Question

If the matrix $A=\begin{bmatrix} \alpha & 0 \\ 2 & \alpha \end{bmatrix}$ $\left(\forall \alpha \in R , \alpha > 0\right)$ and $\left|2 A^{2} - 2 A\right|=144,$ 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)

$\left|2 A^{2} - 2 A\right|=144$
$2^{2}\left|A\right|\left|A - I\right|=144$
$\Rightarrow \left(\alpha \right)^{2}\left(\alpha - 1\right)^{2}=36$
$\alpha \left(\alpha - 1\right)=6$ 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)
$\therefore tr\left(A\right)=2\alpha =6$
$\left|A\right|=\alpha ^{2}=9$
$\left|A\right|+tr\left(A\right)=15$