Question

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

Answer 1

$B=9A^{-1}\Rightarrow B=9\frac{adj\left(A\right)}{|A|}$
$|A|=15-4\lambda$
$adj\left(A\right)=\left[\begin{array}{ccc}12-3\lambda & 3-\lambda & 1\\ 12-4\lambda & 3& -4\\ -3& 3& -4\end{array}\right]$

$b_{33}=\frac{9\times (-4)}{15-4\lambda }$
$\Rightarrow \frac{-9\times 4}{15-4\lambda }=9\times 4$
$\Rightarrow 15-4\lambda =-1$
$\Rightarrow \lambda =4$
We know that $A^{-1}=\frac{adj\left(A\right)}{|A|}$
Therefore, we can conclude $A=A^{-1}=\left[\begin{array}{ccc}0& 1& -1\\ 4& -3& 4\\ 3& -3& 4\end{array}\right]$

Hence $A=A^{-1}\Rightarrow A^2=I$
$B=9A^{-1}\Rightarrow B=9A$
$B^2=81A^2=81I$
$A^2+B^2=82I$
$\text{tr}(A^2+B^2)=82\times 3=246$