Question

If the matrix $A=\left(\begin{array}{cc}0& 2\\ K& -1\end{array}\right)$ satisfies $A\left(A^3+3I\right)=2I$, then the value of $K$ is :

• A. $\frac{1}{2}$
• B. $-\frac{1}{2}$
• C. $-1$
• D. $1$

Answer 1

Answer: (A)

Given matrix $A=\left[\begin{array}{cc}0& 2\\ k& -1\end{array}\right]$
$A^4+3IA=2I$
$\Rightarrow A^4=2I-3A$
Also characteristic equation of $A$ is $|A-\lambda I|=0$
$\Rightarrow \left|\begin{array}{cc}0-\lambda & 2\\ k& -1-\lambda \end{array}\right|=0$
$\Rightarrow \lambda +{\lambda }^2-2K=0$
$\Rightarrow A+A^2=2K.I$
$\Rightarrow A^2=2KI-A$
$\Rightarrow A^4=4K^{2}I+A^2-4AK$
Put $A^2=2KI-A$
and $A^4=2I-3A$
$2I-3A=4K^{2}I+2KI-A-4AK$
$\Rightarrow I\left(2-2K-4K^2\right)=A(2-4K)$
$\Rightarrow -2I\left(2K^2+K-1\right)=2A(1-2K)$
$\Rightarrow -2I(2K-1)(K+1)=2A(1-2K)$
$\Rightarrow (2K-1)(2A)-2I(2K-1)(K+1)=0$
$\Rightarrow (2K-1)[2A-2I(K+1)]=0$
$\Rightarrow K=\frac{1}{2}$