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RD Sharma Solutions Class 12 Mathematics Chapter 6 FBQ

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RD Sharma Solutions Class 12 Mathematics Chapter 6 FBQ

RD Sharma Solutions Class 12 Mathematics Chapter 6 FBQ

Updated on 20 Jan 2022, 02:41 PM IST

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Also Read - RD Sharma Solution for Class 9 to 12 Maths

RD Sharma Class 12 Solutions Chapter 6 FBQ Adjoint & Inverse of Matrix - Other Exercise

Adjoint and Inverese of Matrices Exercise Fill in the blanks Question 1

Answer $\rightarrow I_{n}$
Hint $\rightarrow A^{-1}=adj A/\left | A \right |$
Given $\rightarrow A$ is a unit matrix of order $n$ then $A\left ( adj A \right )=$
Explanation: A is a unit matrix of order $n \Rightarrow \left | A \right |=1$
Now,
$\begin{aligned} &A^{-1}=(\operatorname{adj} A) /|A| \\\\ &A A^{-1}=A(\operatorname{adj} A) /|A| \end{aligned}$
$I_{n \times n}=A(\operatorname{adj} A) /|A|$ $\left[_{\mathrm{as}}|A|=1\right]$
$A(\operatorname{adj} A)=I_{n \times n}$.
Hence,
$A(\operatorname{adj} A)=I_{n \times n}$
Remarks: $A(\operatorname{adj} A)=I_{n \times n}$ where $A$ is a unit matrix of order $n$.

Adjoint and Inverese of Matrices Exercise Fill in the blanks Question 2

Answer$\rightarrow A^{2}$
Hint$\rightarrow$ Determinant of non-singular matrix is non-zero
Given$\rightarrow A$ is a non-singular matrix such that $A^{3}=I$
Explanation$\rightarrow A$ is a non-singular matrix, therefore $\left | A \right |\neq 0$
$\begin{aligned} &A^{3}=I \\ &A \cdot A^{2}=I \end{aligned}$
$A^{-1} A \cdot A^{2}=I \cdot A^{-1} \quad\left[A^{-1} A=A A^{-1}=I\right]$
Hence, $\begin{aligned} A^{2} &=A^{-1} \\ A^{-1} &=A^{2} \end{aligned}$

Adjoint and Inverse of Matrix Excercise Fill in the blanks Question 3

Answer →
Given$\rightarrow A$ and $B$ are square matrices of the same order and
Explanation$\rightarrow AB=3I$
$\begin{aligned} &A^{-1} A B=A^{-1} 3 I \qquad\left[A^{-1} A=A A^{-1}=I\right] \\\\ &I B=3 A^{-1} I \qquad\left[B I=I=I B^{2}\right] \\\\ &B=3 A^{-1} \end{aligned}$
Hence, $A^{-1}=\frac{1}{3}B$

Adjoint and Inverse of Matrix Excercise Fill in the blanks Question 4

Answer
Hint $\rightarrow A$ matrix is non-invertible if and only if $\left | A \right |=0$
Given $\rightarrow A=\begin{bmatrix} 1 & a & 2\\ 1 & 2 & 5\\ 2 & 1 & 1 \end{bmatrix}$is not invertible
Explanation $\rightarrow \left | A \right |=\begin{bmatrix} 1 & a & 2\\ 1 & 2 & 5\\ 2 & 1 & 1 \end{bmatrix}=0$
Expanding along first row
$\begin{aligned} &1(2-5)-a(1-10)+2(1-4)=0 \\\\ &-3+9 a-6=0 \\\\ &9 a=9 \\\\ &a=1 \end{aligned}$
Hence, $a=1$

Adjoint and Inverse Matrix Exercise Fill in the blanks Question 5

Answer$\rightarrow$ zero matrix
Hint $\rightarrow$ Determinant of singular matrix is $0$.
Given$\rightarrow A$ is a singular matrix
Explanation $\rightarrow A^{-1}=\operatorname{adj} A /|A|$
$A(\operatorname{adj} A)=|A| I$ [As $A$ is a singular matrix $\therefore \left | A \right |=0$ ]
$\begin{aligned} &=0 \cdot I \\ &=0 \end{aligned}$
Hence, $A\left ( adj A \right )$ is a null matrix.

Adjoint and Inverse Matrix Exercise Fill in the blanks Question 6

Answer $\rightarrow \left ( 11 \right )^{4}=14,641$
Hint $\rightarrow$ Determinant of cofactors of elements of $A$ is equal to the determinant of adjoint $A$
Given$\rightarrow$ $A$ is a matrix of order $3$ and $\left | A \right |=11$
Explanation $\begin{aligned} &\rightarrow|A|=A(\operatorname{adj} A) \\\\ &|A|=A(C)^{T} \end{aligned}$

Where $C$ represents the cofactor of elements of $A$
$\begin{aligned} &|C|^{T}|A|=\operatorname{det}(|A| I) \\\\ &\operatorname{det}\left(C^{T}\right)=|A|^{3}=|A|^{3-1} \\ \\&\operatorname{det}\left(C^{T}\right)=|A|^{2} \\ & \end{aligned}$
$\! \! \! \! \! \! \! |B|=|A|^{2} \\\\ =(11)^{2} \\\\ =121$
Hence, $\left | B \right |^{2}=\left ( 121 \right )^{2}$
$=14641$

Adjoint and Inverse of a Matrix exercise Fill in the blanks question 7

Answer $\rightarrow 100$
Hint $\rightarrow A(\operatorname{adj} A)=|A| I$
Given $\rightarrow A$ is a square matrix of order $2$ such that $A(\operatorname{adj} A)=\left[\begin{array}{cc} 10 & 0 \\ 0 & 10 \end{array}\right]$
Explanation
$\begin{aligned} &\rightarrow A(\operatorname{adj} A)=|A| I \\ &=\left|\begin{array}{cc} 10 & 0 \\ 0 & 10 \end{array}\right| \\ &=100-0 \\ &=100 \end{aligned}$

Adjoint and Inverse of a Matrix exercise Fill in the blanks question 8

Answer$\rightarrow 9$
Hint $\rightarrow A(\operatorname{adj} A)=\operatorname{det}(A) I_{n \times n}$
Given$\rightarrow A$ is an invertible matrix of order $3$ and $\left | A \right |=3$
Explanation$\rightarrow(\operatorname{adj} A)=|A|^{n-1}$
$\begin{aligned} &=|A|^{3-1} \quad[n=3] \\ &=|A|^{2} \\ &|\operatorname{adj} A|=(3)^{2} \\ &=9 \end{aligned}$

Adjoint and Inverse of a Matrix Exercise fill in the blaks Question 9

Answer $\rightarrow 5A$
Hint $\rightarrow \operatorname{adj}(\operatorname{adj} A)=(\operatorname{det} A)^{n-2} \cdot A$
Given $\rightarrow A$is an invertible matrix of order $3$ and $\left | A \right |=5$
Explanation $\rightarrow A^{-1}=\frac{\operatorname{adj} A}{\operatorname{det}(A)}$ …(i)
$(\operatorname{adj} A)^{-1}=\left(\operatorname{det}(A) A^{-1}\right)^{-1}=A / \operatorname{det}(A)$ …(ii)
In eqn (i) replace A by adj A
$\begin{aligned} &(\operatorname{adj}(A))^{-1}=\frac{\operatorname{adj}(\operatorname{adj}(A))}{\operatorname{det}(\operatorname{adj} A)} \\ &\operatorname{adj}(\operatorname{adj} A)=\operatorname{det}(\operatorname{adj} A)(\operatorname{adj} A)^{-1} \end{aligned}$
$=(\operatorname{det}(A))^{n-1} \times \frac{A}{\operatorname{det}(A)}$ [from (ii)]
$=(\operatorname{det} A)^{n-2} A$
$\operatorname{adj}(\operatorname{adj} A)=|A| A=5 A$ $[n=3]$

Adjoint and Inverse of a Matrix Exercise fill in the blaks Question 10

Answer $\rightarrow 256$
Hint $\rightarrow|\operatorname{adj}(\operatorname{adj} A)|=|A|^{(n-1)^{2}}$
Given$\rightarrow A$ is an invertible matrix of order $3$ and $\left | A \right |=4$
Explanation $\rightarrow A(\operatorname{adj} A)=|A| I$
$|A(\operatorname{adjA})|=\| A|I|$
$|A||\operatorname{adj} A|=|A|^{n}$ $\left[|| A|I|=|A|^{n} \quad \text { and }|A B|=|A||B|\right]$
$\begin{aligned} &|\operatorname{adj} A|=|A|^{n-1} \\\\ &|\operatorname{adj}(\operatorname{adj} A)|=|\operatorname{adj} A|^{n-1} \end{aligned}$
$\begin{aligned} &\left.=| A\right|^{n-1)(n-1)} \\\\ &=|A|^{(n-1)^{2}} \\ \\&=(4)^{(3-1)^{2}} \\\\ &=(4)^{4} \\\\ &=256 \end{aligned}$

Adjoint and Inverse of Matrix Excercise Fill in the Blanks Question 11

Answer$\rightarrow 6^{4}=1296$
Hint$\rightarrow$ In diagonal matrix all elements are zero except diagonal elements and its determinant is equal to product of diagonal elements.
Given $\rightarrow A=\operatorname{diag}(1,2,3)$
Explanation $\rightarrow A(\operatorname{adj} A)=|A| I$
$\begin{aligned} &|\operatorname{adj}(\operatorname{adj} A)|=|A|^{n-1)^{2}} \\\\ &=(6)^{(3-1)^{2}} \mid \\\\ &=(6)^{4} \\\\ &=1296 \end{aligned}$
Hence, $|\operatorname{adj}(\operatorname{adj} A)|=1296$

Adjoint and Inverse of Matrix Excercise Fill in the Blanks Question 12

Answer $\rightarrow 2 / 5$
Hint$\rightarrow \operatorname{det}\left(A^{-1}\right)=1 / \operatorname{det}(A)$
Given$\rightarrow A$ is a square matrix of order $3$ such that $|A|=5 / 2$
Explanation $\rightarrow\left|A A^{-1}\right|=1$
$\begin{aligned} &|A|\left|A^{-1}\right|=1 \\\\ &\left|A^{-1}\right|=\frac{1}{|A|}=\frac{1}{\frac{5}{2}} \\\\ &\left|A^{-1}\right|=2 / 5 \end{aligned}$

Adjoint and Inverse Matrix Exercise Fill in the blanks Question 13

Answer $\rightarrow 10$
Hint $\rightarrow A\left ( adjA \right )=\left | A \right |. I$
Given$\rightarrow A$ is a square matrix such that $\rightarrow A\left ( adjA \right )=10.I$
Explanation $\rightarrow A^{-1}=\frac{1}{|A|}(\operatorname{adj} A)$
$|A|=A(\operatorname{adj} A)$
Hence,$A=10$

Adjoint and Inverse Matrix Exercise Fill in the blanks Question 14

Answer $\rightarrow 25$
Hint $\rightarrow|\operatorname{adj} A|=|A|^{n-1}$
Given$\rightarrow A$ is a square matrix of order $3$ and $B=|A| A^{-1}$ and $|A|=5$
Explanation $\rightarrow A(\operatorname{adj} A)=|A| I$
$\operatorname{adj} A=|A| A^{-1}$
$B=\operatorname{adj} A=|A| A^{-1}$ …(given)
$\begin{aligned} &|B|=|\operatorname{adj} A| \\\\ &=|A|^{n-1} \\\\ &=(5)^{2}=25 \end{aligned}$
Hence, $\left | B \right |=25$

Adjoint and Inverse of a Matrix exercise Fill in the blanks question 15

Answer $\rightarrow k^{2} I$
Hint $\rightarrow \operatorname{adj}(k A)=k^{n-1}(\operatorname{adj} A)$
Given $\rightarrow k$ is a scalar and $I$ is a unit matrix of order $3$
Explanation $\rightarrow \operatorname{adj}(k I)=k^{n-1}(a d j I)$
$\begin{aligned} &\operatorname{adj}(k I)=k^{3-1} a d j(I) \\\\ &=k^{2} \cdot I \end{aligned}$
Hence, $\operatorname{adj}(k I)=k^{2} \cdot I$

Adjoint and Inverse of a Matrix exercise Fill in the blanks question 16

Answer $\rightarrow 1$
Hint $\rightarrow A(\operatorname{adj} A)=|A| I$
Given $\rightarrow A=\left|\begin{array}{cc} \cos x & \sin x \\ -\sin x & \cos x \end{array}\right|$ and $A(\operatorname{adj} A)=\left[\begin{array}{ll} k & 0 \\ 0 & k \end{array}\right]$
Explanation $\rightarrow|A|=\left|\begin{array}{cc} \cos x & \sin x \\ -\sin x & \cos x \end{array}\right|$
$\begin{aligned} &=\cos ^{2} x+\sin ^{2} x \\ &=1 \end{aligned}$
Since we have $\rightarrow A(\operatorname{adj} A)=|A| I$
Applying determinant on both side
$|A(\operatorname{adj} A)|=|| A|I|$ …(i)
Also we have given
$\left | A\left ( adj A \right ) \right |= \begin{bmatrix} k &0 \\ 0 & k \end{bmatrix}= k^{2}\begin{vmatrix} 1 & 0 & 0 &1 \end{vmatrix}$ …(ii)
Comparing (i) and (ii) we get
$\begin{aligned} &|| A|I|=k^{2} \\\\ &|A|^{2}=k^{2} \\\\ &k=1 \end{aligned}$

Adjoint and Inverse of a Matrix Exercise fill in the blaks Question 17

Answer $\rightarrow|A| A$
Given $\rightarrow$ is a non-singular matrix of order $3$
Explanation $\rightarrow|A|=A . \operatorname{adj} A$ …(i)
$(\operatorname{adj} A)^{-1}=\frac{A}{|A|}$
$(\operatorname{adj} A)^{-1}=\frac{\operatorname{adj}(\operatorname{adj} A)}{\operatorname{det}(\operatorname{adj} A)}$ [$A=a d j(a d j(A)) \text { and } \operatorname{det}(\operatorname{adj} A)=(\operatorname{det} \operatorname{det} A)^{n-1}$ ]
$\operatorname{adj}(\operatorname{adj} A)=(\operatorname{adj} A)^{-1} \operatorname{det}(\operatorname{adj} A)$
$=(\operatorname{det} A)^{n-1} \times \frac{A}{\operatorname{det}(A)}$ [$\operatorname{det}(\operatorname{adj} A)=(\operatorname{det} \operatorname{det} A)^{n-1}$]
$=(\operatorname{det} A)^{n-2} A$
Now, $n= 3$
$\begin{aligned} &\operatorname{adj}(\operatorname{adj} A)=(\operatorname{det} A) A \\\\ &=|A| A \end{aligned}$

Adjoint and Inverse of a Matrix Exercise fill in the blaks Question 18

Answer $\rightarrow \frac{1}{9}\left[\begin{array}{ll} 0 & 3 \\ 3 & 1 \end{array}\right]$
Given$\rightarrow A=\left[a_{i j}\right]_{2 \times 2} \text { where } a_{i j}=\left\{i+j \text { if } i \neq j i^{2}-2 j \text { if } i=j\right.$
Hint $\rightarrow A^{-1}=\frac{1}{|A|} \operatorname{adj} A$
Explanation$\rightarrow$ find $a_{11}, a_{12}, a_{21}, a_{22}$
$\begin{aligned} &a_{11}=1^{2}-2 \times 1=-1 \\ &a_{12}=1+2=3 \\ &a_{21}=2+1=3 \\ &a_{22}=2^{2}-2 \times 2=0 \\ &A=\left[\begin{array}{cc} -1 & 3 \\ 3 & 0 \end{array}\right] \end{aligned}$
$\begin{aligned} &|A|=-1 \times 0-3 \times 3=-9 \\ &\operatorname{adj} A=\left[\begin{array}{cc} 0 & -3 \\ -3 & -1 \end{array}\right] \\ &A^{-1}=\frac{1}{-9}\left[\begin{array}{cc} 0 & -3 \\ -3 & -1 \end{array}\right] \end{aligned}$
$A^{-1}=\frac{1}{9}\left[\begin{array}{ll} 0 & 3 \\ 3 & 1 \end{array}\right]$

Adjoint and Inverse Matrix Exercise Fill in the blanks Question 19

Answer$\rightarrow \lambda=-1/6$
Given$\rightarrow A=\left[\begin{array}{ll} 0 & 3 \\ 2 & 0 \end{array}\right], A^{-1}=\lambda(\operatorname{adj} A)$
Explanation $\rightarrow \operatorname{adj} A=\left[\begin{array}{cc} 0 & -3 \\ -2 & 0 \end{array}\right]$
$\lambda(\operatorname{adj} A)=\left[\begin{array}{llll} 0 & -3 \lambda & -2 \lambda & 0 \end{array}\right]$ …(i)
Now,
$A^{-1}=\frac{1}{-6}\begin{vmatrix} 0 &-3 & -20 \end{vmatrix}$
$\Rightarrow \begin{bmatrix} 0 &\frac{1}{2} &\frac{1}{3} & 0 \end{bmatrix}= \lambda\left ( adj A \right )= \begin{vmatrix} 0 & -3\lambda & -2\lambda &0 \end{vmatrix}$ … [from (i)]
$\Rightarrow-3 \lambda=\frac{1}{2} \quad \Rightarrow \lambda=-1 / 6$

Adjoint and Inverse Matrix Exercise Fill in the blanks Question 20

Answer$\rightarrow I$
Given $\rightarrow A A^{T}=A^{T} A \text { and } B=A^{-1} A^{T}$
Explanation $\rightarrow B=A^{-1} A^{T}$
$B^{-1}=\left(A^{-1} A^{T}\right)^{-1}$
$=\left(A^{T}\right)^{-1}\left(A^{-1}\right)^{-1} \qquad\left[(A B)^{-1}=B^{-1} A^{-1}\right]$
Now,
$\begin{aligned} &B B^{-1}=A^{-1} A^{T}\left(A^{T}\right)^{-1}\left(A^{-1}\right)^{-1} \\ &=A^{-1} A^{T}\left(A^{T}\right)^{-1} A \\ &=A^{-1}\left(A A^{-1}\right)^{T} A \\ &=A^{-1} \cdot I \cdot A \end{aligned}$
$=A^{-1} A=1$
$B B^{-1}=I$

Adjoint and Inverse of a Matrix Exercise fill in the blaks Question 21

Answer $\rightarrow A^{2}+B^{2}$
Hint $\rightarrow AB=-BA$
Given $\rightarrow B=-A^{-1}BA$
Explanation$\rightarrow A$ & $B$ are square matrices of same order
$\begin{aligned} &B=-A^{-1} B A \\ &A B=-A A^{-1} B A \end{aligned}$
$AB=-I.BA$
$AB=-BA$ ...(i)
Now,
$(A+B)^{2}=A^{2}+B^{2}+A B+B A$
$=A^{2}+B^{2}-B A+B A$ from (i)
$(A+B)^{2}=A^{2}+B^{2}$

Adjoint and Inverse of a Matrix Exercise fill in the blaks Question 22

Answer $\rightarrow\left(A^{-1}\right)^{3}$
Hint $\rightarrow A$ is a non-singular matrix hence $A$ is non-zero
Given $\rightarrow A$ is a non-singular matrix of order $3\times 3$
Explanation $\rightarrow\left(A^{3}\right)^{-1}=\frac{1}{A^{3}}$
$\begin{aligned} &=\left(\frac{1}{A}\right)^{3} \\ &\left(A^{3}\right)^{-1}=\left(A^{-1}\right)^{3} \end{aligned}$

Adjoint and Inverse of a Matrix exercise Fill in the blanks question 23

Answer $\rightarrow 3\times 3$
Hint $\rightarrow(\operatorname{adj} A)=|A|^{n-1}$
Given$\rightarrow|(\operatorname{adj} A)|=|A|^{2}$
Explanation $\rightarrow \text { adj } A=|A| A^{-1}$
$|\operatorname{adj} A|=|A|^{n-1}$
$|A|^{2}=|A|^{n-1}$ … [ given $|(\operatorname{adj} A)|=|A|^{2}$ ]
$\begin{aligned} &n-1=2 \\ &n=3 \end{aligned}$

Adjoint and Inverse of a Matrix exercise Fill in the blanks question 24

Answer $\rightarrow 79$
Hint $\rightarrow \text { Aadj } A=|A| I$
Given $\rightarrow A=\left[\begin{array}{lll} x & 5 & 2 \\ 2 & y & 3 \\ 1 & 1 & z \end{array}\right] x y z=80,3 x+2 y+10 z=20$
Explanation $\rightarrow|A|=\left|\begin{array}{lll} x & 5 & 2 \\ 2 & y & 3 \\ 1 & 1 & z \end{array}\right|$
$\begin{aligned} &=x(y z-3)-5(2 z-3)+2(2-y) \\ &=x y z-3 x-10 z+15+4-2 y \\ &=x y z-(3 x+2 y+10 z)+19 \\ &=80-20+19 \\ &=79 \end{aligned}$

Adjoint and Inverse of Matrix Excercise Fill in the Blanks Question 25

Answer $\rightarrow \left[\begin{array}{ll} 3 & -4 \\ 1 & -1 \end{array}\right]$
Hint $\rightarrow A^{-1}=\frac{1}{|A|} \operatorname{adj} A$
Given $\rightarrow A=\left[\begin{array}{ll} 3 & -4 \\ 1 & -1 \end{array}\right]$
Explanation $\rightarrow|A|=\left[\begin{array}{ll} 3 & -4 \\ 1 & -1 \end{array}\right]$
$\begin{aligned} &=-3+4=1 \\ &\operatorname{adj} A=\left[\begin{array}{rr} -1 & 4 \\ -1 & 3 \end{array}\right] \end{aligned}$
$\begin{aligned} &A^{-1}=\frac{1}{|A|} \operatorname{adj} A \\ &A^{-1}=\left[\begin{array}{ll} -1 & 4 \\ -1 & 3 \end{array}\right] \end{aligned}$

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