Question

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

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

Answer 1

The correct option is A $\frac{1}{2}$

Given : $A=\left(\begin{array}{cc}0& 2\\ K& -1\end{array}\right)$

characteristic equation is $|A–xI|=0$
$\Rightarrow \left|\begin{array}{cc}0-x& 2\\ K& -1-x\end{array}\right|=0$
$\Rightarrow x(x+1)-2K=0$
$\Rightarrow x^2+x-2K=0$
Every square matrix satisfies its own characteristic equation.
$\therefore A^2+A-2KI=0$
$\Rightarrow A^2=2KI-A$
Now, $A^4=A^2\cdot A^2$
$\Rightarrow A^4=(2KI–A)(2KI–A)$
$\Rightarrow A^4=4K^{2}I-4KA+A^2$
$\Rightarrow A^4=4K^{2}I-4KA+2KI-A$
$\Rightarrow A^4=(4K^2+2K)I-(4K+1)A$
$\Rightarrow A^4+(4K+1)A=(4K^2+2K)I\cdots \left(1\right)$
and given that
$A^4+3A=2I\cdots \left(2\right)$
Comparing the coefficients from $\left(1\right)$ and $\left(2\right),$ we get
$4K+1=3\Rightarrow K=-\frac{1}{2}$
and $4K^2+2K=2$
$\Rightarrow 2K^2+K-1=0$
$\Rightarrow (2K-1)(K+1)=0$
$\Rightarrow K=\frac{1}{2},-1$
$\therefore K=\frac{1}{2}$