Question

If $A=\left|\begin{array}{cc}1& -2\\ 4& 5\end{array}\right|$ and $f\left(t\right)=t^2-3t+7$, then $f\left(A\right)+\left|\begin{array}{cc}3& 6\\ -12& -9\end{array}\right|$ is equal to

• 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

Answer: (B)

$\left|\begin{array}{cc}0& 0\\ 0& 0\end{array}\right|$

Given that,
$A=\left|\begin{array}{cc}1& -2\\ 4& 5\end{array}\right|$ and $f\left(t\right)=t^2-3t+7$
Now, $A^2=\left|\begin{array}{cc}1& -2\\ 4& 5\end{array}\right|\left|\begin{array}{cc}1& -2\\ 4& 5\end{array}\right|$
$=\left|\begin{array}{cc}-7& -12\\ 24& 17\end{array}\right|$
Now, $f\left(A\right)=A^2-3A+7$
$=\left|\begin{array}{cc}-7& -12\\ 24& 17\end{array}\right|-3\left|\begin{array}{cc}1& -2\\ 4& 5\end{array}\right|+7\left|\begin{array}{cc}1& 0\\ 0& 1\end{array}\right|$
$=\left|\begin{array}{cc}-3& -6\\ 12& 9\end{array}\right|$
$\therefore f\left(A\right)+\left|\begin{array}{cc}3& 6\\ -12& -9\end{array}\right|=\left|\begin{array}{cc}-3& -6\\ 12& 9\end{array}\right|+\left|\begin{array}{cc}3& 6\\ -12& -9\end{array}\right|$
$=\left|\begin{array}{cc}0& 0\\ 0& 0\end{array}\right|$