RD Sharma Class 12 Exercise 5.2 Determinant Solutions Maths - Download PDF Free Online

RD Sharma Class 12 Exercise 5.2 Determinant Solutions Maths - Download PDF Free Online

Edited By Kuldeep Maurya | Updated on Jan 25, 2022 07:12 PM IST

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

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|=0where A,B,C are the angles of ΔABC
Hint: 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)fromC_{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}


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: Proved
Hint 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: 0
Hint 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}

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Chapter-wise RD Sharma Class 12 Solutions

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