & b^{2}+a b & -a b \end{array}\right|=(a b+b c+c a)^{3}$
Solution:
$\begin{aligned} &\text { L.H.S }\left|\begin{array}{ccc} -b c & b^{2}+b c & c^{2}+a b \\ a^{2}+a c & -a c & c^{2}+a c \\ a^{2}+a b & b^{2}+a b & -a b \end{array}\right| \\ &R_{1} \rightarrow a R_{1}, R_{2} \rightarrow b R_{2}, R_{3} \rightarrow c R_{3} \\ &=\frac{1}{a b c}\left|\begin{array}{ccc} -a b c & a b^{2}+a b c & a c^{2}+a b c \\ a^{2}+a b c & -a b c & c^{2}+a b c \\ a^{2}+a b c & b^{2} c+a b c & -a b c \end{array}\right| \end{aligned}$
$\begin{aligned} &\text { Common a from } \mathrm{C}_{1}, b \text { from } \mathrm{C}_{2} \text { and } \mathrm{c} \text { from } \mathrm{C}_{3}\\ &=\frac{a b c}{a b c}\left|\begin{array}{ccc} b c & a b+a c & a c+a b \\ a b+b c & -a c & b c+a b \\ a c+b c & b c+a c & -a b \end{array}\right|\\ &\text { Use } R_{1} \rightarrow R_{1}+R_{2}+R_{3}\\ &=\left|\begin{array}{ccc} a b+b c+a c & a b+b c+a c & a b+b c+a c \\ a b+b c & -a c & b c+a b \\ a c+b c & b c+a c & -a b \end{array}\right| \end{aligned}$
$\begin{aligned} &(a b+b c+a c) \text { common from } \mathrm{R}_{1} \\ &=(a b+b c+a c)\left|\begin{array}{ccc} 1 & 1 & 1 \\ a b+b c & -a c & b c+a b \\ a c+b c & b c+a c & -a b \end{array}\right| \\ &\text { Now } \mathrm{C}_{2} \rightarrow C_{2}-C_{1} \text { and } \mathrm{C}_{3} \rightarrow C_{3}-C_{1} \\ &=(a b+b c+a c) \mid \begin{array}{ccc} 1 & 0 & 0 \\ a b+b c & -(a b+b c+a c) & a b+b c+a c \\ a c+b c & 0 & -(a b+b c+a c) \end{array} \end{aligned}$
$\begin{aligned} &(a b+b c+a c) \text { common from } \mathrm{C}_{2} \& C_{3}\\ &=(a b+b c+a c)^{3}\left|\begin{array}{ccc} 1 & 0 & 0 \\ a b+b c & -1 & 1 \\ a c+b c & 0 & -1 \end{array}\right|\\ &\text { Expanding w.r.t } \mathrm{R}_{1}\\ &=(a b+b c+a c)^{3}\{1(1-0)+0\}\\ &=(a b+b c+a c)^{3}\\ &=R \cdot H \cdot S \end{aligned}$
determinants Exercise 5.2 Question 37
Answer:$(5 x+\lambda)(\lambda-x)^{2}$Hint: Use determinant formula
Given: $\left|\begin{array}{ccc} x+\lambda & 2 x & 2 x \\ 2 x & x+\lambda & 2 x \\ 2 x & 2 x & x+\lambda \end{array}\right|=(5 x+\lambda)(\lambda-x)^{2}$
Solution:
$\begin{aligned} &\text { L.H.S }\left|\begin{array}{ccc} x+\lambda & 2 x & 2 x \\ 2 x & x+\lambda & 2 x \\ 2 x & 2 x & x+\lambda \end{array}\right| \\ &\text { Use } \mathrm{C}_{1} \rightarrow C_{1}+C_{2}+C_{3} \\ &=\left|\begin{array}{ccc} 5 x+\lambda & 2 x & 2 x \\ 5 x+\lambda & x+\lambda & 2 x \\ 5 x+\lambda & 2 x & x+\lambda \end{array}\right| \end{aligned}$
$\begin{aligned} &(5 x+\lambda) \text { common from } \mathrm{C}_{1}\\ &=(5 x+\lambda)\left|\begin{array}{ccc} 1 & 2 x & 2 x \\ 1 & x+\lambda & 2 x \\ 1 & 2 x & x+\lambda \end{array}\right|\\ &\text { Now } \mathrm{R}_{2} \rightarrow R_{2}-R_{1} \text { and } \mathrm{R}_{3} \rightarrow R_{3}-R_{1}\\ &=(5 x+\lambda)\left|\begin{array}{ccc} 1 & 2 x & 2 x \\ 0 & x-\lambda & 0 \\ 0 & 0 & x-\lambda \end{array}\right|\\ &=(5 x+\lambda)(-1)(-1)\left|\begin{array}{ccc} 1 & 2 x & 2 x \\ 0 & -x+\lambda & 0 \\ 0 & 0 & -x+\lambda \end{array}\right| \end{aligned}$
$\text { Expanding w.r.t } \mathrm{R}_{1}$
$\begin{aligned} &=(5 x+\lambda)\left[1(\lambda-x)^{2}-0+0\right] \\ &=(5 x+\lambda)(\lambda-x)^{2} \\ &=R \cdot H . S \end{aligned}$
deteminants Exercise 5.2 Question 38
Answer:$(5 x+4)(4-x)^{2}$Hint Use determinant formula
Given:$\left|\begin{array}{ccc} x+4 & 2 x & 2 x \\ 2 x & x+4 & 2 x \\ 2 x & 2 x & x+4 \end{array}\right|=(5 x+4)(4-x)^{2}$
Solution:
$\begin{aligned} &\text { L.H.S }\left|\begin{array}{ccc} x+4 & 2 x & 2 x \\ 2 x & x+4 & 2 x \\ 2 x & 2 x & x+4 \end{array}\right| \\ &\text { Use C }_{1} \rightarrow C_{1}+C_{2}+C_{3} \\ &=\left|\begin{array}{ccc} 5 x+4 & 2 x & 2 x \\ 5 x+4 & x+4 & 2 x \\ 5 x+4 & 2 x & x+4 \end{array}\right| \end{aligned}$
$\begin{aligned} &(5 x+4) \text { common from } \mathrm{C}_{1} \\ &=(5 x+4)\left|\begin{array}{ccc} 1 & 2 x & 2 x \\ 1 & x+4 & 2 x \\ 1 & 2 x & x+4 \end{array}\right| \end{aligned}$
$\begin{aligned} &\text { Now } \mathrm{R}_{2} \rightarrow R_{2}-R_{1} \text { and } \mathrm{R}_{3} \rightarrow R_{3}-R_{1} \\ &=(5 x+4)\left|\begin{array}{ccc} 0 & x-4 & 0 \\ 0 & 4+x-2 x & 2 x-x-4 \\ 1 & 2 x & x+4 \end{array}\right| \end{aligned}$
$\begin{aligned} &(x-4) \text { from } \mathrm{R}_{1} \text { common }\\ &(x-4)(5 x+4)\left|\begin{array}{ccc} 0 & x-4 & 0 \\ 0 & 4+x-2 x & 2 x-x-4 \\ 1 & 2 x & x+4 \end{array}\right| \end{aligned}$
Expanding w.r.t $C_{1}$
$\begin{aligned} &=(x-4)(5 x+4)[(x-4)(x-4)-0] \\ &=(5 x+4)(x-4)^{2} \\ &=R . H . S \end{aligned}$
deteminants Exercise 5.2 Question 39
Answer:$4 x y z$Hint Use determinant formula
Given: $\left|\begin{array}{ccc} y+z & z & y \\ z & z+x & x \\ y & x & x+y \end{array}\right|=4 x y z$
Solution:
$\begin{aligned} &\text { L.H.S }\left|\begin{array}{ccc} y+z & z & y \\ z & z+x & x \\ y & x & x+y \end{array}\right| \\ &\text { Apply C }_{1} \rightarrow C_{1}+C_{2}+C_{3} \\ &=\left|\begin{array}{ccc} 2(y+z) & z & y \\ 2(x+z) & z+x & x \\ 2(x+y) & x & x+y \end{array}\right| \end{aligned}$
$\begin{aligned} &2 \text { common from } \mathrm{C}_{1}\\ &=2\left|\begin{array}{ccc} (y+z) & z & y \\ (x+z) & z+x & x \\ (x+y) & x & x+y \end{array}\right|\\ &\begin{aligned} &\text { Now } \mathrm{C}_{1} \rightarrow C_{1}-C_{2} \\ &=2\left|\begin{array}{ccc} y & z & y \\ 0 & z+x & x \\ y & x & x+y \end{array}\right| \end{aligned} \end{aligned}$
$\begin{aligned} &\text { Now } \mathrm{C}_{3} \rightarrow C_{3}-C_{1} \\ &=2\left|\begin{array}{ccc} y & z & 0 \\ 0 & z+x & x \\ y & x & x \end{array}\right| \\ &{ }^{\text {Now }} \mathrm{R}_{3} \rightarrow R_{3}-R_{1} \\ &=2\left|\begin{array}{ccc} y & z & 0 \\ 0 & z+x & x \\ 0 & x-z & x \end{array}\right| \end{aligned}$
Expanding w.r.t $C_{1}$
$\begin{aligned} &=2 y(z+x) x-x(x-z)-0+0 \\ &=2 y\left(z x+x^{2}-x^{2}+x z\right) \\ &=4 x y z \\ &=R \cdot H . S \end{aligned}$
determinants Exercise 5.2 Question 40
Answer: $a b c\left(a^{2}+b^{2}+c^{2}\right)^{3}$Hint Use determinant formula
Given: $\left|\begin{array}{ccc} -a\left(b^{2}+c^{2}-a^{2}\right) & 2 b^{3} & 2 c^{3} \\ 2 a^{3} & -b\left(c^{2}+a^{2}-b^{2}\right) & 2 c^{3} \\ 2 a^{3} & 2 b^{3} & -c\left(a^{2}+b^{2}-c^{2}\right) \end{array}\right|=a b c\left(a^{2}+b^{2}+c^{2}\right)^{3}$
Solution:
$\begin{aligned} &\text { L.H.S }\left|\begin{array}{ccc} -a\left(b^{2}+c^{2}-a^{2}\right) & 2 b^{3} & 2 c^{3} \\ 2 a^{3} & -b\left(c^{2}+a^{2}-b^{2}\right) & 2 c^{3} \\ 2 a^{3} & 2 b^{3} & -c\left(a^{2}+b^{2}-c^{2}\right) \end{array}\right|\\ &\text { Common a from } \mathrm{C}_{1}, b \text { from } \mathrm{C}_{2} \text { and c from } \mathrm{C}_{3}\\ &=a b c\left|\begin{array}{ccc} -\left(b^{2}+c^{2}-a^{2}\right) & 2 b^{2} & 2 c^{2} \\ 2 a^{2} & -\left(c^{2}+a^{2}-b^{2}\right) & 2 c^{2} \\ 2 a^{2} & 2 b^{2} & -\left(a^{2}+b^{2}-c^{2}\right) \end{array}\right| \end{aligned}$
$\begin{aligned} &\text { Apply } \mathrm{C}_{1} \rightarrow C_{1}+C_{2}+C_{3} \\ &=a b c\left|\begin{array}{ccc} -b^{2}-c^{2}+a^{2}+2 b^{2}+2 c^{2} & 2 b^{2} & 2 c^{2} \\ 2 a^{2}+2 c^{2}-c^{2}-a^{2}+b^{2} & -c^{2}+a^{2}-b^{2} & 2 c^{2} \\ 2 a^{2}+2 b^{2}-a^{2}-b^{2}+c^{2} & 2 b^{2} & -a^{2}+b^{2}-c^{2} \end{array}\right| \\ &=a b c\left|\begin{array}{ccc} a^{2}+b^{2}+c^{2} & 2 b^{2} & 2 c^{2} \\ a^{2}+b^{2}+c^{2} & -c^{2}+a^{2}-b^{2} & 2 c^{2} \\ a^{2}+b^{2}+c^{2} & 2 b^{2} & -a^{2}+b^{2}-c^{2} \end{array}\right| \\ &=a b c\left(a^{2}+b^{2}+c^{2}\right)\left|\begin{array}{ccc} 1 & 2 b^{2} & 2 c^{2} \\ 1 & -c^{2}+a^{2}-b^{2} & 2 c^{2} \\ 1 & 2 b^{2} & -a^{2}+b^{2}-c^{2} \end{array}\right| \end{aligned}$
$\begin{aligned} &\text { Now } \mathrm{R}_{2} \rightarrow R_{2}-R_{1}, R_{3} \rightarrow R_{3}-R_{1} \\ &=a b c\left(a^{2}+b^{2}+c^{2}\right)\left|\begin{array}{ccc} 1 & 2 b^{2} & 2 c^{2} \\ 0 & -\left(a^{2}+b^{2}+c^{2}\right) & 0 \\ 0 & 0 & -\left(a^{2}+b^{2}+c^{2}\right) \end{array}\right| \end{aligned}$
Expanding w.r.t $C_{1}$
$\begin{aligned} &=a b c\left(a^{2}+b^{2}+c^{2}\right)\left(a^{2}+b^{2}+c^{2}\right)^{2} \\ &=a b c\left(a^{2}+b^{2}+c^{2}\right)^{3} \\ &=R \cdot H \cdot S \end{aligned}$
Confused between CGPA and Percentage?
Get your results instantly with our calculator!
💡 Conversion Formula used is: Percentage = CGPA × 9.5
Question:41.1
determinants Exercise 5. 2 Question 41
Answer: $a^{3}+3 a^{2}$Hint Use determinant formula
Given:$\left|\begin{array}{ccc} 1+a & 1 & 1 \\ 1 & 1+a & 1 \\ 1 & 1 & 1+a \end{array}\right|=a^{3}+3 a^{2}$
Solution:
$\begin{aligned} &\text { L.H.S }\left|\begin{array}{ccc} 1+a & 1 & 1 \\ 1 & 1+a & 1 \\ 1 & 1 & 1+a \end{array}\right| \\ &\quad \text { Apply } \mathrm{C}_{1} \rightarrow C_{1}+C_{2}+C_{3} \\ &=\left|\begin{array}{ccc} a+3 & 1 & 1 \\ a+3 & 1+a & 1 \\ a+3 & 1 & 1+a \end{array}\right| \end{aligned}$
$\begin{aligned} &(a+3) \text { common from } \mathrm{C}_{1}\left[C_{1} \rightarrow \frac{C_{1}}{a+3}\right] \\ &=(a+3)\left|\begin{array}{ccc} 1 & 1 & 1 \\ 1 & 1+a & 1 \\ 1 & 1 & 1+a \end{array}\right| \end{aligned}$
$\begin{aligned} &\text { Now } \mathrm{C}_{3} \rightarrow C_{3}-C_{1} \\ &=(a+3)\left|\begin{array}{ccc} 1 & 1 & 0 \\ 1 & 1+a & 0 \\ 1 & 1 & a \end{array}\right| \end{aligned}$
Expanding w.r.t $C_{3}$
$\begin{aligned} &=(a+3)\left\{a \mid \begin{array}{cc} 1 & 1 \\ 1 & 1+a \end{array}\right\} \\ &=(a+3)\{a(1+a-1)\} \\ &=(a+3) a^{2} \\ &=a^{3}+3 a^{2} \\ &=R H . S \end{aligned}$
determinants Exercise 5.2 Question 41 sub question 2
Answer:
Answer:$(a-1)^{3}$Hint Use determinant formula
Given: $\left|\begin{array}{ccc} a^{2}+2 a & 2 a+1 & 1 \\ 2 a+1 & a+2 & 1 \\ 3 & 3 & 1 \end{array}\right|=(a-1)^{3}$
Solution:
$\text { L.H.S }\left|\begin{array}{ccc} a^{2}+2 a & 2 a+1 & 1 \\ 2 a+1 & a+2 & 1 \\ 3 & 3 & 1 \end{array}\right|$
$\begin{aligned} &\text { Apply } \mathrm{C}_{1} \rightarrow C_{1}-C_{2} \& C_{2} \rightarrow C_{2}-C_{3}\\ &=\left|\begin{array}{ccc} a^{2}-1 & 2 a & 1 \\ a-1 & a+1 & 1 \\ 0 & 2 & 1 \end{array}\right|\\ &(a-1) \text { common from } \mathrm{C}_{1}\\ &=(a-1)\left|\begin{array}{ccc} a+1 & 2 a & 1 \\ 1 & a+1 & 1 \\ 0 & 2 & 1 \end{array}\right| \end{aligned}$
$\begin{aligned} &\text { Apply } \mathrm{R}_{1} \rightarrow R_{1}-R_{2} \& R_{2} \rightarrow R_{2}-R_{3}\\ &=(a-1)\left|\begin{array}{ccc} a & (a-1) & 0 \\ 1 & a-1 & 0 \\ 0 & 2 & 1 \end{array}\right|\\ &\text { Expanding w.r.t } \mathrm{C}_{3}\\ &=(a-1)\left(1\left(a^{2}-a-a+1\right)\right)\\ &=(a-1)\left(a^{2}-2 a+1\right)\\ &=(a-1)(a-1)^{2}\\ &=(a-1)^{3}\\ &=R . H \cdot S \end{aligned}$
deteminants Exercise 5.2 Question 42
Answer: $(x+y+z)^{3}$Hint Use determinant formula
Given:$\left|\begin{array}{ccc} 2 y & y-z-x & 2 y \\ 2 z & 2 z & z-x-y \\ x-y-z & 2 x & 2 x \end{array}\right|=(x+y+z)^{3}$
$\text { L.H.S }\left|\begin{array}{ccc} 2 y & y-z-x & 2 y \\ 2 z & 2 z & z-x-y \\ x-y-z & 2 x & 2 x \end{array}\right|$
Solution:
$\begin{aligned} &\text { Apply } \mathrm{R}_{1} \rightarrow R_{1}+R_{2}+R_{3} \\ &=\left|\begin{array}{ccc} x+y+z & x+y+z & x+y+z \\ 2 z & 2 z & z-x-y \\ x-y-z & 2 x & 2 x \end{array}\right| \\ &(x+y+z) \text { common from } \mathrm{C}_{1} \\ &=(x+y+z)\left|\begin{array}{ccc} 1 & 1 & 1 \\ 2 z & 2 z & z-x-y \\ x-y-z & 2 x & 2 x \end{array}\right| \end{aligned}$
$\begin{aligned} &\text { Apply } C_{2} \rightarrow C_{2}-C_{1} \& \mathrm{C}_{3} \rightarrow C_{3}-C_{1} \\ &=(x+y+z)\left|\begin{array}{ccc} 1 & 0 & 1 \\ 2 z & 0 & -(x+y+z) \\ x-y-z & x+y+z \quad & x+y+z \end{array}\right| \\ &(x+y+z) \text { common from } \mathrm{C}_{2} \& C_{3} \\ &=(x+y+z)^{3}\left|\begin{array}{ccc} 1 & 0 & 1 \\ 2 z & 0 & -1 \\ x-y-z & 1 & 1 \end{array}\right| \end{aligned}$
Expanding w.r.t $C_{3}$
$=(x+y+z)^{3}\left[1\left|\begin{array}{cc} 0 & -1 \\ 1 & 1 \end{array}\right|-0(1+0) 1\right]$
$\begin{aligned} &=(x+y+z)^{3}\left|\begin{array}{cc} 0 & -1 \\ 1 & 1 \end{array}\right| \\ &=(x+y+z)^{3}(0(1)-(-1) 1) \\ &=(x+y+z)^{3} \\ &=R \cdot H . S \end{aligned}$
determinants Exercise 5.2 Question 43
Answer:$(x+y+z)(x-z)^{2}$Hint Use determinant formula
Given:$\left|\begin{array}{lll} y+z & x & y \\ z+x & z & x \\ x+y & y & z \end{array}\right|=(x+y+z)(x-z)^{2}$
Solution:
$\begin{aligned} &\text { L.H.S }\left|\begin{array}{ccc} y+z & x & y \\ z+x & z & x \\ x+y & y & z \end{array}\right| \\ &\text { Apply } \mathrm{R}_{1} \rightarrow R_{1}+R_{2}+R_{3} \\ &=\left|\begin{array}{ccc} 2(x+y+z) & x+y+z & x+y+z \\ z+x & z & x \\ x+y & y & z \end{array}\right| \end{aligned}$
$\begin{aligned} &(x+y+z) \text { common from } \mathrm{R}_{1}\\ &=(x+y+z)\left|\begin{array}{ccc} 2 & 1 & 1 \\ z+x & z & x \\ x+y & y & z \end{array}\right|\\ &\text { Apply } \mathrm{R}_{1} \rightarrow R_{1}-R_{2}\\ &=(x+y+z)\left|\begin{array}{lll} 1 & 1 & 1 \\ x & z & x \\ x & y & z \end{array}\right|\\ &\text { Apply } \mathrm{C}_{2} \rightarrow C_{2}-C_{1} \& \mathrm{C}_{3} \rightarrow C_{3}-C_{1}\\ &=(x+y+z)\left|\begin{array}{ccc} 1 & 0 & 0 \\ x & z-x & 0 \\ x & y-x & z-x \end{array}\right| \end{aligned}$
Expanding w.r.t $C_{3}$
$\begin{aligned} &=(x+y+z)\left((z-x)^{2}-0\right) \\ &=(x+y+z)(x-z)^{2} \\ &=R \cdot H \cdot S \end{aligned}$
determinants Exercise 5.2 Question 44
Answer:$a^{2}(a+x+y+z)$Hint Use determinant formula
Given:$\left|\begin{array}{ccc} a+x & y & z \\ x & a+y & z \\ x & y & a+z \end{array}\right|=a^{2}(a+x+y+z)$
Solution:
$\begin{aligned} &\text { L.H.S }\left|\begin{array}{ccc} a+x & y & z \\ x & a+y & z \\ x & y & a+z \end{array}\right| \\ &\text { Apply } \mathrm{C}_{1} \rightarrow C_{1}+C_{2}+C_{3} \\ &=\left|\begin{array}{ccc} a+x+y+z & y & z \\ a+x+y+z & a+y & z \\ a+x+y+z & y & a+z \end{array}\right| \end{aligned}$
$(a+x+y+z) \text { common from } \mathrm{C}_{1}$
$\begin{aligned} &=(a+x+y+z)\left|\begin{array}{ccc} 1 & y & z \\ 1 & a+y & z \\ 1 & y & a+z \end{array}\right|\\ &\text { Apply } \mathrm{R}_{1} \rightarrow R_{1}-R_{2} \& R_{2} \rightarrow R_{2}-R_{3}\\ &=(a+x+y+z)\left|\begin{array}{ccc} 0 & -a & 0 \\ 0 & a & -a \\ 1 & y & a+z \end{array}\right|\\ &\text { Expanding w.r.t } \mathrm{C}_{3}\\ &=(a+x+y+z)\left(1\left(a^{2}\right)\right)\\ &=(a+x+y+z) a^{2}\\ &=R \cdot H . S \end{aligned}$
deteminants Exercise 5.2 Question 45
Answer: $2(a-b)(b-c)(c-a)(a+b+c)$Hint Use determinant formula
Given:$\left|\begin{array}{lll} a^{3} & 2 & a \\ b^{3} & 2 & b \\ c^{3} & 2 & c \end{array}\right|=2(a-b)(b-c)(c-a)(a+b+c)$
Solution:
$\begin{aligned} &\text { L.H.S }\left|\begin{array}{ccc} a^{3} & 2 & a \\ b^{3} & 2 & b \\ c^{3} & 2 & c \end{array}\right| \\ &\text { Apply } \mathrm{R}_{2} \rightarrow R_{2}-R_{1} \& R_{3} \rightarrow R_{3}-R_{1} \\ &=\left|\begin{array}{ccc} a^{3} & 2 & a \\ b^{3}-a^{3} & 0 & b-a \\ c^{3}-a^{3} & 0 & c-a \end{array}\right| \\ &\text { Use } a^{3}-b^{3}=(a-b)\left(a^{2}+a b+b^{2}\right) \\ &=\left|\begin{array}{ccc} a^{3} & 2 & a \\ (b-a)\left(b^{2}+a b+a^{2}\right) & 0 & b-a \\ (c-a)\left(c^{2}+a c+a^{2}\right) & 0 & c-a \end{array}\right| \end{aligned}$
$\begin{aligned} &=(-1)(-1)\left|\begin{array}{ccc} a^{3} & 2 & a \\ (a-b)\left(b^{2}+a b+a^{2}\right) & 0 & a-b \\ (c-a)\left(c^{2}+a c+a^{2}\right) & 0 & c-a \end{array}\right|\\ &(a-b)(c-a) \text { common from } \mathrm{R}_{2} \& R_{3}\\ &=(a-b)(c-a)\left|\begin{array}{ccc} a^{3} & 2 & a \\ \left(b^{2}+a b+a^{2}\right) & 0 & 1 \\ \left(c^{2}+a c+a^{2}\right) & 0 & 1 \end{array}\right|\\ &\text { Expanding w.r.t } \mathrm{C}_{2}\\ &=(a-b)(c-a)\left[2\left(b^{2}+a b+a^{2}-c^{2}-a c-a^{2}\right)\right]\\ &=2(a-b)(c-a)[(b+c)(b-c)+a(b-c)]\\ &=2(a-b)(b-c)(c-a)(a+b+c)\\ &=R . H . S \end{aligned}$
determinants Exercise 5.2 Question 46
Answer: ProvedHint Use determinant formula
Given: $\left|\begin{array}{lll} a & b & c \\ x & y & z \\ p & q & r \end{array}\right|=\left|\begin{array}{lll} x & y & z \\ p & q & r \\ a & b & c \end{array}\right|=\left|\begin{array}{lll} y & b & q \\ x & a & p \\ z & c & r \end{array}\right|$
Solution:
$\begin{aligned} &\left|\begin{array}{lll} a & b & c \\ x & y & z \\ p & q & r \end{array}\right|=\left|\begin{array}{lll} x & y & z \\ p & q & r \\ a & b & c \end{array}\right| \\ &R_{1} \leftrightarrow R_{3}, R_{2} \leftrightarrow R_{1}, R_{3} \leftrightarrow R_{1} \\ &=(-)(-)\left|\begin{array}{lll} x & y & z \\ a & b & c \\ p & q & r \end{array}\right| \end{aligned}$
For interchanging R2↔R3
negative sign will be added
Now convert rows in columns and column in row
$\begin{aligned} &\left|\begin{array}{lll} x & y & z \\ p & q & r \\ a & b & c \end{array}\right|=\left|\begin{array}{lll} x & p & a \\ y & q & b \\ z & r & c \end{array}\right| \\ &\text { Now } R_{1} \leftrightarrow R_{2}, C_{2} \leftrightarrow C_{3} \\ &(-)(-)\left|\begin{array}{lll} y & q & b \\ x & p & a \\ z & r & c \end{array}\right|=\left|\begin{array}{lll} y & b & q \\ x & a & p \\ z & c & r \end{array}\right| \end{aligned}$
deteminants Exercise 5.2 Question 47
Answer: 0Hint $a, b, c \text { are in AP }$
Given: $\left|\begin{array}{lll} x+1 & x+2 & x+a \\ x+2 & x+3 & x+b \\ x+3 & x+4 & x+c \end{array}\right|=0 \text { where } a, b, c \text { are in } \mathrm{AP}$
Solution:
$\begin{aligned} &\text { L.H.S }\left|\begin{array}{ccc} x+1 & x+2 & x+a \\ x+2 & x+3 & x+b \\ x+3 & x+4 & x+c \end{array}\right| \\ &\text { Now } R_{3} \rightarrow-2 R_{2}+R_{1}+R_{3} \\ &=\left|\begin{array}{ccc} x+1 & x+2 & x+a \\ x+2 & x+3 & x+b \\ 0 & 0 & a+c-2 b \end{array}\right| \end{aligned}$
As $a b_{2} c \text { are in } \mathrm{AP}, 2 \mathrm{~b}=\mathrm{a}+\mathrm{c}$
Substituting the value in the denominator,
$\begin{aligned} &=\left|\begin{array}{ccc} x+1 & x+2 & x+a \\ x+2 & x+3 & x+b \\ 0 & 0 & a+c-a-c \end{array}\right| \\ &=\left|\begin{array}{ccc} x+1 & x+2 & x+a \\ x+2 & x+3 & x+b \\ 0 & 0 & 0 \end{array}\right| \end{aligned}$
Expanding along $R_{3}$
$=0$
deteminants Exercise 5.2 Question 48
Answer:$0$Hint $\alpha, \beta, \gamma \text { are in AP }$
Given:$\left|\begin{array}{lll} x-3 & x-4 & x-\alpha \\ x-2 & x-3 & x-\beta \\ x-1 & x-2 & x-\gamma \end{array}\right|=0 \text { where } \alpha, \beta, \gamma \text { are in } \mathrm{AP}$
Solution:
$$$ \begin{aligned} &\alpha, \beta, \gamma \text { are in } \mathrm{AP}\\ &a_{2}-a_{1}=a_{3}-a_{1}\\ &\beta-\gamma=\gamma-\beta\\ &\beta+\beta=\gamma+\alpha\\ &2 \beta=\gamma+\alpha\\ &\gamma+\alpha-2 \beta=0 \end{aligned}$ ...(1)
$$$ \begin{aligned} &\text { L.H.S }\left|\begin{array}{ccc} x-3 & x-4 & x-\alpha \\ x-2 & x-3 & x-\beta \\ x-1 & x-2 & x-\gamma \end{array}\right| \\ &\text { Now } R_{1} \rightarrow R_{1}+R_{3}-2 R_{2} \\ &=\left|\begin{array}{ccc} 0 & 0 & 2 \beta-(\alpha+\gamma) \\ x-2 & x-3 & x-\beta \\ x-1 & x-2 & x-\gamma \end{array}\right| \end{aligned}$
$$$ \begin{aligned} &=\left|\begin{array}{ccc} 0 & 0 & 2 \beta-2 \beta \\ x-2 & x-3 & x-\beta \\ x-1 & x-2 & x-\gamma \end{array}\right| \ldots \ldots \ldots \text { from(1) } \\ &=\left|\begin{array}{ccc} 0 & 0 & 0 \\ x-2 & x-3 & x-\beta \\ x-1 & x-2 & x-\gamma \end{array}\right| \end{aligned}$
Expanding along $R_{1}$,
$=0$ ..hence proved
determinants Exercise 5.2 Question 49
Answer: $a+b+c=0 \& a=b=c$Hint Use determinant formula
Given: $$$ \text { If } a, b, c \text { are real no. such that }\left|\begin{array}{ccc} b+c & c+a & a+b \\ c+a & a+b & b+c \\ a+b & b+c & c+a \end{array}\right|=0 \text { show that either } a+b=c \text { or } a=b=c$
Solution:
$$$ \begin{aligned} &\text { L.H.S }\left|\begin{array}{lll} b+c & c+a & a+b \\ c+a & a+b & b+c \\ a+b & b+c & c+a \end{array}\right| \\ &\text { Apply } R_{1} \rightarrow R_{1}+R_{2}+R_{3} \\ &=\left|\begin{array}{ccc} 2(a+b+c) & 2(a+b+c) & 2(a+b+c) \\ c+a & a+b & b+c \\ a+b & b+c & c+a \end{array}\right| \end{aligned}$
$$$ \begin{aligned} &2(a+b+c) \text { common from } \mathrm{R}_{1} \\ &=2(a+b+c)\left|\begin{array}{ccc} 1 & 1 & 1 \\ c+a & a+b & b+c \\ a+b & b+c & c+a \end{array}\right| \end{aligned}$
$$$ \begin{aligned} &\text { Apply C }_{1} \rightarrow C_{1}-C_{2} \& \mathrm{C}_{2} \rightarrow C_{2}-C_{3} \\ &=2(a+b+c)\left|\begin{array}{ccc} 0 & 0 & 1 \\ c-b & a-c & b+c \\ a-c & b-a & c+a \end{array}\right| \end{aligned}$
$$$ \begin{aligned} &\text { Expand from } \mathrm{R}_{1}\\ &=2(a+b+c)\left\{1 \mid \begin{array}{ll} c-b & a-c \\ a-c & b-a \end{array}\right\}\\ &(a+b+c)=0 \quad \text { or } \quad b c-b^{2}-a c+a b-(a-c)^{2}=0\\ &b c-b^{2}-a c+a b=a^{2}+c^{2}-2 a c\\ &a^{2}+b^{2}+c^{2}-a^{2}-a b-a c=0\\ &2\left(a^{2}+b^{2}+c^{2}\right)-2 a c-2 a b-2 b c=0\\ &(a-b)^{2}+(b-c)^{2}+(c-a)^{2}=0\\ &a-b=0, \quad b-c=0, \quad c-a=0\\ &a=b, b=c, \quad c=a \end{aligned}$
Hence,$a+b+c=0 \& a=b=c$
deteminants Exercise 5.2 Question 50
Answer: $2$Hint Use determinant formula
Given: $$$ \left|\begin{array}{lll} p & b & c \\ q & q & c \\ a & b & r \end{array}\right|=0 \text { Find value of } \frac{p}{p-a}+\frac{q}{q-b}+\frac{r}{r-c} \& p \neq a, q \neq b, r \neq c$
Solution:
$$$ \begin{aligned} &\text { L.H.S }\left|\begin{array}{ccc} p & b & c \\ q & q & c \\ a & b & r \end{array}\right| \\ &\text { Apply } \mathrm{R}_{1} \rightarrow R_{1}-R_{3} \& R_{2} \rightarrow R_{2}-R_{3} \\ &\left|\begin{array}{ccc} p-a & 0 & c-p \\ 0 & q-b & c-p \\ a & 0 & r \end{array}\right|=0 \end{aligned}$
$$$ \begin{aligned} &\text { Expand from } \mathrm{R}_{1}\\ &=p-a\left|\begin{array}{cc} q-b & c-p \\ b & r \end{array}\right|+(c-r)\left|\begin{array}{cc} 0 & q-b \\ a & b \end{array}\right|=0\\ &=(p-a)[r(q-b)-b(c-r)+(c-r)(-a(q-b))]=0\\ &=\frac{r(q-b)(p-a)}{(p-a)(q-b)(r-c)}-\frac{b(p-a)(c-r)}{(p-a)(q-b)(r-c)}-\frac{a(c-r)(q-b)}{(p-a)(q-b)(r-c)}=0 \end{aligned}$
$$$ \begin{aligned} &=\frac{r}{r-c}+\frac{b}{q-b}+\frac{a}{p-a}=0 \\ &=\frac{r}{r-c}+\frac{b+q-q}{q-b}+\frac{a+p-p}{p-a}=0 \\ &=\frac{r}{r-c}+\frac{b-q}{q-b}+\frac{q}{q-b}+\frac{a-p}{p-a}-\frac{p}{p-a}=0 \\ &=\frac{r}{r-c}-1+\frac{q}{q-b}-1-\frac{p}{p-a}=0 \\ &=\frac{p}{p-a}+\frac{q}{q-b}+\frac{r}{r-c}=2 \end{aligned}$
determinants Exercise 5.2 Question 51
Answer: $$$ x=1,2,-3$Hint Use determinant formula
Given: $$$ \text { S.T } x=2 \text { is root of equation }\left|\begin{array}{ccc} x & -6 & -1 \\ 2 & 3 x & x-3 \\ -3 & 2 x & x+3 \end{array}\right|=0$
Solution:
$$$ \left|\begin{array}{ccc} x & -6 & -1 \\ 2 & 3 x & x-3 \\ -3 & 2 x & x+3 \end{array}\right|=0$
$$$ \begin{aligned} &\text { Expand from } \mathrm{C}_{1}\\ &=x\left(-3 x^{2}-6 x-2 x^{2}+6 x\right)+(2 x+4+3 x-9)+30 x-30+5 x=0\\ &=-5 x^{2}+30 x-30+5 x=0\\ &=-5\left(x^{3}+7 x+6\right)=0\\ &=x^{3}+7 x+6=0\\ &=(x-1)\left(x^{2}+x-6\right)=0\\ &=(x-1)\left(x^{2}+3 x-2 x-6\right)=0\\ &=(x-1)(x+3)(x-2)=0\\ &x=1,2,-3 \end{aligned}$
determinants Exercise 5.2 Question 52
Answer: $0$Hint Use determinant formula
Given: $$$ \left|\begin{array}{ccc} x+a & b & c \\ a & x+b & c \\ a & b & x+c \end{array}\right|=0$
Solution:
$$$ \begin{aligned} &\text { Apply C }_{1} \rightarrow C_{1}+C_{2}+C_{3} \\ &\left|\begin{array}{lcc} x+a+b+c & b & c \\ x+a+b+c & x+b & c \\ x+a+b+c & b & x+c \end{array}\right| \\ &(x+a+b+c) \text { common from } \mathrm{C}_{1} \\ &=(x+a+b+c)\left|\begin{array}{ccc} 1 & b & c \\ 1 & x+b & c \\ 1 & b & x+c \end{array}\right| \end{aligned}$
$$$ \begin{aligned} &\text { Apply } \mathrm{R}_{2} \rightarrow R_{2}-R_{1} \& R_{3} \rightarrow R_{3}-R_{1}\\ &=(x+a+b+c)\left|\begin{array}{lll} 1 & b & c \\ 0 & x & 0 \\ 0 & 0 & x \end{array}\right|\\ &\text { Expand from } \mathrm{C}_{1}\\ &=(x+a+b+c)\left(\left(x^{2}-0\right)-0+0\right)=0\\ &x+a+b+c=0, x^{2}=0\\ &x=-(a+b+c), x=0 \end{aligned}$
deteminants Exercise 5.2 Question 52 sub question 2
Answer:$x=\frac{-a}{3}$Hint: Use determinant formula
Given: $$$ \left|\begin{array}{ccc} x+a & x & x \\ x & x+a & x \\ x & x & x+a \end{array}\right|=0$
Solution:
$$$ \begin{aligned} &\text { Apply C }_{1} \rightarrow C_{1}+C_{2}+C_{3} \\ &\left|\begin{array}{ccc} 3 x+a & x & x \\ 3 x+a & x+a & x \\ 3 x+a & x & x+a \end{array}\right| \\ &(3 x+a) \text { common from } C_{1} \\ &=(3 x+a)\left|\begin{array}{ccc} 1 & x & x \\ 1 & x+a & x \\ 1 & x & x+c \end{array}\right| \end{aligned}$
$$$ \begin{aligned} &\text { Apply } \mathrm{R}_{2} \rightarrow R_{2}-R_{1} \& R_{3} \rightarrow R_{3}-R_{1}\\ &=(3 x+a)\left|\begin{array}{ccc} 1 & x & x \\ 0 & a & 0 \\ 0 & 0 & a \end{array}\right|\\ &\text { Expand from } \mathrm{C}_{1}\\ &=(3 x+a) a^{2}=0\\ &3 x+a=0, a^{2}=0\\ &3 x=-a, a=0\\ &x=\frac{-a}{3}, a=0 \end{aligned}$
deteminants Exercise 5.2 Question 52 sub question 3
Answer: $x=\frac{2}{3}, \frac{11}{3}, \frac{11}{3}$Hint: Use determinant formula
Given: $$$ \left|\begin{array}{ccc} 3 x-8 & 3 & 3 \\ 3 & 3 x-8 & 3 \\ 3 & 3 & 3 x-8 \end{array}\right|=0$
Solution:
$$$ \text { L.H.S }\left|\begin{array}{ccc} 3 x-8 & 3 & 3 \\ 3 & 3 x-8 & 3 \\ 3 & 3 & 3 x-8 \end{array}\right|$
$$$ \begin{aligned} &\text { Apply C }_{1} \rightarrow C_{1}+C_{2}+C_{3} \\ &\left|\begin{array}{ccc} 3 x-2 & 3 & 3 \\ 3 x-2 & 3 x-8 & 3 \\ 3 x-2 & 3 & 3 x-8 \end{array}\right| \end{aligned}$
$$$ \begin{aligned} &(3 x-2) \text { common from } \mathrm{C}_{1}\\ &=(3 x-2)\left|\begin{array}{ccc} 1 & 3 & 3 \\ 1 & 3 x-8 & 3 \\ 1 & 3 & 3 x-8 \end{array}\right|\\ &\text { Apply } \mathrm{R}_{2} \rightarrow R_{2}-R_{1} \& R_{3} \rightarrow R_{3}-R_{1}\\ &=(3 x-2)\left|\begin{array}{ccc} 1 & 3 & 3 \\ 0 & -3 x-11 & 0 \\ 0 & 0 & -3 x-11 \end{array}\right| \end{aligned}$
$\begin{aligned} &\text { Expand from } \mathrm{C}_{1}\\ &=(3 x-2)(3 x-11)^{2}=0\\ &3 x-2=0,3 x-11=0\\ &3 x=2,3 x=11\\ &x=\frac{2}{3}, x=\frac{11}{3}, \frac{11}{3}\\ &x=\frac{2}{3}, \frac{11}{3}, \frac{11}{3} \end{aligned}$
deteminants Exercise 5.2 Question 52 sub question 4
Answer: $x=a, b$Hint Use determinant formula
Given:$\left|\begin{array}{lll} 1 & x & x^{2} \\ 1 & a & a^{2} \\ 1 & b & b^{2} \end{array}\right|=0$
Solution:
$$$ L \cdot H S\left|\begin{array}{lll} 1 & x & x^{2} \\ 1 & a & a^{2} \\ 1 & b & b^{2} \end{array}\right|$
$$$ \begin{aligned} &\text { Apply } \mathrm{R}_{2} \rightarrow R_{2}-R_{1} \& R_{3} \rightarrow R_{3}-R_{1} \\ &\left|\begin{array}{ccc} 1 & x & x^{2} \\ 0 & a-x & a^{2}-x^{2} \\ 0 & b-x & b^{2}-x^{2} \end{array}\right| \end{aligned}$
$$$ \begin{aligned} &(a-x)(b-x) \text { common from } \mathrm{R}_{2} \& R_{3} \\ &=(a-x)(b-x)\left|\begin{array}{ccc} 1 & x & x^{2} \\ 0 & 1 & a+x \\ 0 & 1 & b+x \end{array}\right| \end{aligned}$
Expand from $C_{1}$
$\begin{aligned} &(a-x)(b-x)[b+x-a+x]=0 \\ &(a-x)(b-x)[-a+b]=0 \\ &a=x \\ &b=x \\ &a=b-\text { neglect } \end{aligned}$
deteminants Exercise 5.2 Question 52 sub question 5
Answer:
Answer: $x=1,1,-9$Hint: Use determinant formula
Given: $\left|\begin{array}{ccc} x+1 & 3 & 5 \\ 2 & x+2 & 5 \\ 2 & 3 & x+4 \end{array}\right|=0$
Solution:
$\begin{aligned} &\text { Apply } \mathrm{C}_{1} \rightarrow C_{1}+C_{2}+C_{3} \\ &\left|\begin{array}{ccc} x+9 & 3 & 5 \\ x+9 & x+2 & 5 \\ x+9 & 3 & x+4 \end{array}\right|=0 \end{aligned}$
$\begin{aligned} &(x+9) \text { common from } \mathrm{C}_{1}\\ &=(x+9)\left|\begin{array}{ccc} 1 & 3 & 5 \\ 1 & x+2 & 5 \\ 1 & 3 & x+4 \end{array}\right|\\ &\text { Apply } \mathrm{R}_{2} \rightarrow R_{2}-R_{1} \& R_{3} \rightarrow R_{3}-R_{1}\\ &=(x+9)\left|\begin{array}{ccc} 1 & 3 & 5 \\ 0 & x-1 & 0 \\ 0 & 0 & x-1 \end{array}\right| \end{aligned}$
$\begin{aligned} &\text { Expand from } \mathrm{C}_{1}\\ &(x+9)(x-1)^{2}=0\\ &x+9=0,(x-1)^{2}=0\\ &x=-9, x=1,1\\ &x=1,1,-9 \end{aligned}$
deteminants Exercise 5.2 Question 52 Sub Question 6
Answer: $x=b, c,-(b+c)$Hint: Use determinant formula
Given:$\left|\begin{array}{lll} 1 & x & x^{3} \\ 1 & b & b^{3} \\ 1 & c & c^{3} \end{array}\right|=0$
Solution:
$\text { LHS }\left|\begin{array}{lll} 1 & x & x^{3} \\ 1 & b & b^{3} \\ 1 & c & c^{3} \end{array}\right|$
$\begin{aligned} &\text { Apply } \mathrm{R}_{2} \rightarrow R_{2}-R_{1} \& R_{3} \rightarrow R_{3}-R_{1} \\ &\begin{array}{lcc} 1 & x & x^{3} \\ 0 & b-x & b^{3}-x^{3} \\ 0 & c-x & c^{3}-x^{3} \end{array} \mid \\ &\text { Use } b^{3}-x^{3}=(b-x)\left(b^{2}+b x+x^{2}\right) \\ &\left|\begin{array}{ccc} 1 & x & x^{3} \\ 0 & b-x & (b-x)\left(b^{2}+b x+x^{2}\right) \\ 0 & c-x & (c-x)\left(c^{2}+c x+x^{2}\right) \end{array}\right| \end{aligned}$
$\begin{aligned} &(b-x)(c-x) \text { common from } \mathrm{R}_{2} \& R_{3} \\ &=(b-x)(c-x)\left|\begin{array}{ccc} 1 & x & x^{3} \\ 0 & 1 & \left(b^{2}+b x+x^{2}\right) \\ 0 & 1 & \left(c^{2}+c x+x^{2}\right) \end{array}\right| \end{aligned}$
$\begin{aligned} &\text { Expand from } \mathrm{C}_{1}\\ &(b-x)(c-x)\left(c^{2}+c x+x^{2}-b^{2}-b x-x^{2}\right)=0\\ &b-x=0, c-x=0, c^{2}-b^{2}-b x+c x=0\\ &x=b, c=x,(c+b)(c-b)-x(c-b)=0 \end{aligned}$
deteminants Exercise 5.2 Question 52 Sub Question 7
Answer: $x=4$Hint: Use determinant formula
Given: $\left|\begin{array}{ccc} 15-2 x & 11-3 x & 7-x \\ 11 & 17 & 14 \\ 10 & 16 & 13 \end{array}\right|=0$
Solution:
$\begin{aligned} &\text { Apply } R_{3} \rightarrow R_{3}-R_{2} \\ &\begin{array}{ccc} 15-2 x & 11-3 x & 7-x \\ 11 & 17 & 14 \\ 1 & 1 & 1 \end{array} \mid=0 \\ &\text { Now } \mathrm{C}_{1} \rightarrow C_{1}-C_{2} \& C_{2} \rightarrow C_{2}-C_{3} \\ &\left|\begin{array}{ccc} 4+x & 4-2 x & 7-x \\ -6 & 3 & 14 \\ 0 & 0 & 1 \end{array}\right|=0 \end{aligned}$
$\begin{aligned} &\text { Expand from } \mathrm{R}_{3}\\ &0\left|\begin{array}{cc} 4+x & 4-2 x \\ -6 & 3 \end{array}\right|-0\left|\begin{array}{cc} 4+x & 7-x \\ -6 & 14 \end{array}\right|+1\left|\begin{array}{cc} 4+x & 4-2 x \\ -6 & 3 \end{array}\right|=0\\ &1(3(4+x)-(-6)(4-2 x))=0\\ &12+3 x+24-12 x=0\\ &36=9 x\\ &x=4 \end{aligned}$
deteminants Exercise 5.2 Question 52 Sub Question 8
Answer:$x=2,1$Hint: Use determinant formula
Given: $\left|\begin{array}{ccc} 1 & 1 & x \\ p+1 & p+1 & p+x \\ 3 & x+1 & x+2 \end{array}\right|$
Solution:
$\begin{aligned} &\text { Apply } \mathrm{C}_{1} \rightarrow C_{1}-C_{2}\\ &\left|\begin{array}{ccc} 0 & 1 & x \\ 0 & p+1 & p+x \\ 2-x & x+1 & x+2 \end{array}\right|=0\\ &\text { Expand from } \mathrm{C}_{1}\\ &(2-x)(p+x-x(p+1))=0\\ &(2-x)(p+x-p x-x)=0\\ &(2-x)(1-x)=0\\ &2-x=0,1-x=0\\ &x=2, x=1 \end{aligned}$
deteminants Exercise 5.2 Question 52 Sub Question 9
Answer: $\theta=n \pi+\frac{(-1)^{n} \pi}{6}, n \in z \text { or } n \pi$Hint: Use determinant formula
Given:$\left|\begin{array}{ccc} 3 & -2 & \sin 3 \theta \\ -7 & 8 & \cos 2 \theta \\ -11 & 14 & 2 \end{array}\right|=0$
Solution:
$\begin{aligned} &\text { Apply C }_{1} \rightarrow C_{1}+C_{2} \\ &\begin{array}{lcc} 1 & -2 & \sin 3 \theta \\ 1 & 8 & \cos 2 \theta \\ 3 & 14 & 2 \end{array} \mid=0 \\ &\text { Apply } \mathrm{R}_{2} \rightarrow R_{2}-R_{3} \& R_{3} \rightarrow R_{3}-R_{1} \\ &\left|\begin{array}{ccc} 1 & -2 & \sin 3 \theta \\ 0 & 10 & \cos 2 \theta-\sin 3 \theta \\ 0 & 20 & 2-3 \sin 3 \theta \end{array}\right|=0 \end{aligned}$
$\begin{aligned} &\text { Expand from } \mathrm{C}_{1}\\ &1[10(2-3 \sin \theta)-20(\cos 2 \theta-\sin 3 \theta)]=0\\ &2-3 \sin 3 \theta-2(\cos 2 \theta-\sin 3 \theta)=0\\ &2-3 \sin 3 \theta-2 \cos 2 \theta+2 \sin 3 \theta=0\\ &2-3 \sin 3 \theta-2 \cos 2 \theta=0\\ &2=3 \sin \theta-4 \sin ^{3} \theta+2\left(1-2 \sin ^{2} \theta\right)\\ &2=3 \sin \theta-4 \sin ^{3} \theta+2-4 \sin ^{2} \theta\\ &0=\sin \theta-\left(3-4 \sin ^{2} \theta-4 \sin \theta\right)\\ &-4 \sin ^{2} \theta-4 \sin \theta+3=0\\ &4 \sin ^{2} \theta+4 \sin \theta-3=0\\ &\sin \theta=x\\ &4 x^{2}+4 x-3=0 \end{aligned}$
$\begin{aligned} &4 x^{2}+6 x-2 x-3=0 \\ &2 x(2 x+3)-1(2 x+3)=0 \\ &2 x+3=0,2 x-1=0,2 x=0 \\ &x=\frac{-3}{2}, x=\frac{1}{2}, x=0, \\ &\sin \theta=\frac{-3}{2}, \sin \theta=\frac{1}{2}, \sin \theta=0 \\ &\text { As } \sin \theta=0, \theta=n \pi \& \text { as } \sin \theta=\frac{1}{2}, \theta=n \pi+\frac{(-1)^{n} \pi}{6}, n \in z \end{aligned}$
deteminants Exercise 5.2 Question 52.1
Answer: $x=-12,0$Hint: Use determinant formula
Given: $\left|\begin{array}{lll} 4-x & 4+x & 4+x \\ 4+x & 4-x & 4+x \\ 4+x & 4+x & 4-x \end{array}\right|=0$
Solution:
$\begin{aligned} &\text { Apply C }_{1} \rightarrow C_{1}+C_{2}+C_{3} \\ &\left|\begin{array}{lll} 12+x & 4+x & 4+x \\ 12+x & 4-x & 4+x \\ 12+x & 4+x & 4-x \end{array}\right|=0 \end{aligned}$
$\begin{aligned} &(12+x) \text { common from } \mathrm{C}_{1}\\ &(12+x)\left|\begin{array}{ccc} 1 & 4+x & 4+x \\ 1 & 4-x & 4+x \\ 1 & 4+x & 4-x \end{array}\right|=0 \end{aligned}$
$\begin{aligned} &\text { Apply } \mathrm{R}_{2} \rightarrow R_{2}-R_{1} \& R_{3} \rightarrow R_{3}-R_{1}\\ &(12+x)\left|\begin{array}{ccc} 1 & 4+x & 4+x \\ 0 & -2 x & 0 \\ 0 & 0 & -2 x \end{array}\right|=0\\ &\text { Expand from } \mathrm{C}_{1}\\ &(12+x)\left(4 x^{2}\right)=0\\ &12+x=0,4 x^{2}=0\\ &x=-12, x=0 \end{aligned}$
determinants Exercise 5.2 Question 53
Answer: $1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0$Hint Use determinant formula
Given:$\left|\begin{array}{ccc} 1+a & 1 & 1 \\ 1 & 1+b & 1 \\ 1 & 1 & 1+c \end{array}\right|=0 \text { p.t } \frac{1}{a}+\frac{1}{b}+\frac{1}{c}+1=0$
Solution:
$\begin{aligned} &\text { Divide a to } \mathrm{C}_{1}, \mathrm{~b} \text { to } \mathrm{C}_{2} \text { and } \mathrm{c} \text { to } \mathrm{C}_{3}\\ &(a \times b \times c)\left|\begin{array}{ccc} \frac{1}{a}+1 & \frac{1}{b} & \frac{1}{c} \\ \frac{1}{a} & \frac{1}{b}+1 & \frac{1}{c} \\ \frac{1}{a} & \frac{1}{b} & \frac{1}{c}+1 \end{array}\right|=0 \end{aligned}$
$\begin{aligned} &\text { Apply } \mathrm{C}_{1} \rightarrow C_{1}+C_{2}+C_{3} \\ &\text { (abc) }\left|\begin{array}{lll} \frac{1}{a}+\frac{1}{b}+\frac{1}{c}+1 & \frac{1}{b} & \frac{1}{c} \\ \frac{1}{a}+\frac{1}{b}+\frac{1}{c}+1 & \frac{1}{b}+1 & \frac{1}{c} \\ \frac{1}{a}+\frac{1}{b}+\frac{1}{c}+1 & \frac{1}{b} & \frac{1}{c}+1 \end{array}\right|=0 \end{aligned}$
$\begin{aligned} &\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+1\right) \text { common from } \mathrm{C}_{1}\\ &(a b c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+1\right)\left|\begin{array}{lll} 1 & \frac{1}{b} & \frac{1}{c} \\ 1 & \frac{1}{b}+1 & \frac{1}{c} \\ 1 & \frac{1}{b} & \frac{1}{c}+1 \end{array}\right|=0\\ &\text { Apply } \mathrm{R}_{2} \rightarrow R_{2}-R_{1} \& R_{3} \rightarrow R_{3}-R_{1}\\ &a b c\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+1\right)\left|\begin{array}{lcc} 1 & \frac{1}{b} & \frac{1}{c} \\ 0 & \frac{1}{b}+1-\frac{1}{b} & \frac{1}{c}-\frac{1}{c} \\ 0 & \frac{1}{b}-\frac{1}{b} & \frac{1}{c}+1-\frac{1}{c} \end{array}\right|\\ &a b c\left|\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+1\right|\left|\begin{array}{ccc} 1 & \frac{1}{b} & \frac{1}{c} \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right| \end{aligned}$
$\begin{aligned} &\text { Expand from } \mathrm{R}_{3}\\ &a b c\left|\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+1\right|=0\\ &a, b, c \neq 0, \frac{1}{a}+\frac{1}{b}+\frac{1}{c}+1=0\\ &a b c \neq 0 \end{aligned}$
determinants Exercise 5.2 Question 5
Answer: $2$Hint: Use determinant formula
Given:$\left|\begin{array}{ccc} a & b-y & c-z \\ a-x & b & c-z \\ a-x & b-y & c \end{array}\right|=0 \text { then using property of determinant. Find } \frac{a}{x}+\frac{b}{y}+\frac{c}{z} \text { where } x, y, z=0$
Solution:
$\begin{aligned} &\text { Take } x, y, z \text { common from } C_{1}, C_{2}, C_{3}\\ &(x y z)\left|\begin{array}{ccc} \frac{a}{x} & \frac{b}{y}-1 & \frac{c}{z}-1 \\ a-\frac{1}{x} & \frac{b}{y} & \frac{c}{z}-1 \\ \frac{a}{x}-1 & \frac{b}{y}-1 & \frac{c}{z} \end{array}\right|=0 \end{aligned}$
$\begin{aligned} &\text { Apply } \mathrm{C}_{1} \rightarrow C_{1}+C_{2}+C_{3} \\ &\mid \frac{a}{x}+\frac{b}{y}+\frac{c}{z}-2 & \frac{b}{y}-1 & \frac{c}{z}-1 \\ &\frac{a}{x}+\frac{b}{y}+\frac{c}{z}-2 & \frac{b}{y} & \frac{c}{z}-1 \\ &\frac{a}{x}+\frac{b}{y}+\frac{c}{z}-2 & \frac{b}{y}-1 & \frac{c}{z} \end{aligned} \mid=0$
$\left(\frac{a}{x}+\frac{b}{y}+\frac{c}{z}-2\right)\left|\begin{array}{ccc} 1 & \frac{b}{y}-1 & \frac{c}{z}-1 \\ 1 & \frac{b}{y} & \frac{c}{z}-1 \\ 1 & \frac{b}{y}-1 & \frac{c}{z} \end{array}\right|=0$
$\begin{aligned} &\text { Apply } \mathrm{R}_{2} \rightarrow R_{2}-R_{1} \& R_{3} \rightarrow R_{3}-R_{1} \\ &\left(\frac{a}{x}+\frac{b}{y}+\frac{c}{z}-2\right)\left|\begin{array}{ccc} 1 & \frac{b}{y}-1 & \frac{c}{z}-1 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right|=0 \end{aligned}$
$\left(\frac{a}{x}+\frac{b}{y}+\frac{c}{z}-2\right) \cdot 1=0 \text { or } \frac{a}{x}+\frac{b}{y}+\frac{c}{z}=2$
determinants Exercise 5.2 Question 55
Answer: $9(3 x y z+x y+y z+z x)$
Hint: Use determinant formula
Given: $\left|\begin{array}{ccc} 1 & 1 & 1+3 x \\ 1+3 y & 1 & 1 \\ 1 & 1+3 z & 1 \end{array}\right|=9(3 x y z+x y+y z+z x)$
Solution:
$\begin{aligned} &\text { L.H.S }\left|\begin{array}{ccc} 1 & 1 & 1+3 x \\ 1+3 y & 1 & 1 \\ 1 & 1+3 z & 1 \end{array}\right| \\ &\text { Apply } \mathrm{R}_{1} \rightarrow R_{1}-R_{3} \& R_{2} \rightarrow R_{2}-R_{3} \\ &\left|\begin{array}{ccc} 0 & -3 z & 3 x \\ 3 y & -3 z & 0 \\ 1 & 1+3 z & 1 \end{array}\right| \end{aligned}$
$\begin{aligned} &3 \text { common from } \mathrm{R}_{1} \& R_{2}\\ &3 \times 3\left|\begin{array}{ccc} 0 & -z & x \\ y & -z & 0 \\ 1 & 1+3 z & 1 \end{array}\right| \end{aligned}$
$\begin{aligned} &\text { Expand from } \mathrm{R}_{1}\\ &9[z(y)+x(y+3 z y+z)\\ &9(3 x y z+x y+y z+z x) \end{aligned}$
The class 12 chapter 5, Determinants, is where the tendency to score marks easily as possible. The students need to make use of this opportunity to the best level. The concepts present in the RD Sharma class 12th exercise 5.2, are to evaluate the determinants, prove the determinant's value is zero, Arithmetic Progression in Determinants, and prove using the properties of determinants. A total of eighty-eight questions, including its subparts, are present in this exercise. It would take much time to solve or recheck the answers. Hence a copy of the RD Sharma Class 12 Chapter 5 Exercise 5.2 solutions book is a must.
Every answer here is provided and verified by experts. And the solutions follow the NCERT method, which gives a good reason for the CBSE students to adapt to it. An abundance of students has scored high marks in this chapter by preparing with the help of the Class 12 RD Sharma Class 12 Solutions Determinant Ex 5.2. Not only can students prepare for exams, but they can also use it to complete their daily homework and assignments.
For the welfare of the students, they can download the RD Sharma class 12th exercise 5.2 solution books for free of cost from the Career360 website. So they need not pay even a single penny for it. All you need is to visit the Career 360 website, search for the Class 12 RD Sharma Chapter 5 Exercise 5.2 Solution, and click the download button. It saves you hundreds to thousands of rupees that you would spend on other solution books.
If you strive to cross your benchmark score in the Determinants chapter, download the RD Sharma Class 12 Solutions Chapter 5 ex 5.2 and start preparing your exams with it. As the problems given in the RD Sharma books are solved in various methods, the students get the opportunity to decide which way they can follow. It clears the air for the people who are confused and also helps in rechecking their solutions.
Chapter-wise RD Sharma Class 12 Solutions
- Chapter 1 - Relations
- Chapter 2 - Functions
- Chapter 3 - Inverse Trigonometric Functions
- Chapter 4 - Algebra of Matrices
- Chapter 5 - Determinants
- Chapter 6 - Adjoint and Inverse of a Matrix
- Chapter 7 - Solution of Simultaneous Linear Equations
- Chapter 8 - Continuity
- Chapter 9 - Differentiability
- Chapter 10 - Differentiation
- Chapter 11 - Higher Order Derivatives
- Chapter 12 - Derivative as a Rate Measurer
- Chapter 13 - Differentials, Errors and Approximations
- Chapter 14 - Mean Value Theorems
- Chapter 15 - Tangents and Normals
- Chapter 16 - Increasing and Decreasing Functions
- Chapter 17 - Maxima and Minima
- Chapter 18 - Indefinite Integrals
- Chapter 19 - Definite Integrals
- Chapter 20 - Areas of Bounded Regions
- Chapter 21 - Differential Equations
- Chapter 22 - Algebra of Vectors
- Chapter 23 - Scalar Or Dot Product
- Chapter 24 - Vector or Cross Product
- Chapter 25 - Scalar Triple Product
- Chapter 26 - Direction Cosines and Direction Ratios
- Chapter 27 - Straight Line in Space
- Chapter 28 - The Plane
- Chapter 29 - Linear programming
- Chapter 30- Probability
- Chapter 31 - Mean and Variance of a Random Variable
Articles
Upcoming School Exams
Applications for Admissions are open.
Study at a world-renowned UK university in India | Admissions open for UG & PG programs.
Victoria University, Delhi NCR
ApplyApply for UG & PG programmes from Victoria University, Delhi NCR Campus
Illinois Tech Mumbai
ApplyAdmissions open for UG & PG programs at Illinois Tech Mumbai
University of Aberdeen Mumbai
ApplyApply for UG & PG courses at University of Aberdeen, Mumbai Campus
University of York, Mumbai
ApplyUG & PG Admissions open for CS/AI/Business/Economics & other programmes.
University of Bristol, Mumbai Enterprise Campus
ApplyBristol's expertise meets Mumbai's innovation. Admissions open for UG & PG programmes