Question

Let $A$ and $B$ be $3\times 3$ square matrices such that $AB=9I$ where $A=\left[\begin{array}{ccc}0& 1& -1\\ 4& -3& 4\\ 3& -3& \lambda \end{array}\right].\text{If}{B}_{33}=9A_{23},$ then the value of $Tr\left(\frac{A^2+B^2}{82}\right)+\lambda$ is

Answer 1

$B=9A^{-1}$
$=\frac{9\left[\begin{array}{ccc}-& -& -\\ -& -& -\\ -& -& -4\end{array}\right]}{15-4\lambda }36=\frac{-36}{15-4\lambda }\lambda =4$

Now,
$A^2=I\Rightarrow B^2=81ITr\left(\frac{A^2+B^2}{82}\right)+\lambda =7$