#### Explain solution RD Sharma class 12 Chapter 6 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}

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