Question

Let $A={\left[a_{ij}\right]}_{3\times 3}$ be a square matrix such that $A{A}^T=4I,\left|A\right|<0$ . If $\left|\begin{array}{ccc}{a}_{11}+4& a_{12}& a_{13}\\ a_{21}& a_{22}+4& a_{23}\\ a_{31}& a_{32}& a_{33}+4\end{array}\right|=5\lambda \left|A+I\right|$ . Then, $\lambda$ is equal to

• A. $\frac{4}{5}$
• B. $-\frac{4}{5}$
• C. $\frac{8}{5}$
• D. $-\frac{8}{5}$

Answer 1

Answer: (D)

$A{A}^T=4I\Rightarrow \left|A{A}^T\right|=\left|4I\right|\Rightarrow {\left|A\right|}^2=64$
$\Rightarrow \left|A\right|=-8$ or $8$
As, $\left|A\right|<0\Rightarrow \left|A\right|=-8$
$\left|\begin{array}{ccc}{a}_{11}+4& a_{12}& a_{13}\\ a_{21}& a_{22}+4& a_{23}\\ a_{31}& a_{32}& a_{33}+4\end{array}\right|=\left|A+4I\right|=\left|A+A{A}^T\right|$
$=\left|A\right|\left|I+A^T\right|$
$\Rightarrow -8\left|A+I\right|=5\lambda \left|A+I\right|$
$\Rightarrow \lambda =-\frac{8}{5}$