Question

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

Answer: (D)

Given : $A=\left[\begin{array}{cc}1& 2\\ 4& -3\end{array}\right]$
$\therefore A^2=A.A=\left[\begin{array}{cc}1& 2\\ 4& -3\end{array}\right]\left[\begin{array}{cc}1& 2\\ 4& -3\end{array}\right]$
$=\left[\begin{array}{cc}1+8& 2-6\\ 4-12& 8+9\end{array}\right]=\left[\begin{array}{cc}9& -4\\ -8& 17\end{array}\right]$
Now, $f\left(x\right)=x^2+4x-5$
$\therefore f\left(A\right)=A^2+4A-5$
$=A^2+4A-5I$ (I is a $2\times 2$unit matrix)
$=\left[\begin{array}{cc}9& -4\\ -8& 17\end{array}\right]+4\left[\begin{array}{cc}1& 2\\ 4& -3\end{array}\right]-5\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$
$=\left[\begin{array}{cc}9& -4\\ -8& 17\end{array}\right]+\left[\begin{array}{cc}4& 8\\ 16& -12\end{array}\right]+\left[\begin{array}{cc}-5& 0\\ 0& -5\end{array}\right]$
$=\left[\begin{array}{cc}8& 4\\ 8& 0\end{array}\right]$