Careers360 Logo
RD Sharma Solutions Class 12 Mathematics Chapter 5 FBQ

RD Sharma Solutions Class 12 Mathematics Chapter 5 FBQ

Edited By Kuldeep Maurya | Updated on Jan 25, 2022 06:54 PM IST

The RD Sharma class 12th exercise FBQ solution tops the list of NCERT solutions as per the opinion of students. Since CBSE schools require students to practice and master their NCERT book, they need to answer the given questions correctly and learn the concepts clearly. RD Sharma solutions Mathematics is a rather complex subject that can only be mastered with self-practice. Therefore, the RD Sharma class 12th exercise FBQ will be a good guide in your endeavor to score high in exams and ace the paper.

RD Sharma Class 12 Solutions Chapter 5 FBQ Determinants - Other Exercise

Determinants Excercise:FBQ

Determinants exercise Fill in the blanks question 1

Answer: \left | A \right |=6
Hint: Here, we use basic concept of diagonal and determinant of matrix
Given: \begin{bmatrix} 1 &0 &0 \\ 0&2 &0 \\ 0&0 & 3 \end{bmatrix}
Solution:
det(A)=\left | A \right |=\begin{bmatrix} 1 &0 &0 \\ 0&2 &0 \\ 0&0 & 3 \end{bmatrix}
=1\begin{bmatrix} 2 & 0\\ 0&3 \end{bmatrix}-0\begin{bmatrix} 0 &0 \\ 0&3 \end{bmatrix}+0\begin{bmatrix} 0 &0 \\ 2&0 \end{bmatrix}
=1(6-0)-0(0-0)+0(0-0)
=1(6)-0+0
\left | A \right |=6

Determinants exercise Fill in the blanks question 2

Answer:\left | A \right |=0
Hint: Here, we use basic concept of skew symmetric matrix and determinant of matrix
Given: A is 3\times 3 order skew symmetric matrix
Solution: Here, A_{\left [ 3\times 3 \right ]} and A=-A^{T} ( Skew Symmetric )
So, A=-A^{T}
\left | A \right |=-\left | A^{T} \right | [Taking determinant both side]
\left | A \right |=-\left | A \right |\; \; \; \; \; \; \; \; \left [ \left | A^{T} \right |=\left | A \right | \right ]
\left | A \right |+\left | A \right |=0
2\left | A \right |=0
\left | A \right |=0
So, here determinant of 3 3\times 3 skew symmetric matrix always be zero.

Determinants exercise Fill in the blanks question 3

Answer: x = 4
Hint: Here, we use basic concept of singular matrix
Given:A=\begin{bmatrix} 1 &3 &x+2 \\ 2& 4 & 8\\ 3& 5& 10 \end{bmatrix} is singular
Solution: Here, A is regular
So, \left | A \right | is zero
\left | A \right |=0
So,
\begin{bmatrix} 1 &3 &x+2 \\ 2& 4 & 8\\ 3& 5& 10 \end{bmatrix}=0
1\begin{bmatrix} 4 & 8\\ 5& 10 \end{bmatrix}-3\begin{bmatrix} 2 & 8\\ 3& 10 \end{bmatrix}+(x+2)\begin{bmatrix} 2 & 4\\ 3& 5 \end{bmatrix}=0
1(40-40)-3(20-24)+(x+2)(10-12)=0
0-3(-4)+(x+2)(-2)=0
12+\left [ -2x-4 \right ]=0
8-2x=0
8=2x
x=\frac{8}{2}=4

Determinants exercise Fill in the blanks question 4

Answer: K^{2}
Hint: Here, we use basic concept of singular matrix
Given: A and B non singular square matrix of order ‘n’
Such that A = KB
Solution: Here, A and B are non singular
And both are n order of singular matrix
Here
A=KB
\left | A \right |=\left | K \right |\left | B \right | (Let's take both side determinate)
\left | A \right |=K^{n}\left | B \right |
\frac{\left | A \right |}{\left |B \right |}=K^{n} [A and B are n\times n matrix]
So, \frac{\left | A \right |}{\left |B \right |}=K^{n}

Determinants exercise Fill in the blanks question 6

Answer:\left | 5A \right |=125
Hint: Here, we use basic concept \left | KA \right |=K^{n}\left | A \right |
Given: Here, A is 2\times 2 matrix
So, n = 2
\left | A \right |=5
Solution: Here, n=2 and \left | A \right |=5
So, \left |5 A \right |=?
So, here 5 is just like k
\left |K A \right |=K^{n}\left | A \right |
\left |5 A \right |=5^{2}\times 5\; \; \; \; \; \; \; \; \left [ K=5,\left | A \right |=5,n=2 \right ]
\left |5 A \right |=125
\left |5 A \right | is equal to 125.

Determinants exercise Fill in the blanks question 5

Answer: R-\left | 1 \right |
Hint: Here, we use basic concept of non-singular matrix
Given:A=\begin{bmatrix} a & 2\\ 2&4 \end{bmatrix} is non-singular
Solution: Here, A is non-singular
So, \left | A \right |\neq 0
\begin{bmatrix} a & 2\\ 2&4 \end{bmatrix}\neq 0
(a(4))-(2)(2)\neq 0
4a-4\neq 0
4a\neq 4
a\neq 1
So, a should be in R-\left | 1 \right |

Determinants exercise Fill in the blanks question 7

Answer: \left | 3AB \right |=-81
Hint: Here, we use basic concept of \left | KA \right |=K^{n}\left | A \right |
Given: A and B are square matrix of order 3. So, n = 3
\left | A \right |=(-1), \; \; \left | B \right |=3
Solution: Here, \left | 3AB \right |
Here 3 is just like K. So,
\left | KAB \right |=K^{n}\left | A \right |\left | B \right |
\left | 3AB \right |=3^{3}\times -1(3)\; \; \; \; \; \; \; \left [ n=3,\left | A \right | =-1,\left | B \right |=3\right ]
\left | 3AB \right |=-81
So, here \left | 3AB \right |=-81

Determinants exercise Fill in the blanks question 8

Answer: \partial =\pm 3
Hint: Here, we use basic concept of determinant of matrix
Given: \left | A \right |^{3}=125, A=\begin{bmatrix} \partial & 2\\ 2& \partial \end{bmatrix}
Solution: Here, A=\begin{bmatrix} \partial & 2\\ 2& \partial \end{bmatrix}
So, \left | A \right |=\partial \times \partial -2\times 2
\left | A \right |=\partial^{2} -4 ........(1)
And,\left | A \right |^{3}=125
\left ( \left | A \right | \right )^{3}=125
\left ( \left | A \right | \right )^{3}=(5)^{3}
\left | A \right |=5 ...........(2)
From equation (1) and equation (2)
\partial ^{2}-4=5
\partial ^{2}=4+5
\partial ^{2}=9
\partial=\pm 3

Determinants exercise Fill in the blanks question 9

Answer: x=e^{2}
Hint: Here, we use basic concept of determinant of matrix
Given: A=\left[\begin{array}{cc} \log x & -1 \\ -\log x & 2 \end{array}\right]
det (A) = 2
Solution: Here, A=\left[\begin{array}{cc} \log x & -1 \\ -\log x & 2 \end{array}\right]
\begin{aligned} &|A|=\log x \times 2-(-1 \times-\log x) \\ &=2 \log x-\log x \\ &|A|=\log x \end{aligned}
\begin{aligned} &\text { And }|A|=2 \text { (Given) }\\ &\text { So, }|A|=\log x=2 \end{aligned}
\begin{aligned} &e^{\log x}=e^{2} \\ &x=e^{2}\left[e^{\log x}=x\right] \end{aligned}

Determinants exercise Fill in the blanks question 10

Answer: 1
Hint: Here, we use basic concept of identity matrix and determinant of matrix
Given: Here I is order of 10 and identity matrix
Solution: Here, I is identity matrix and it’s order is 10.
Any order of identity matrix is determinant of matrix be 1
Let I is 2\times 2 matrix
\begin{aligned} &I=\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right] \\ &|I|=\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]=1-0=1 \\ &S o,\left|I_{10}\right|=1 \end{aligned}

Determinants exercise Fill in the blanks question 11

Answer: 8\left | A \right |
Hint: Here, we use basic concept of determinant of matrix \left | KA \right |=K^{n}\left | A \right |
Given: A is 3\times 3 matrix. So, n = 3
Solution: Here A is 3\times 3 matrix. So, n = 3
\left | -2A \right |=-2A is constant. So,K=(-2)
Let’s put value of K in below, Formula,
\begin{aligned} &|K A|=K^{n}|A| \\ &|-2 A|=(-2)^{n}|A| \end{aligned} \quad[K=-2]
\begin{aligned} &|-2 A|=(-2)^{3}|A| \quad[n=3] \\ &|-2 A|=-8|A| \end{aligned}
So, here8\left | A \right | is our answer.

Determinants exercise Fill in the blanks question 12

Answer: 5
Hint: Here, we use basic concept of determinant of matrix
Given: A is 3\times 3 matrix. So, n = 3
|A|=5 \; \; C_{i j}=\operatorname{cofactor} \text { of } a_{i j} . \text { So, let } C_{i j}=A_{i j}, C=A
Solution: A is \left | a_{ij} \right | be a square matrix
So, A=\left(\begin{array}{lll} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{array}\right) 1 \leq i \leq 3
\begin{aligned} &|A|=5 \\ &|A|=a_{11}\left[\begin{array}{cc} a_{22} & a_{23} \\ a_{32} & a_{33} \end{array}\right]-a_{12}\left[\begin{array}{cc} a_{21} & a_{23} \\ a_{31} & a_{33} \end{array}\right]+a_{13}\left[\begin{array}{cc} a_{21} & a_{22} \\ a_{31} & a_{32} \end{array}\right] \\ &|A|=a_{11}\left[\begin{array}{cc} a_{22} & a_{23} \\ a_{32} & a_{33} \end{array}\right]+a_{12}\left[\begin{array}{cc} a_{23} & a_{21} \\ a_{33} & a_{31} \end{array}\right]+a_{13}\left[\begin{array}{ll} a_{21} & a_{22} \\ a_{31} & a_{32} \end{array}\right] \end{aligned}
[Interchange column of a12]
\begin{aligned} &\Rightarrow|A|=a_{11} A_{11}+a_{12} A_{12}+a_{13} A_{13} \\ &5=a_{11} A_{11}+a_{12} A_{12}+a_{13} A_{13} \\ &5=a_{11} C_{11}+a_{12} C_{12}+a_{13} C_{13} \quad[A=C] \end{aligned}

Determinants exercise Fill in the blanks question 13

Answer: 0
Hint: Here we use basic concept of determinant of matrix
Given: A is 3\times 3 matrix. So, n = 3
\begin{aligned} &|A|=5\\ &C_{i y} \text { is cofactor of } C_{11 j} \end{aligned}
Solution:
\begin{aligned} &A_{i j}=\left(\begin{array}{lll} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{array}\right) \\ &\text { and cofactor }\left(A_{i j}\right)=(-1)^{i+j}|\operatorname{mij}| \end{aligned}
So, here
\begin{aligned} &a_{11} A_{21}+a_{12} A_{22}+a_{13} A_{23} \\ &A_{21}=(-1)^{2+1}\left[a_{12} a_{33}-a_{13} a_{32}\right] \\ &A_{22}=(-1)^{2+2}\left[a_{11} a_{33}-a_{13} a_{31}\right] \\ &A_{23}=(-1)^{2+3}\left[a_{11} a_{32}-a_{12} a_{31}\right] \end{aligned}
\begin{aligned} &=a_{11} A_{21}+a_{12} A_{22}+a_{13} A_{23} \\ &=\left(-a_{11} a_{12} a_{33}\right)+a_{11} a_{13} a_{32}+a_{11} a_{12} a_{33}-a_{12} a_{13} a_{31}-a_{12} a_{13} a_{32}+a_{12} a_{13} a_{31} \\ &=\left[a_{11} a_{12} a_{33}-a_{11} a_{12} a_{33}\right]+\left[a_{11} a_{13} a_{32}-a_{11} a_{13} a_{32}\right]+\left[a_{12} a_{13} a_{31}-a_{12} a_{13} a_{31}\right] \\ &\text { So } 0+0+0 \\ &=0 \end{aligned}

Determinants exercise Fill in the blanks question 14

Answer: 576
Hint: Here, we use basic concept of diagonal matrix and determinant of matrix
Given: A= diag(2,34)
Solution: diag(2,34)
\text { So, } A=\left[\begin{array}{lll} 2 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 4 \end{array}\right]
Let’s find determinant of A
\begin{aligned} &|A|=2\left[\begin{array}{ll} 3 & 0 \\ 0 & 4 \end{array}\right]-0\left[\begin{array}{ll} 0 & 4 \\ 0 & 0 \end{array}\right]+0\left[\begin{array}{ll} 0 & 3 \\ 0 & 0 \end{array}\right] \\ &=2(12-0)-0(0-0)+0(0-0) \\ &=2(12)-0+0 \\ &|A|=24 \end{aligned}
So,
\begin{aligned} &\left|A^{2}\right|=|A|^{2}=(24)^{2} \\ &\left|A^{2}\right|=576 \end{aligned}

Determinants exercise Fill in the blanks question 15

Answer: |C|=|A|^{2}=2^{2}=4
Hint: Here, we use basic concept of diagonal matrix and determinant of matrix
Given: \begin{aligned} &A=\left[c_{i j}\right]_{3 \times 3}|A|=2 \text { and } c_{i j} \text { be cofactor of } a_{i j}\\ &C=\left[c_{i j}\right] \end{aligned}
Solution: c is cofactor matrix
\begin{aligned} &A d j A=(C)^{T}\\ &\text { But Adj } A=\text { Det. of } C \end{aligned}
Because both are transpose of each other,
|\operatorname{adj} A|=|C|=|A|^{n-1}
Here n = 3 because A is 3\times 3 matrix,
\begin{aligned} &|C|=|A|^{3-1} \\ &|C|=|A|^{2} \end{aligned}
Here,
\begin{aligned} &|A|=2 \\ &|C|=(2)^{2}=4 \end{aligned}

Determinants exercise Fill in the blanks question 16

Answer: 0
Hint: Here, we use basic concept of determinant of matrix.
Given: \Delta=\left[\begin{array}{ccc} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 3 x & 6 x & 9 x \end{array}\right]
Solution:
\Delta=\left[\begin{array}{ccc} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 3 x & 6 x & 9 x \end{array}\right]
\begin{aligned} &=1\left[\begin{array}{cc} 5 & 6 \\ 6 x & 9 x \end{array}\right]-2\left[\begin{array}{cc} 4 & 6 \\ 3 x & 9 x \end{array}\right]+3\left[\begin{array}{cc} 4 & 5 \\ 3 x & 6 x \end{array}\right] \\ &=1(45 x-36 x)-2(36 x-18 x)+3(24 x-15 x) \\ &=9 x-36 x+27 x \\ &=0 \end{aligned}

Determinants exercise Fill in the blanks question 17

Answer: 0
Hint: Here, we use basic concept of determinant of matrix
Given: \Delta=\left[\begin{array}{ccc} \sec ^{2} \theta & \tan ^{2} \theta & 1 \\ \tan ^{2} \theta & \sec ^{2} \theta & -1 \\ 12 & 10 & 2 \end{array}\right]
Solution: We know that
\sec ^{2} \theta-\tan ^{2} \theta=1 ..............(1)
Applying C_{2}\rightarrow C_{2}-C_{1}
\begin{aligned} \Delta &=\left[\begin{array}{ccc} \sec ^{2} \theta & \sec ^{2} \theta-\tan ^{2} \theta & 1 \\ \tan ^{2} \theta & \tan ^{2} \theta-\sec ^{2} \theta & -1 \\ 12 & 12-10 & 2 \end{array}\right] \\ \Delta &=\left[\begin{array}{ccc} \sec ^{2} \theta & 1 & 1 \\ \tan ^{2} \theta & -1 & -1 \\ 12 & 2 & 2 \end{array}\right] \end{aligned}[From equation (1)]
\Delta =0 because according to the property of determinant, if any 2 columns are same then \Delta =0

Determinants exercise Fill in the blanks question 18

Answer: 0
Hint: Here, we use basic concept of determinant of matrix
Given: \Delta=\left[\begin{array}{ccc} 0 & x-y & y-z \\ y-x & 0 & z-x \\ z-y & x-z & 0 \end{array}\right]
Solution: Here \Delta is skew symmetric matrix.
Here, clearly we can see that \Delta is skew symmetric and its diagonal are 0
So, \Delta=0

Determinants exercise Fill in the blanks question 19

Answer: 120
Hint: Here, we use basic concept of determinant of matrix
Given:
\begin{aligned} &b_{i 1}=2 a_{i 1} \\ &b_{i 2}=3 a_{i 2} \\ &b_{i 3}=4 a_{i 3} i, i=1,2,3 \end{aligned}
Let \left | A \right |=\left [ a_{ij} \right ] and B=b_{ij} be a square matrix of order 3 and \left | A \right |=5
Solution:
\begin{aligned} &|B|=\left(\begin{array}{lll} b_{11} & b_{12} & b_{13} \\ b_{21} & b_{22} & b_{23} \\ b_{31} & b_{32} & b_{33} \end{array}\right) \\ &|B|=\left(\begin{array}{lll} 2 a_{11} & 3 a_{12} & 4 a_{13} \\ 2 a_{21} & 3 a_{22} & 4 a_{23} \\ 2 a_{31} & 3 a_{32} & 4 a_{33} \end{array}\right) \end{aligned}
Let’s transfer \left | B \right |=\left | B^{T} \right | So,
\begin{aligned} &|B|=\left(\begin{array}{ccc} 2 a_{11} & 2 a_{12} & 2 a_{13} \\ 3 a_{21} & 3 a_{22} & 3 a_{23} \\ 4 a_{31} & 4 a_{32} & 4 a_{33} \end{array}\right) \\ &|B|=2 \times 3 \times 4\left(\begin{array}{lll} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{array}\right) \end{aligned}
\begin{aligned} &=24 \times\left|A^{T}\right| \\ &=24 \times 5 \quad\left[|A|=\left|A^{T}\right|=5\right] \\ &=120 \end{aligned}

Determinants exercise Fill in the blanks question 20

Answer: -64
Hint: Here, we use basic concept of determinant
|K A B|=K^{n}|A||B|
Given:n = 3
\left | A \right |=-2
\left | B \right |=4
Solution: A and B are 3\times 3 order of matrix
So, n = 3
Here, \left | 2AB \right |\; \; \; \; \; \; \; So,K=2
So, now
|K A B|=K^{n}|A||B|
|2 A B|=2^{3} \times-2 \times 4 \; \; \; \; \; \; \; \; \; \quad[n=3,|A|=-2,|B|=4]
\begin{aligned} &|2 A B|=8 \times-2 \times 4 \\ &|2 A B|=8 \times-8 \\ &|2 A B|=-64 \end{aligned}

Determinants exercise Fill in the blanks question 21

Answer: y
Hint: Here, we use basic concept of determinant of matrix
Given: \left[\begin{array}{ccc} \cos (x+y) & -\sin (x+y) & \cos 2 y \\ \sin x & \cos x & \sin y \\ -\cos x & \sin x & \sin y \end{array}\right]
Solution: Formula expanding,
\begin{aligned} &{\left[\begin{array}{ccc} \cos x \cos y-\sin x \sin y & -\sin x \cos y+\cos x \sin y & \cos ^{2} y-\sin ^{2} y \\ \sin x & \cos x & \sin y \\ -\cos x & \sin x & \sin y \end{array}\right]} \\ &R_{1} \rightarrow R_{1}+R_{2}(\sin y)+R_{3}(\cos y) & \end{aligned}
\left[\begin{array}{ccc} \cos x \cos y-\sin x \sin y+\sin y \sin x-\cos y \cos x & -\sin x \cos y-\cos x \sin y+\sin y \cos x+\cos y \sin x & \cos ^{2} y-\sin ^{2} y+\sin ^{2} y-\cos ^{2} y \\ \sin x & \cos x & \sin y \\ -\cos x & \sin x & -\sin y \end{array}\right]\text { So }\left[\begin{array}{ccc} 0 & 0 & 0 \\ \sin x & \cos x & \sin x \\ -\cos x & \sin x & -\cos x \end{array}\right]=0
Here, we get answer is zero which is obtained by multiplying 2nd row with sin y and 3rd row with cosy.
So, Value of determinant depends on y.

Determinants exercise Fill in the blanks question 22

Answer: zero
Hint: Here, we use basic concept of determinant of matrix
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: Here, clearly 2 column has same value
According to property of determinant, if two column has same value then determinant of matrix should be zero.

Determinants exercise Fill in the blanks question 23

Answer: zero
Hint: Here, we use basic concept of determinant.
Given: \left[\begin{array}{lll} (1+x)^{17} & (1+x)^{19} & (1+x)^{23} \\ (1+x)^{23} & (1+x)^{24} & (1+x)^{34} \\ (1+x)^{41} & (1+x)^{43} & (1+x)^{47} \end{array}\right]
A+B x+c x^{2}+\ldots \ldots \ldots \text { then find } A=?
Solution: Let’s take something common from 3 rows
(1+x)^{17}\left(1+x^{23}\right)\left(1+x^{41}\right)\left[\begin{array}{ccc} 1 & (1+x)^{2} & (1+x)^{6} \\ 1 & (1+x)^{6} & (1+x)^{11} \\ 1 & (1+x)^{2} & (1+x)^{6} \end{array}\right]
Here, we clearly see that 1st and 3rd row has same value
So, f(x) = 0
So, here x = 0 and
\begin{aligned} &A+B x+C x^{2} \ldots \ldots \ldots .=0 \\ &S o, A=0 \end{aligned}

Determinants exercise Fill in the blanks question 24

Answer:\frac{1}{2}
Hint: Here, we use basic concept of determinant of matrix.
Given: \cos 2 \theta=0
\left[\begin{array}{ccc} 0 & \cos \theta & \sin \theta \\ \cos \theta & \sin \theta & 0 \\ \sin \theta & 0 & \cos \theta \end{array}\right]^{2}=?
Solution:
\cos 2 \theta=0
\begin{aligned} &\cos 2 \theta=\cos 90^{\circ} \quad\left(\cos 90^{\circ}=0\right) \\ &2 \theta=90^{\circ} \\ &\theta=45^{\circ} \end{aligned}
Let’s put \theta=45^{\circ} in matrix
\left[\begin{array}{ccc} 0 & \cos 45^{\circ} & \sin 45^{\circ} \\ \cos 45^{\circ} & \sin 45^{\circ} & 0 \\ \sin 45^{\circ} & 0 & \cos 45^{\circ} \end{array}\right]^{2}
\left[\begin{array}{ccc} 0 & \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} & 0 \\ \frac{1}{\sqrt{2}} & 0 & \frac{1}{\sqrt{2}} \end{array}\right]^{2} \quad\left[\sin 45^{\circ}=\frac{1}{\sqrt{2}}, \cos 45^{\circ}=\frac{1}{\sqrt{2}}\right]
Let’s take \frac{1}{\sqrt{2}} in common
=\frac{1}{\sqrt{2}} \times \frac{1}{\sqrt{2}} \times \frac{1}{\sqrt{2}}\left[\begin{array}{lll} 0 & 1 & 1 \\ 1 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]
\begin{aligned} &=\left[\frac{1}{2 \sqrt{2}}[-1 \times(1-0)+1(0-1)]\right]^{2} \\ &=\frac{1}{8} \times[-1-1]^{2} \\ &=\frac{1}{8} \times(-2)^{2} \\ &=\frac{1}{2} \end{aligned}

Determinants exercise Fill in the blanks question 25

Answer: 9
Hint: Here, we use basic concept of minors in matrix
Given: Here A is 3\times 3 matrix
Solution: In any matrix minors are equal to total number of elements
So, here total number of elements of A=3\times 3=9
So, total minors in A is 9

Determinants exercise Fill in the blanks question 26

Answer: 2,7
Hint: Here, we use basic concept of determinant of matrix
Given:x=-y and \left[\begin{array}{lll} x & 3 & 7 \\ 2 & x & 2 \\ 7 & 6 & x \end{array}\right]=0
Solution:
\left[\begin{array}{lll} x & 3 & 7 \\ 2 & x & 2 \\ 7 & 6 & x \end{array}\right]
\begin{aligned} &=x\left[\begin{array}{ll} x & 2 \\ 6 & x \end{array}\right]-3\left[\begin{array}{cc} 2 & 2 \\ 7 & x \end{array}\right]+7\left[\begin{array}{cc} 2 & x \\ 7 & 6 \end{array}\right] \\ &=x\left(x^{2}-12\right)-3(2 x-14)+7(12-7 x) \\ &f(x)=x^{3}-12 x-6 x+42+84-49 x=0 \\ &f(x)=x^{3}-67 x+126=0 \end{aligned}
Let put x = 2 and x = 7
\begin{aligned} &f(2)=(2)^{3}-67(2)+126 \\ &=8-134+126 \\ &=0 \\ &f(7)=(7)^{3}-67(7)+126 \\ &=49 \times 7-67+126 \\ &=0 \end{aligned}
So, 2 and 7 are other roots.

Determinants exercise Fill in the blanks question 27

Answer: (y-z)(z-x)[y-x+x y z]
Hint: Here, we use basic concept of determinant of matrix
Given: \left[\begin{array}{ccc} 0 & x y z & x-z \\ y-x & 0 & y-z \\ z-x & z-y & 0 \end{array}\right]
Solution:
0\left[\begin{array}{cc} 0 & y-z \\ z-y & 0 \end{array}\right]-x y z\left[\begin{array}{cc} y-x & y-z \\ z-x & 0 \end{array}\right]+x-z\left[\begin{array}{cc} y-x & 0 \\ z-x & z-y \end{array}\right]
\begin{aligned} &=0(0-(y-z)) \times(z-y)-x y z(0-(y-z)(z-x))+(x-z)((y-x)(z-y)-0) \\ &=x y z(y-z)(z-x)+(z-x)(y-z)(y-z) \\ &=(y-z)(z-x)[x y z+(y-x)] \\ &=(y-z)(z-x)[y-x+x y z] \end{aligned}

Determinants exercise Fill in the blanks question 28

Answer: 675
Hint: Here, we use basic concept of determinant of matrix |K A B|=K^{n}|A||B|
Given: A is 3\times 3 matrix. So, n = 3
|A|=5,|B|=5
Solution: |3 A B|
Here 3 is constant K. So,
\begin{aligned} &|K A B|=K^{n}|A||B| \\ &|3 A B|=3^{n}|A||B| \quad(K=3) \\ &|3 A B|=3^{3} \times 5 \times 5[n=3,|A|=5,|B|=5] \\ &=27 \times 25 \\ &|3 A B|=675 \end{aligned}

Determinants exercise Fill in the blanks question 29

Answer: AP
Hint: Here, we use basic concept of determinant of matrix.
Given:\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]=0
Solution:
So, Let’s
\begin{aligned} &R_{2} \rightarrow R_{2}-R_{1} \\ &R_{3} \rightarrow R_{3}-R_{1} \\ &{\left[\begin{array}{ccc} x+1 & x+2 & x+a \\ 1 & 1 & b-a \\ 2 & 2 & c-a \end{array}\right]=0} \end{aligned}
Let’s perform differentiate
\begin{aligned} &=x+a[0+0]-1[(0-a)-2(b-a)]+(x+a) 0=0 \\ &c+a-2 b=0 \\ &2 b=a+c \end{aligned}
So a,b,c are in A.P

Determinants exercise Fill in the blanks question 30

Answer: Zero
Hint: Here, we use basic concept of determinant of matrix.
Given:\left[\begin{array}{ccc} \sin A & \cos A & \sin A+\cos B \\ \sin B & \cos A & \sin B+\cos B \\ \sin C & \cos A & \sin C+\cos B \end{array}\right]=0
Solution: Let’s take sin A, sin B and sin C common from respectively 1st , 2nd and 3rd row
\sin A \sin B \sin C\left[\begin{array}{lll} 1 & \cos A & 1+\cos B \\ 1 & \cos A & 1+\cos B \\ 1 & \cos A & 1+\cos B \end{array}\right]
Here, all rows are same
According to properties of determinants, if there are 2 or more rows are same then, value of determinants is zero.
So, Answer is zero.

Determinants exercise Fill in the blanks question 31

Answer: 8
Hint: Here, we use basic concept of determinant of matrix.
Given: \left[\begin{array}{lll} x+a & p+u & l+f \\ y+b & q+v & m+g \\ z+c & r+w & n+h \end{array}\right] K=?
Solution: Here, we use the property of determinant if elements of row and column of determinant are expressed as sum of two (or more) terms, then it can be expressed as sum of two ( or more) determinant.
Let’s take \Delta of determinant
\Delta=\left[\begin{array}{ccc} x+a & p+u & l+f \\ y+b & q+v & m+g \\ z+c & r+w & n+h \end{array}\right]
Let’s use property for C1
\begin{aligned} &\Delta=\left[\begin{array}{ccc} x & p+u & l+f \\ y & q+v & m+g \\ z & r+w & n+h \end{array}\right] \\ &\Delta=\left[\begin{array}{ccc} x & p+u & l+f \\ y & q+v & m+g \\ z & r+w & n+h \end{array}\right]+\left[\begin{array}{lll} a & p+u & l+f \\ b & q+v & m+g \\ c & r+w & n+h \end{array}\right] \end{aligned}
Property using for C2
\Delta=\left[\begin{array}{ccc} x & p+u & l+f \\ y & q+v & m+g \\ z & r+w & n+h \end{array}\right]+\left[\begin{array}{ccc} x & p+u & l+f \\ y & q+v & m+g \\ z & r+w & n+h \end{array}\right]+\left[\begin{array}{ccc} a & p+u & l+f \\ b & q+v & m+g \\ c & r+w & n+h \end{array}\right]+\left[\begin{array}{ccc} a & p+u & l+f \\ b & q+v & m+g \\ c & r+w & n+h \end{array}\right]
Property using for C3
\Delta=\left[\begin{array}{ccc} x & p+u & l+f \\ y & q+v & m+g \\ z & r+w & n+h \end{array}\right]+\left[\begin{array}{ccc} x & p+u & l+f \\ y & q+v & m+g \\ z & r+w & n+h \end{array}\right]+\left[\begin{array}{ccc} x & p+u & l+f \\ y & q+v & m+g \\ z & r+w & n+h \end{array}\right]
+\left[\begin{array}{ccc} x & p+u & l+f \\ y & q+v & m+g \\ z & r+w & n+h \end{array}\right]+\left[\begin{array}{ccc} x & p+u & l+f \\ y & q+v & m+g \\ z & r+w & n+h \end{array}\right]+\left[\begin{array}{ccc} x & p+u & l+f \\ y & q+v & m+g \\ z & r+w & n+h \end{array}\right]
+\left[\begin{array}{ccc} a & p+u & l+f \\ b & q+v & m+g \\ c & r+w & n+h \end{array}\right]+\left[\begin{array}{ccc} a & p+u & l+f \\ b & q+v & m+g \\ c & r+w & n+h \end{array}\right]+\left[\begin{array}{ccc} a & p+u & l+f \\ b & q+v & m+g \\ c & r+w & n+h \end{array}\right]
So, here we split in 8 parts.
So, k = 8

Determinants exercise Fill in the blanks question 32

Answer: \frac{1}{2}
Hint: Here we use basic concept of determinant of matrix.
Given: \left[\begin{array}{ccc} 1 & 1 & 1 \\ 1 & 1+\sin \theta & 1 \\ 1 & 1 & 1+\cot \theta \end{array}\right]
Solution:
\left[\begin{array}{ccc} 1 & 1 & 1 \\ 1 & 1+\sin \theta & 1 \\ 1 & 1 & 1+\cot \theta \end{array}\right]
\begin{aligned} &C_{2} \rightarrow C_{2}-C_{1}, C_{3} \rightarrow C_{3}-1 \\ &\Delta=\left[\begin{array}{ccc} 1 & 0 & 0 \\ 1 & \sin \theta & 0 \\ 1 & 0 & \cot \theta \end{array}\right] \\ &1\left[\begin{array}{cc} \sin \theta & 0 \\ 0 & \cot \theta \end{array}\right]-0\left[\begin{array}{cc} 1 & 0 \\ 1 & \cot \theta \end{array}\right]+0\left[\begin{array}{cc} 1 & \sin \theta \\ 1 & 0 \end{array}\right] \end{aligned}
\begin{aligned} &1(\sin \theta \times \cot \theta)-0(0)+0(0) \\ &\sin \theta \times \cot \theta-0+0 \\ &\sin \theta \times \frac{\cos \theta}{\sin \theta}=\cos \theta \\ &\Delta=\cos \theta \end{aligned}
Let's\Delta =0
\cos \; \theta =0=\frac{\pi }{2} \left [ So,maximum\; value\; is\; \frac{1}{2}\; of \; \cos \theta \right ]

Determinants exercise Fill in the blanks question 34

Answer: 0
Hint: Here, we use basic concept of determinant of matrix
Given:D=\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: Let’s use separation property of determinant
\begin{aligned} &{\left[\begin{array}{ccc} \sqrt{23} & \sqrt{5} & \sqrt{5} \\ \sqrt{15} & 5 & \sqrt{10} \\ 3 & \sqrt{15} & 5 \end{array}\right]+\left[\begin{array}{ccc} \sqrt{3} & \sqrt{5} & \sqrt{5} \\ \sqrt{15} & 5 & \sqrt{10} \\ 3 & \sqrt{15} & 5 \end{array}\right]} \\ &{\left[\begin{array}{lll} \sqrt{23} & \sqrt{5} & \sqrt{5} \\ \sqrt{2 \times 23} & \sqrt{5 \times 5} & \sqrt{2 \times 5} \\ \sqrt{5 \times 23} & \sqrt{5 \times 3} & \sqrt{5 \times 5} \end{array}\right]+\left[\begin{array}{ccc} \sqrt{3} & \sqrt{5} & \sqrt{5} \\ \sqrt{3 \times 5} & \sqrt{5 \times 5} & \sqrt{2 \times 5} \\ \sqrt{3 \times 3} & \sqrt{5 \times 3} & \sqrt{5 \times 5} \end{array}\right]} \end{aligned}
\begin{aligned} C_{1} & \rightarrow \frac{C_{1}}{\sqrt{3}} \\ C_{2} & \rightarrow \frac{C_{2}}{\sqrt{5}} \text { and } C_{3} \rightarrow \frac{C_{3}}{\sqrt{5}} \end{aligned}
\sqrt{23} \sqrt{5} \sqrt{5}\left[\begin{array}{ccc} 1 & 1 & 1 \\ \sqrt{2} & \sqrt{5} & \sqrt{2} \\ \sqrt{5} & \sqrt{3} & \sqrt{5} \end{array}\right]+\sqrt{3} \sqrt{5} \sqrt{5}\left[\begin{array}{ccc} 1 & 1 & 1 \\ \sqrt{5} & \sqrt{5} & \sqrt{2} \\ \sqrt{3} & \sqrt{3} & \sqrt{5} \end{array}\right]
C_{1} \rightarrow C_{1}-C_{3} \text { and } C_{1} \rightarrow C_{1}-C_{2}
5 \sqrt{23}\left[\begin{array}{ccc} 0 & 1 & 1 \\ 0 & \sqrt{5} & \sqrt{2} \\ 0 & \sqrt{3} & \sqrt{5} \end{array}\right]+5 \sqrt{23}\left[\begin{array}{ccc} 0 & 1 & 1 \\ 0 & \sqrt{5} & \sqrt{2} \\ 0 & \sqrt{3} & \sqrt{5} \end{array}\right]
Here, both column has 0 value of each column
So, determinant value is zero.

Determinants exercise Fill in the blanks question 35

Answer: 0
Hint: Here, we use basic concept of determinant of matrix
Given: \left[\begin{array}{ccc} \sin ^{2} 23^{\circ} & \sin ^{2} 67^{\circ} & \cos 180^{\circ} \\ -\sin ^{2} 67^{\circ} & -\sin ^{2} 23^{\circ} & \cos ^{2} 180^{\circ} \\ -1 & \sin ^{2} 23^{\circ} & \sin ^{2} 67^{\circ} \end{array}\right]
Solution:
\left[\begin{array}{ccc} \sin ^{2} 23^{\circ} & \sin ^{2} 67^{\circ} & \cos 180^{\circ} \\ -\sin ^{2} 67^{\circ} & -\sin ^{2} 23^{\circ} & \cos ^{2} 180^{\circ} \\ -1 & \sin ^{2} 23^{\circ} & \sin ^{2} 67^{\circ} \end{array}\right]
\begin{aligned} &{\left[\begin{array}{ccc} \sin ^{2} 23^{\circ} & \sin ^{2}\left(90^{\circ}-23^{\circ}\right) & -1 \\ -\sin ^{2}\left(90^{\circ}-23^{\circ}\right) & -\sin ^{2} 23^{\circ} & 1 \\ -1 & \sin ^{2} 23^{\circ} & \cos ^{2}\left(90^{\circ}-23^{\circ}\right) \end{array}\right]} \\ &{\left[\begin{array}{ccc} \sin ^{2} 23^{\circ} & \cos ^{2} 23 & -1 \\ -\cos ^{2} 23 & -\sin ^{2} 23^{\circ} & 1 \\ -1 & \sin ^{2} 23^{\circ} & \cos ^{2} 23^{\circ} \end{array}\right]} \end{aligned}
Let's apply C_{1} \rightarrow C_{1}+C_{2}
\begin{aligned} &{\left[\begin{array}{ccc} \sin ^{2} 23^{\circ}+\cos ^{2} 23^{\circ} & \cos ^{2} 23 & -1 \\ -\cos ^{2} 23-\sin ^{2} 23^{\circ} & -\sin ^{2} 23^{\circ} & 1 \\ -1+\sin ^{2} 23^{\circ} & \sin ^{2} 23^{\circ} & \cos ^{2} 23^{\circ} \end{array}\right]} \\ &{\left[\begin{array}{ccc} 1 & 1 & -1 \\ -1 & -\sin ^{2} 23^{\circ} & 1 \\ -\cos ^{2} 23^{\circ} & \sin ^{2} 23^{\circ} & \cos ^{2} 23^{\circ} \end{array}\right]} \\ &(-1)\left[\begin{array}{ccc} 1 & 1 & -1 \\ -1 & -\sin ^{2} 23^{\circ} & 1 \\ -\cos ^{2} 23^{\circ} & \sin ^{2} 23^{\circ} & \cos ^{2} 23^{\circ} \end{array}\right] \end{aligned}
So, here two columns become same. So, determinant value becomes zero.
So, our answer is zero.

Determinants exercise Fill in the blanks question 36

Answer: 0,-6
Hint: Here we use basic determinant
Given: \left[\begin{array}{ccc} 2-x & 2+x & 2+x \\ 2+x & 2-x & 2+x \\ 2+x & 2+x & 2-x \end{array}\right]=0
Solution: Here, firstly
2-x\left[\begin{array}{ll} 2-x & 2+x \\ 2+x & 2-x \end{array}\right]-2+x\left[\begin{array}{cc} 2+x & 2+x \\ 2+x & 2-x \end{array}\right]+2+x\left[\begin{array}{cc} 2+x & 2-x \\ 2+x & 2+x \end{array}\right]
\begin{aligned} &0=(2-x)\left[(2-x)^{2}-(2+x)^{2}\right]-[2+x]\left[2^{2}-x^{2}[2+x]^{2}\right]+ &(2+x)\left[(2+x)^{2}-\left[\left(2^{2}-x^{2}\right)\right]\right. \end{aligned}
\begin{aligned} &0=8 x^{2}+16 x+8 x+8 x^{2}+8 x^{2}+4 x^{3}\end{aligned}
\begin{aligned} &0=8 x^{2}+8 x^{2}+8 x^{2}+4 x^{3} \end{aligned}
\begin{aligned} &0=24 x^{2}+4 x^{3} \end{aligned}
Let’s take 4x2 common
\begin{aligned} &4 x^{2}(6+x)=0 \\ &4 x^{2}=0 \text { and } 6+x=0 \\ &x^{2}=0 \text { and } x=-6 \end{aligned}

Determinants exercise Fill in the blanks question 37

Answer: -4
Hint: Here, we use basic component of determinant
Given: A is square matrix of order 3
So, n = 3
\left | A \right |=2
Solution:
Here, \left |-A A^{T} \right |
So, \left | -A \right |\times \left |A^{T} \right |
Here, \left |A^{T} \right |=\left | A \right | because both are transpose to each other
So, \left | -A \right |\times \left | A \right |
Let’s take -1 common
\begin{aligned} &-1([A]) \times|A| \\ &-1(2) \times 2[|A|=2] \\ &S o,\left|-A A^{T}\right|=-4 \end{aligned}

Determinants exercise Fill in the blanks question 33

Answer: zero
Hint: Here, we use basic concept of determinant of matrix
Given: D=\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: Let’s perform some row operations
\begin{aligned} &R_{1} \rightarrow R_{1}-R_{3} \\ &R_{2} \rightarrow R_{2}-R_{3} \\ &{\left[\begin{array}{ccc} \sin ^{2} A-\sin ^{2} C & \cot A-\cot c & 0 \\ \sin ^{2} B-\sin ^{2} C & \cot B-\cot c & 0 \\ \sin ^{2} C & \cot C & 1 \end{array}\right]} \end{aligned}
=\sin (A-C) \times \sin (B-C)\left[\begin{array}{ccc} \sin B & \frac{1}{\sin A \sin C} & 0 \\ \sin A & \frac{1}{\sin B \sin C} & 0 \\ \sin ^{2} C & \cot c & 1 \end{array}\right]
\begin{aligned} &=\sin (A-C) \sin (B-C)\left[\frac{1}{\sin C}-\frac{1}{\sin C}\right] \\ &=\sin (A-C) \sin (B-C) \times 0 \\ &=0 \end{aligned}

The class 12 RD Sharma chapter 5 exercise FBQ solution is highly trusted by students and teachers in the entire country. The answers provided in the RD Sharma class 12th exercise FBQ book are expert-created, making them accurate and easy to understand. You will also find some exceptional tips in the book that might not be taught in class.

The class 12 RD Sharma chapter 5 exercise FBQ solution includes the chapter Determinants, which explores linear equations that need to be solved using determinants. The FBQ section is specifically essential because it contains questions from the entire chapter like symmetric matrix, skew-symmetric matrix, singular matrix, order of a matrix, minor, cofactor, and determinant value. Thus, there are 38 questions that students must answer.

The RD Sharma class 12 solutions Determinants ex FBQ can be an excellent guide for students due to the following reasons:

  • The RD Sharma Class 12 Chapter 5 Exercise FBQ solutions are trusted by thousands of students in India.

  • You might find common questions in your board paper if you practice the RD Sharma solutions thoroughly and diligently.

  • RD Sharma class 12 solutions Determinants ex FBQ has the latest and updated syllabus, which is created to correspond to the NCERT textbooks used by teachers.

  • Teachers will often use the RD Sharma class 12 solutions chapter 5 ex FBQ to give homework and test students' progress. This means using the RD Sharma solutions will help you to solve your homework no matter how difficult it is.

  • RD Sharma class 12 solutions Determinants ex FBQ can be used by students for self-practice. They can use these solutions to compare answers and mark themselves at home.

  • RD Sharma class 12th exercise FBQ Solutions can be downloaded from the Career360 website, for all subjects and classes.

  • RD Sharma class 12 solutions Determinants ex FBQ book can be downloaded free of cost from Career360. So you won't need to pay anything to access or download the pdf.

RD Sharma Chapter wise Solutions


Frequently Asked Questions (FAQs)

1. Where can I download the class 12 RD Sharma chapter 5 exercise FBQ solution?

Students will be able to download the RD Sharma solutions from the Career360 website. They are available for free. 

2. Can I use RD Sharma class 12th exercise FBQ for JEE Mains preparations?

Yes, students can use RD Sharma class 12th exercise FBQ for their JEE mains preparations. The answers will help them test their knowledge and enhance their skills.

3. Does RD Sharma class 12 solutions chapter 5 ex FBQ has the latest syllabus?

RD Sharma class 12 solutions chapter 5 ex FBQ is regularly updated to correspond to the syllabus of NCERT Books. You can download the latest version of the pdf by entering your school year. 


4. Can I use RD Sharma class 12 chapter 5 exercise FBQ for solving homework?

Teachers like to test students' progress by assigning homework from RD Sharma class 12 chapter 5 exercise FBQ. Therefore,  you can use these solutions to solve your homework. 

5. How can I use class 12 RD Sharma chapter 5 exercise FBQ solution for self-practice?

You can solve questions from your NCERT textbooks and compare answers with class 12 RD Sharma chapter 5 exercise FBQ solution to mark yourself.

Articles

Upcoming School Exams

Application Date:09 September,2024 - 14 November,2024

Application Date:09 September,2024 - 14 November,2024

Application Date:17 September,2024 - 14 October,2024

Application Date:17 September,2024 - 14 October,2024

View All School Exams
Get answers from students and experts
Back to top