If A = [ cos α − sin α sin α cos α ] and A adj A = [ k 0 0 k ] ,then k is equal to

Question

If A = [ cos α − sin α ​ sin α cos α ​ ] and A adj A = [ k 0 ​ 0 k ​ ] ,then k is equal to, 0, 1, sin α cos α, cos 2 α

MARKS App · Accepted Answer

$\begin{aligned} & A=\left[\begin{array}{cc} \cos \theta & \sin \theta \ -\sin \theta & \cos \theta \end{array}\right] \ & \therefore|A|=\left|\begin{array}{cc} \cos \theta & \sin \theta \ -\sin \theta & \cos \theta \end{array}\right|=\cos ^2 \theta+\sin ^2 \theta=1 \neq 0 \end{aligned}$ Thus, $A^{-1}$ exists. Now, $\begin{aligned} & \Rightarrow A^{-1}=\operatorname{adj} A \ & \Rightarrow A A^{-1}=\operatorname{Aadj} A \ & \Rightarrow A A^{-1}=\left[\begin{array}{ll} k & 0 \ 0 & k \end{array}\right] \ & \Rightarrow\left[\begin{array}{ll} 1 & 0 \ 0 & 1 \end{array}\right]=\left[\begin{array}{ll} k & 0 \ 0 & k \end{array}\right]\left[\because A A^{-1}=I\right] \ & \Rightarrow k=1 \end{aligned}$

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