Question

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

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

Answer 1

The correct option is D $\left[\begin{array}{cc}8& 4\\ 8& 0\end{array}\right]$

$f\left(x\right)=x^2-4x-5$

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

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

$=\left[\begin{array}{cc}1+8& 2-6\\ 4-12& 8+9\end{array}\right]=\left[\begin{array}{cc}9& -4\\ -8& 17\end{array}\right]$

$4\cdot A=\left[\begin{array}{cc}4& 8\\ 16& -12\end{array}\right]$

$5I=\left[\begin{array}{cc}5& 0\\ 0& 5\end{array}\right]$

$f\left(A\right)=A^2+4A-5I=\left[\begin{array}{cc}9& -4\\ -8& 17\end{array}\right]+\left[\begin{array}{cc}4& 8\\ 16& -12\end{array}\right]-\left[\begin{array}{cc}5& 0\\ 0& 5\end{array}\right]=\left[\begin{array}{cc}13& 4\\ 8& 5\end{array}\right]-\left[\begin{array}{cc}5& 0\\ 0& 5\end{array}\right]$

$f\left(A\right)=\left[\begin{array}{cc}8& 4\\ 8& 0\end{array}\right]$