Mathematics - Matrices Question with Solution | TestHub

Given, $A=\left[\begin{array}{cc}2& -3\\ -4& 1\end{array}\right]$

$\Rightarrow A^2=\left[\begin{array}{cc}2& -3\\ -4& 1\end{array}\right] \left[\begin{array}{cc}2& -3\\ -4& 1\end{array}\right]$

$\Rightarrow A^2=\left[\begin{array}{cc}4+12& -6-3\\ -8-4& 12+1\end{array}\right]$

$\Rightarrow A^2=\left[\begin{array}{cc}16& -9\\ -12& 13\end{array}\right]$

$\Rightarrow 3A^2=\left[\begin{array}{cc}48& -27\\ -36& 39\end{array}\right]$ and  $12A=\left[\begin{array}{cc}24& -36\\ -48& 12\end{array}\right]$

$\therefore \left(3A^2+12A\right)=\left[\begin{array}{cc}72& -63\\ -84& 51\end{array}\right]$

$\mathrm{Adj}\left(3A^2+12A\right)={\left[\begin{array}{cc}51& 84\\ 63& 72\end{array}\right]}^T=\left[\begin{array}{cc}51& 63\\ 84& 72\end{array}\right]$