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Adjoint and Inverse of a Matrix — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsAdjoint and Inverse of a Matrix3 Marks
Question
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.
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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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