Question

If the matrix $A=\left[\begin{array}{cc}\alpha & 0\\ 2& \alpha \end{array}\right]$ $\left(\forall \alpha \in R,\alpha >0\right)$ and $\left|2A^2-2A\right|=144,$ then the value of $\left|A\right|+tr\left(A\right)$ is equal to (where $t_r\left(A\right)$ represent the trace of the matrix $A$ i.e. the sum of all the principal diagonal elements of the matrix $A$ and $\left|A\right|$ is the determinant value of $A$ )

• A. $8$
• B. $15$
• C. $20$
• D. $24$

Answer 1

Answer: (B)

$\left|2A^2-2A\right|=144$
$2^2\left|A\right|\left|A-I\right|=144$
$\Rightarrow {\left(\alpha \right)}^2{\left(\alpha -1\right)}^2=36$
$\alpha \left(\alpha -1\right)=6$ or $-6$
${\alpha }^2-\alpha -6=0$ or ${\alpha }^2-\alpha +6=0$
$\left(\alpha -3\right)\left(\alpha +2\right)=0$ or $\alpha =\frac{1\pm \sqrt{1-24}}{2\times 1}$
$\alpha =3$ or $-2$ (rejected) or $\frac{1}{2}\pm i\frac{\sqrt{23}}{2}$ (rejected)
$\therefore tr\left(A\right)=2\alpha =6$
$\left|A\right|={\alpha }^2=9$
$\left|A\right|+tr\left(A\right)=15$