Question

If A = $\left[\begin{array}{cc}\cos x& \sin x\\ -\sin x& \cos x\end{array}\right]$ and A (adj A) = $\left[\begin{array}{cc}k& 0\\ 0& k\end{array}\right]$, then k = _________________.

Answer 1

Given:
A = $\left[\begin{array}{cc}\cos x& \sin x\\ -\sin x& \cos x\end{array}\right]$
A(adj A) = $\left[\begin{array}{cc}k& 0\\ 0& k\end{array}\right]$


Now,
$$\begin{gathered} A=\left[\begin{array}{cc}\cos x& \sin x\\ -\sin x& \cos x\end{array}\right] \\ \Rightarrow \left|A\right|=\left|\begin{array}{cc}\cos x& \sin x\\ -\sin x& \cos x\end{array}\right| \\ \Rightarrow \left|A\right|={\cos }^{2}x+{\sin }^{2}x \\ \Rightarrow \left|A\right|=1 \end{gathered}$$


As we know,
$$\begin{gathered} A\left(\mathrm{adj}A\right)=\left|A\right|I \\ \Rightarrow \left[\begin{array}{cc}k& 0\\ 0& k\end{array}\right]=\left|A\right|I \\ \Rightarrow k\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]=\left|A\right|I \\ \Rightarrow kI=\left|A\right|I \\ \Rightarrow k=\left|A\right| \\ \Rightarrow k=1\left(\because \left|A\right|=1\right) \end{gathered}$$


Hence, $k=\underline{1}$.