Question

If $f\left(x\right)=x^2+4x-5$ and $A=\left[\begin{array}{cc}1& 2\\ 4& -3\end{array}\right]$ then $f\left(A\right)$ equals

• A. $\left[\begin{array}{cc}2& -2\\ 4& 1\end{array}\right]$
• B. $\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$
• C. $\left[\begin{array}{cc}-8& -4\\ -8& 0\end{array}\right]$
• D. $\left[\begin{array}{cc}8& 4\\ 8& 0\end{array}\right]$

Answer 1

Answer: (D)

$\left[\begin{array}{cc}8& 4\\ 8& 0\end{array}\right]$

Given $A=\left[\begin{array}{cc}1& 2\\ 4& -3\end{array}\right]$
$f\left(x\right)=x^2+4x-5$
$\Rightarrow f\left(A\right)=A^2+4A-5I$
Now, $A^2=\left[\begin{array}{cc}1& 2\\ 4& -3\end{array}\right]\left[\begin{array}{cc}1& 2\\ 4& -3\end{array}\right]$
$A^2=\left[\begin{array}{cc}9& -4\\ -8& 17\end{array}\right]$
So, $f\left(A\right)=\left[\begin{array}{cc}9& -4\\ -8& 17\end{array}\right]+\left[\begin{array}{cc}4& 8\\ 16& -12\end{array}\right]+\left[\begin{array}{cc}-5& 0\\ 0& -5\end{array}\right]$
$\Rightarrow f\left(A\right)=\left[\begin{array}{cc}8& 4\\ 8& 0\end{array}\right]$