Question

lf $A=\left[\begin{array}{cc}1& -2\\ 4& 5\end{array}\right]$ and $f\left(t\right)=t^2-3t+7$ then $f\left(A\right)+\left[\begin{array}{cc}3& 6\\ -12& -9\end{array}\right]=$

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

Answer 1

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

Given, $A=\left[\begin{array}{cc}1& -2\\ 4& 5\end{array}\right]$
$f\left(t\right)=t^2-3t+7$
$\Rightarrow f\left(A\right)=A^2-3A+7I$

Now, $A^2-3A+7I=\left[\begin{array}{cc}1& -2\\ 4& 5\end{array}\right]\left[\begin{array}{cc}1& -2\\ 4& 5\end{array}\right]-3\left[\begin{array}{cc}1& -2\\ 4& 5\end{array}\right]+7\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$
$=\left[\begin{array}{cc}-7& -12\\ 24& 17\end{array}\right]+\left[\begin{array}{cc}-3& 6\\ -12& -15\end{array}\right]+\left[\begin{array}{cc}7& 0\\ 0& 7\end{array}\right]$
$\Rightarrow f\left(A\right)=\left[\begin{array}{cc}-3& -6\\ 12& 9\end{array}\right]$
Now, $f\left(A\right)+\left[\begin{array}{cc}3& 6\\ -12& -9\end{array}\right]$
$=\left[\begin{array}{cc}-3& -6\\ 12& 9\end{array}\right]+\left[\begin{array}{cc}3& 6\\ -12& -9\end{array}\right]$
$=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]$