Question

If A= $\left[\begin{array}{cc}2& -1\\ -1& 2\end{array}\right]$, then the general solution of $sin\theta =|A^2-4A+3I|$ is

• A. $n\pi$
• B. $2n+1\frac{\pi }{2}$
• C. $n\pi +(-1{)}^n\frac{\pi }{2}$
• D. $2n\pi ,nϵz$

Answer 1

The correct option is A $n\pi$

$A=\left[\begin{array}{cc}2& -1\\ -1& 2\end{array}\right]$
$A^2=\left[\begin{array}{cc}2& -1\\ -1& 2\end{array}\right]\left[\begin{array}{cc}2& -1\\ -1& 2\end{array}\right]=\left[\begin{array}{cc}5& -4\\ -4& 5\end{array}\right]$
$A^2-4A+3I=\left[\begin{array}{cc}5& -4\\ -4& 5\end{array}\right]+\left[\begin{array}{cc}-8& -4\\ -4& -8\end{array}\right]+\left[\begin{array}{cc}3& 0\\ 0& 3\end{array}\right]=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]$
$\Rightarrow |A^2-4A+3I|=0$
So,$\sin \theta =0$
Hence, the general solution is $\theta =n\pi$