Question

If $A=\left[\begin{array}{cc}2& -3\\ -4& 1\end{array}\right]$, then adj $\left(3A^2+12A\right)$ is equal to :

• A. $\left[\begin{array}{cc}51& 63\\ 84& 72\end{array}\right]$
• B. $\left[\begin{array}{cc}51& 84\\ 63& 72\end{array}\right]$
• C. $\left[\begin{array}{cc}72& -63\\ -84& 51\end{array}\right]$
• D. $\left[\begin{array}{cc}72& -84\\ -63& 51\end{array}\right]$

Answer 1

Answer: (A)

$A=\left[\begin{array}{cc}2& -3\\ -4& 1\end{array}\right]$
$\left|A-\lambda l\right|=\left[\begin{array}{cc}2-\lambda & -3\\ -4& 1-\lambda \end{array}\right]$
$=\left(2-2\lambda -\lambda +{\lambda }^2\right)-12$
$f\left(\lambda \right)={\lambda }^2-3\lambda -10$
$\because$ A satisfies $f\left(\lambda \right)$
$\therefore A^2-3A-10l=0$
$A^2-3A=10l$
$3A^2-9A=30l$
$3A^2+12A=30l+21A$
$=\left[\begin{array}{cc}30& 0\\ 0& 30\end{array}\right]+\left[\begin{array}{cc}42& -63\\ -84& 21\end{array}\right]$
$=\left[\begin{array}{cc}72& -63\\ -84& 51\end{array}\right]$
$adj\left(3A^2+12A\right)=\left[\begin{array}{cc}51& 63\\ 84& 72\end{array}\right]$