Question

If $A=\left[\begin{array}{cc}2& 3\\ -1& 2\end{array}\right]$, find $3A^2-4A+I$.

Answer 1

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

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

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

Therefore,

$3A^2-4A+I=\left[\begin{array}{cc}1& 12\\ -4& 1\end{array}\right]-\left[\begin{array}{cc}8& 12\\ -4& 8\end{array}\right]+\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$

$3A^2-4A+I=\left[\begin{array}{cc}-6& 0\\ 0& -6\end{array}\right]=-6I$

$\therefore 3A^2-4A+I=-6I$