Question

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

• 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

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

Explanation for the correct option:

Step 1. Find the value of $f\left(A\right)+\left[\begin{array}{cc}3& 6\\ -12& 9\end{array}\right]$

Given, $A=\left[\begin{array}{cc}1& -2\\ 4& 5\end{array}\right]$

$f\left(t\right)=t^2-3t+7$ ..….(1)

Step 2. Put $A$ at the place of $t$ in equation (1):

$f\left(A\right)=A^2-3A+7I$ ……(2)

$\begin{array}{rcl}{A}^2& =& \left[\begin{array}{cc}1& -2\\ 4& 5\end{array}\right]\left[\begin{array}{cc}1& -2\\ 4& 5\end{array}\right]\\ & =& \left[\begin{array}{cc}1\times 1+\left(-2\right)\times 4& 1\times -2+\left(-2\right)\times 5\\ 4\times 1+5\times 4& 4\times \left(-2\right)+5\times 5\end{array}\right]\\ & =& \left[\begin{array}{cc}-7& -12\\ 24& 17\end{array}\right]\end{array}$

$\begin{array}{rcl}3A& =& 3\left[\begin{array}{cc}1& -2\\ 4& 5\end{array}\right]\\ & =& \left[\begin{array}{cc}3& -6\\ 12& 15\end{array}\right]\end{array}$

Step 3. Put the value of $A^2$ and $3A$ in equation (2):

$\begin{array}{rcl}{A}^2-3A+7I& =& \left[\begin{array}{cc}-7& -12\\ 24& 17\end{array}\right]-\left[\begin{array}{cc}3& -6\\ 12& 15\end{array}\right]+7\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]\\ & =& \left[\begin{array}{cc}-7-3& -12+6\\ 24-12& 17-15\end{array}\right]+\left[\begin{array}{cc}7& 0\\ 0& 7\end{array}\right]\\ & =& \left[\begin{array}{cc}-10& -6\\ 12& 2\end{array}\right]+\left[\begin{array}{cc}7& 0\\ 0& 7\end{array}\right]\\ & =& \left[\begin{array}{cc}-10+7& -6+0\\ 12+0& 2+7\end{array}\right]\\ & =& \left[\begin{array}{cc}-3& -6\\ 12& 9\end{array}\right]\end{array}$

Step 4. Put the value of $f\left(A\right)$ in given expression:

$\begin{array}{rcl}\therefore f\left(A\right)+\left[\begin{array}{cc}3& 6\\ -12& -9\end{array}\right]& =& \left[\begin{array}{cc}-3& -6\\ 12& 9\end{array}\right]+\left[\begin{array}{cc}3& 6\\ -12& 9\end{array}\right]\\ & =& \left[\begin{array}{cc}-3+3& -6+6\\ 12-12& -9+9\end{array}\right]\\ & =& \left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]\end{array}$

Hence, Option ‘B’ is Correct.