Question

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

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

Answer 1

The correct option is D $\left[\begin{array}{cc}51& 63\\ 84& 72\end{array}\right]$

Solution:-

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

$A^2=\left[\begin{array}{cc}2& -3\\ -4& 1\end{array}\right]\times \left[\begin{array}{cc}2& -3\\ -4& 1\end{array}\right]=\left[\begin{array}{cc}16& -9\\ -12& 13\end{array}\right]$

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

$12A=12\times \left[\begin{array}{cc}2& -3\\ -4& 1\end{array}\right]=\left[\begin{array}{cc}24& -36\\ -48& 12\end{array}\right]$

$\therefore 3A^2+12A=\left[\begin{array}{cc}48& -27\\ -36& 39\end{array}\right]+\left[\begin{array}{cc}24& -36\\ -48& 12\end{array}\right]=\left[\begin{array}{cc}72& -63\\ -84& 51\end{array}\right]$

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