Question

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

• A. $\left[\begin{array}{cc}1& 1\\ 1& 0\end{array}\right]$
• B. $\left[\begin{array}{cc}-14& 51\\ -17& -14\end{array}\right]$
• C. $\left[\begin{array}{cc}-14& -51\\ -17& -14\end{array}\right]$
• D. $\left[\begin{array}{cc}-1& -1\\ -1& 0\end{array}\right]$

Answer 1

The correct option is B $\left[\begin{array}{cc}-14& 51\\ -17& -14\end{array}\right]$

$A=\left(\begin{array}{cc}2& 3\\ -1& 2\end{array}\right)$
$\therefore A^2=\left(\begin{array}{cc}1& 12\\ -4& 1\end{array}\right),A^3=\left(\begin{array}{cc}-10& 27\\ -9& -10\end{array}\right)$
$\therefore A^3+3A^2-4A+t=\left(\begin{array}{cc}-10& 27\\ -9& -10\end{array}\right)+\left(\begin{array}{cc}3& 36\\ -12& 3\end{array}\right)+\left(\begin{array}{cc}-8& -12\\ 4& -8\end{array}\right)+\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right)$
$=\left(\begin{array}{cc}-14& 51\\ -17& -14\end{array}\right)$
Hence (b) is correct choice.