Question

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

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

Answer 1

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

We have, $A=\left[\begin{array}{cc}2& -3\\ -4& 1\end{array}\right]\therefore A^2=A.A\left[\begin{array}{cc}2& -3\\ -4& 1\end{array}\right]\left[\begin{array}{cc}2& -3\\ -4& 1\end{array}\right]=\left[\begin{array}{cc}4+12& -6-3\\ -8-4& 12+1\end{array}\right]=\left[\begin{array}{cc}16& -9\\ -12& 13\end{array}\right]Now,3A^2+12A=3\left[\begin{array}{cc}16& -9\\ -12& 13\end{array}\right]+12\left[\begin{array}{cc}2& -3\\ -4& 1\end{array}\right]=\left[\begin{array}{cc}48& -27\\ -36& 39\end{array}\right]+\left[\begin{array}{cc}24& -36\\ 48& 12\end{array}\right]=\left[\begin{array}{cc}72& -63\\ -84& 51\end{array}\right]\therefore adj(3A^2+12A)=\left[\begin{array}{cc}51& 63\\ 84& 72\end{array}\right]$