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Maths 3 Marks Question

Adjoint and Inverse of a Matrix · Uttarakhand Board (UBSE) STD 12 Science (Uttarakhand - English Medium). 15 questions.

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Question 13 Marks
If $\text{adj A}=\begin{bmatrix} 2 & 3 \\ 4 & -1 \end{bmatrix} \text{and B}=\begin{bmatrix} 1 & -2 \\ -3 & 1 \end{bmatrix},$ find adj AB.
Answer
If A and B are non-singular square matrices of the same order, then adj (AB) = (adj B)(adj A) Now, ATQ$\text{adj A}=\begin{bmatrix} 2 & 3 \\ 4 & -1 \end{bmatrix}$
$\text{adj B}=\begin{bmatrix} 1 & -2 \\ -3 & 1 \end{bmatrix}$
So, adj(AB) = (adj B)(adj A)$=\begin{bmatrix} 1 & -2 \\ -3 & 1 \end{bmatrix}\begin{bmatrix} 2 & 3 \\ 4 & -1 \end{bmatrix}$
$=\begin{bmatrix} -6 & 5 \\ -2 & -10 \end{bmatrix}$
Hence, $\text{adj (AB)}=\begin{bmatrix} -6 & 5 \\ -2 & -10 \end{bmatrix}$
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Question 23 Marks
Find the adjoint of the following matrices:$\begin{bmatrix}-3 & 5 \\ 2 & 4 \end{bmatrix}$
Verify that (adjoint A) A = |A|I = A (adjoint A) for the above matrices.
Answer
$\text{A}=\begin{bmatrix}-3 & 5 \\ 2 & 4 \end{bmatrix}$$\text{adjoint A}=\begin{bmatrix} 4 & -5 \\ -2 & -3 \end{bmatrix}$
$\text{adjoint A}=\begin{bmatrix} -22 & 0 \\ 0 & -22 \end{bmatrix}$
$|\text{A}|=-22$
$|\text{A}|\text{I}=\begin{bmatrix} -22 & 0 \\ 0 & -22 \end{bmatrix}$
$\text{A(adjoint A)}=\begin{bmatrix} -22& 0 \\ 0 & -22 \end{bmatrix}$
$\therefore\ \text{(adjoint A)A}=|\text{A}|\text{I}=\text{A(adjoint A)}$
Hence verified.
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Question 33 Marks
Find the inverse of the following matrices:$\begin{bmatrix}\cos\theta & \sin\theta \\-\sin\theta & \cos\theta \end{bmatrix}$
Answer
$\text{A}=\begin{bmatrix}\cos\theta & \sin\theta \\-\sin\theta & \cos\theta \end{bmatrix}$$|\text{A}|=\cos^3\theta+\sin^2\theta=1\neq0$
A is a singular matrix; therefore, it is invertible Let $C_{ij}$ be a cofactor of $a_{ij}$ in A.
Now,
$\text{C}_{11}=\cos\theta$
$\text{C}_{12}=\sin\theta$
$\text{C}_{21}=-\sin\theta$
$\text{C}_{22}=\cos\theta$
$\text{adj A}=\begin{bmatrix}\cos\theta & \sin\theta \\-\sin\theta & \cos\theta \end{bmatrix}^\text{T}=\begin{bmatrix}\cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{bmatrix}$
$\therefore\ \text{A}^{-1}=\frac{1}{|\text{A}|}\text{ adj A}=\begin{bmatrix}\cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{bmatrix}$
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Question 43 Marks
Find the inverse of the following matrices:$\begin{bmatrix}2 & 5 \\ -3 & 1 \end{bmatrix}$
Answer
$\text{D}=\begin{bmatrix}2 & 5 \\ -3 & 1 \end{bmatrix}$$|\text{D}|=2+15=17\neq0$
D is a singular matrix; therefore, it is invertible
Let $C_{ij}$ be a cofactor of $d_{ij}$ in D.
Now,
$C_{11} = 1$
$C_{12} = 3$
$C_{21} = -5$
$C_{22} = 2$
$\text{adj D}=\begin{bmatrix}1 & 3 \\ -5 & 2 \end{bmatrix}^\text{T}=\begin{bmatrix}1 & -5 \\ 3 & 2 \end{bmatrix}$
$\therefore\ \text{D}^{-1}=\frac{1}{|\text{D}|}\text{ adj D}=\frac{1}{17}\begin{bmatrix}1 & -5 \\ 3 & 2 \end{bmatrix}$
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Question 53 Marks
Find the inverse of the following matrices:$\begin{bmatrix} \text{a} & \text{b} \\ \text{c} & \frac{1+\text{bc}}{\text{a}} \end{bmatrix}$
Answer
$\text{C}=\begin{bmatrix} \text{a} & \text{b} \\ \text{c} & \frac{1+\text{bc}}{\text{a}} \end{bmatrix}$$|\text{C}|=1+\text{bc}-\text{bc}=1\neq0$
C is a singular matrix; therefore, it is invertible.
Let $C_{ij}$ be a cofactor of $C_{ij}$ in C.
Now,
$\text{C}_{11}=\frac{1+\text{bc}}{\text{a}}$
$\text{C}_{12}=-\text{C}$
$\text{C}_{21}=-\text{b}$
$\text{C}_{22}=\text{a}$
$\text{adj C}=\begin{bmatrix}\frac{1+\text{bc}}{\text{a}} & -\text{c} \\ -\text{b} & \text{a} \end{bmatrix}=\begin{bmatrix}\frac{1+\text{bc}}{\text{a}} & -\text{b} \\ -\text{c} & \text{a} \end{bmatrix}$
$\therefore\ \text{C}^{-1}=\frac{1}{|\text{C}|}\text{adj C}=\begin{bmatrix}\frac{1+\text{bc}}{\text{a}} & -\text{b} \\ -\text{c} & \text{a} \end{bmatrix}$
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Question 63 Marks
Let $\text{F}(\alpha)=\begin{bmatrix}\cos\alpha & -\sin\alpha & 0 \\ \sin\alpha & \cos\alpha & 0 \\ 0 & 0 & 1\end{bmatrix}$ and$\text{G}(\beta)=\begin{bmatrix} \cos\beta & 0 & \sin\beta \\ 0 & 1 & 0 \\ -\sin\beta & 0 & \cos\beta \end{bmatrix}$
Show that$\big[\text{G}(\beta)\big]^{-1}=\text{G}(-\beta)$
Answer
$\text{G}(\beta)=\begin{bmatrix} \cos\beta & 0 & \sin\beta \\ 0 & 1 & 0 \\ -\sin\beta & 0 & \cos\beta \end{bmatrix}\Rightarrow\ |\text{G}(\beta)|=\cos^2\beta+\sin^2$$\text{C}_{11}=\cos\beta, \text{C}_{21}=0,\text{C}_{31}=\sin\beta$
$\text{C}_{12}=0,\text{C}_{22}=1,\text{C}_{32}=0$
$\text{C}_{13}=\sin\beta, \text{C}_{23}=0,\text{C}_{33}=\cos\beta$
$\big[\text{G}(\beta)\big]^{-1}=\frac{\text{adj}(\text{G}(\beta))}{|\text{G}(\beta)|}=\frac{1}{1}\begin{bmatrix} \cos\beta & 0 & \sin\beta \\ 0 & 1 & 0 \\ -\sin\beta & 0 & \cos\beta \end{bmatrix}\ .....\text{(i)}$
Now, $\text{G}(-\beta)=\begin{bmatrix} \cos(-\beta) & 0 & \sin(-\beta) \\ 0 & 1 & 0 \\ -\sin(-\beta) & 0 & \cos(-\beta) \end{bmatrix}$
$=\begin{bmatrix} \cos\beta & 0 & -\sin\beta \\ 0 & 1 & 0 \\ \sin\beta & 0 & \cos\beta \end{bmatrix}\ .....\text{(ii)}$
From (i) & (ii)
$\big[\text{G}(\beta)\big]^{1}=\text{G}(-\beta)$
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Question 73 Marks
If $\text{A}=\begin{bmatrix}-4 & -3 & -3 \\1 & 0 & 1 \\ 4 & 4 & 3 \end{bmatrix},$ show that adj A = A.
Answer
$\text{A}=\begin{bmatrix}-4 & -3 & -3 \\1 & 0 & 1 \\ 4 & 4 & 3 \end{bmatrix}$Now,
$\text{C}_{11}=\begin{vmatrix}0 & 1 \\4 & 3 \end{vmatrix}=-4,\ \text{C}_{12}=\begin{vmatrix}1 & 1 \\4 & 3 \end{vmatrix}=1$
$\text{and C}_{13}=\begin{bmatrix}1 & 0 \\4 & 4 \end{bmatrix}=4$
$\text{C}_{21}=\begin{vmatrix}-3 & -3 \\4 & 3 \end{vmatrix}=-3,\ \text{C}_{22}=\begin{vmatrix}-4 & -3 \\4 & 3 \end{vmatrix}=0$
$\text{and C}_{23}=\begin{bmatrix}-4 & -3 \\4 & 4 \end{bmatrix}=4$
$\text{C}_{31}=\begin{vmatrix}-3 & -3 \\0 & 1 \end{vmatrix}=-3,\ \text{C}_{32}=\begin{vmatrix}-4 & -3 \\1 & 1 \end{vmatrix}=1$
$\text{and C}_{33}=\begin{bmatrix}-4 & -3 \\1 & 0 \end{bmatrix}=3$
$\therefore\ \text{adj A}=\begin{bmatrix}-4 & 1 & 4\\-3 & 0 & 4 \\ -3 & 1 & 3 \end{bmatrix}^\text{T}=\begin{bmatrix}-4 & -3 & -3 \\1 & 0 & 1 \\ 4 & 4 & 3 \end{bmatrix}=\text{A}$
Hence proved.
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Question 83 Marks
Find A (adjoint A) for the matrix $\text{A}=\begin{bmatrix}1 & -2 & 3 \\0 & 2 & -1 \\ -4 & 5 & 2 \end{bmatrix}$.
Answer
$\text{A}=\begin{bmatrix}1 & -2 & 3 \\0 & 2 & -1 \\ 4 & 5 & 2 \end{bmatrix}$Now, $\text{C}_{11}=\begin{bmatrix}2 & -1 \\5 & 2 \end{bmatrix}=9,\text{C}_{12}=-\begin{bmatrix}0 & -1 \\-4 & 2 \end{bmatrix}=4$
$\text{and C}_{13}=\begin{bmatrix}0 & 2 \\-4 & 5 \end{bmatrix}=8$
$\text{C}_{21}=\begin{bmatrix}-2 & 3 \\5 & 2 \end{bmatrix}=19,\text{C}_{22}=\begin{bmatrix}1 & 3 \\-4 & 2 \end{bmatrix}=14$
$\text{and C}_{23}=\begin{bmatrix}1 & -2 \\-4 & 5 \end{bmatrix}=3$
$\text{C}_{31}=\begin{bmatrix}-2 & 3 \\ 2 & -1 \end{bmatrix}=4,\text{C}_{32}=-\begin{bmatrix}1 & 3 \\0 & -1 \end{bmatrix}=1$
$\text{and C}_{33}=\begin{bmatrix}1 & -2 \\0 & 2 \end{bmatrix}=2$
$\text{adj A}=\begin{bmatrix}9 & 4 & 8 \\19 & 14 & 3 \\ -4 & 1 & 2 \end{bmatrix}^\text{T}=\begin{bmatrix}9 & 19 & -4 \\4 & 14 & 1 \\ 8 & 3 & 2 \end{bmatrix}$
$\therefore\ \text{A(adj A)}=\begin{bmatrix}25 & 0 & 0 \\0 & 25 & 0 \\ 0 & 0 & 25 \end{bmatrix}$
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Question 93 Marks
If A is a square matrix of order 3 such that |A| = 3, then write the value of adj (adj A).
Answer
If A is a non-singular matrix of order n, then$|\text{adj (adj A)}|=|\text{A}|^{(\text{n}-1)^{2}}$
Now, ATQ$|\text{A}|=3$
$\text{n}=3$
$|\text{adj (adj A)}|=|3|^{(3-1)^{2}}$
$=(3)^{2^3}$
$=(3)^4$
$=81$
Hence,$|\text{adj (adj A)}|=81$
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Question 103 Marks
If $\text{A}=\begin{bmatrix} \cos\theta & \sin\theta \\ -\sin\theta & \cos\theta \end{bmatrix}\text{ and A (adj A =)}\begin{bmatrix} \text{k} & 0 \\ 0 & \text{k} \end{bmatrix},$ then find the value of $k$.
Answer
$\text{A}=\begin{bmatrix} \cos\theta & \sin\theta \\ -\sin\theta & \cos\theta \end{bmatrix}$$\therefore|\text{A}|=\begin{bmatrix} \cos\theta & \sin\theta \\ -\sin\theta & \cos\theta \end{bmatrix}$
$=\cos^2\theta+\sin^2\theta=1\neq0$
Thus, $A^{-1}$ exists.
Now,
$\text{A}^{-1}=\frac{\text{adj A}}{|\text{A}|}=\begin{bmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{bmatrix}$
$\Rightarrow\text{A}^{-1}=\text{adj A}$
$\Rightarrow\text{AA}^{-1}=\text{A adj A}$
$\Rightarrow\text{AA}^{-1}=\begin{bmatrix} \text{k} & 0 \\ 0 & \text{k} \end{bmatrix}$
$\Rightarrow\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}=\begin{bmatrix} \text{k} & 0 \\ 0 & \text{k} \end{bmatrix} \ \big[\because\ \text{AA}^{-1}=\text{I}\big]$
$\Rightarrow\text{k}=1$
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Question 113 Marks
Given $\text{A}=\begin{bmatrix}2 & -3 \\-4 & 7 \end{bmatrix},$ compute $A^{-1}$ and show that $2A^{-1} = 9I - A$.
Answer
We have, $\text{A}=\begin{bmatrix}2 & -3 \\-4 & 7 \end{bmatrix}$ Now, $\text{adj (A)}=\begin{bmatrix}7 & 3 \\ 4 & 2 \end{bmatrix}$ and |A| = 2$\therefore\ \text{A}^{-1}=\frac{1}{2}\begin{bmatrix}7 & 3 \\ 4 & 2 \end{bmatrix}$
Now, $2A^{-1} = 9I - A ~\text{L.H.S}=2\text{A}^{-1}=\begin{bmatrix}7 & 3 \\ 4 & 2 \end{bmatrix}$
$\text{R.H.S}=9\text{I}-\text{A}=9\begin{bmatrix}1 & 0 \\0 & 1 \end{bmatrix}-\begin{bmatrix}2 & -3 \\-4 & 7 \end{bmatrix}$
$=\begin{bmatrix}7 & 3 \\4 & 2 \end{bmatrix}=\text{L.H.S}$
Hence proved.
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Question 123 Marks
If $\text{A}=\begin{bmatrix}2 & 3 \\ 5 & -2 \end{bmatrix}$ be sech that $A^{-1} = kA$, then find the value of $k$.
Answer
$\text{A}=\begin{bmatrix}2 & 3 \\ 5 & -2 \end{bmatrix}$$\therefore|\text{A}|=\begin{vmatrix}2 & 3 \\ 5 & -2 \end{vmatrix}=-14-15=-19$
The value is non-zero, so $A^{-1}$ exists.
By definition, we have
$A^{-1} A = I$ [I is the identity matrix]
$\Rightarrow kAA = I$ [Substituting $A^{-1} = kA]$
$\Rightarrow\text{k}\begin{bmatrix}2 & 3 \\ 5 & -2 \end{bmatrix}\begin{bmatrix}2 & 3 \\ 5 & -2 \end{bmatrix}=\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$
$\Rightarrow\text{k}\begin{bmatrix} 4+15 & 6-6 \\ 10-10 & 15+4 \end{bmatrix}=\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$
$\Rightarrow\text{k}\begin{bmatrix} 19 & 0 \\ 0 & 19 \end{bmatrix}=\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$
$\Rightarrow\text{k}=\frac{1}{19}$
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Question 133 Marks
Find the inverse of the following matrices:$\begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}$
Answer
$\text{A}=\begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}$Now $|\text{A}|=-1\neq0$
Hence $A^{-1}$ exists.
Cofactors of a are:
$C_{11} = 0\ C_{12} = -1$
$C_{21} = -1\ C_{22} = 0$
$\therefore\ \text{adj A}=\begin{bmatrix}\text{C}_{11} & \text{C}_{12} \\ \text{C}_{21} & \text{C}_{22} \end{bmatrix}$
$=\begin{bmatrix}0 & -1 \\ -1 & 0 \end{bmatrix}$
Also, $\text{A}^{-1}=\frac{1}{|\text{A}|}(\text{adj A})$
$\text{A}^{-1}=\frac{1}{(-1)}\begin{bmatrix}0 & -1 \\ -1 & 0 \end{bmatrix}$
$\text{A}^{-1}=\begin{bmatrix}0 & 1 \\ 1 & 0 \end{bmatrix}$
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Question 143 Marks
If $\text{A}=\begin{bmatrix}4 & 5 \\2 & 1 \end{bmatrix},$ then show that $A - 3I = 2 (I + 3A^{-1})$.
Answer
We have, $\text{A}=\begin{bmatrix}4 & 5 \\2 & 1 \end{bmatrix}$ Now,$\text{adj (A)}\begin{bmatrix}1 & -5 \\-2 & 4 \end{bmatrix}$
and |A| = -6$\therefore\ \text{A}^{-1}=-\frac{1}{6}\begin{bmatrix}1 & -5 \\-2 & 4 \end{bmatrix}$
Now, $A - 3I = I + 3A^{-1} \text{L.H.S}=\text{A}-3\text{I}=\begin{bmatrix}4 & 5 \\2 & 1 \end{bmatrix}-3\begin{bmatrix}1 & 0 \\0 & 1 \end{bmatrix}=\begin{bmatrix}1 & 5 \\2 & -2 \end{bmatrix}$
$\text{R.H.S}=2(\text{I}+3\text{A}^{-1})=2\left\{\begin{bmatrix}1 & 0 \\0 & 1 \end{bmatrix}-3\times\frac{1}{3}\begin{bmatrix}1 & -5 \\-2 & 4 \end{bmatrix}\right\}$
$=2\begin{bmatrix}0.5 & 2.5 \\1 & -1 \end{bmatrix}=\begin{bmatrix}1 & 5 \\2 & -2 \end{bmatrix}\text{L.H.S}$
Hence proved.
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Question 153 Marks
If $\text{A}=\begin{bmatrix}-1 & -2 & -2 \\2 & 1 & -2 \\ 2 & -2 & 1 \end{bmatrix},$ show that $A = 3A^T$.
Answer
Here $\text{A}=\begin{bmatrix}-1 & -2 & -2 \\2 & 1 & -2 \\ 2 & -2 & 1 \end{bmatrix},$ Cofactor of a are: $C_{11} = -3, C_{21} = 6, C_{31} = 6 C_{12}= -6, C_{22} = 3, C_{32} = -6 C_{13} = -6, C_{23} = -6, C_{33} = 3$
$\therefore\ \text{adj A}=\begin{bmatrix}\text{C}_{11} & \text{C}_{12} & \text{C}_{13} \\ \text{C}_{21} & \text{C}_{22} & \text{C}_{23} \\ \text{C}_{31} & \text{C}_{32} & \text{C}_{33} \end{bmatrix}$
$=\begin{bmatrix}-3 & -6 & -6 \\ 6 & 3 & -6 \\ 6 & -6 & 3 \end{bmatrix}$
$\therefore\ \text{adj A}=\begin{bmatrix}-3 & -6 & -6 \\ -6 & 3 & -6 \\ -6 & -6 & 3 \end{bmatrix}\ .....\text{(i)}$
Now, $3\text{A}^\text{T}=3\begin{bmatrix}-1& 2 & 2 \\-2 & 1 & -2 \\ -2 & -2 & 1 \end{bmatrix}=\begin{bmatrix}-3 & -6 & -6 \\ -6 & 3 & -6 \\ -6 & -6 & 3 \end{bmatrix}\ .....\text{(ii)}$$\therefore\ \text{adj A}=3\text{A}^\text{T}$
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