Question

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

• A. $A^2+7A-5I=0$
• B. $A^2-7A+5I=0$
• C. $A^2+5A-7I=0$
• D. $A^2-5A+7I=0$

Answer 1

The correct option is D

$A^2-5A+7I=0$

The explanation of the correct option :

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

$A^2=\left[\begin{array}{cc}3& 1\\ -1& 2\end{array}\right]\left[\begin{array}{cc}3& 1\\ -1& 2\end{array}\right]=\left[\begin{array}{cc}8& 5\\ -5& 3\end{array}\right]$

$5A=5\left[\begin{array}{cc}3& 1\\ -1& 2\end{array}\right]=\left[\begin{array}{cc}15& 5\\ -5& 10\end{array}\right]$

$7I=7\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]=\left[\begin{array}{cc}7& 0\\ 0& 7\end{array}\right]$

Now, $A^2-5A+7I=\left[\begin{array}{cc}8& 5\\ -5& 3\end{array}\right]-\left[\begin{array}{cc}15& 5\\ -5& 10\end{array}\right]+\left[\begin{array}{cc}7& 0\\ 0& 7\end{array}\right]$

$$\begin{gathered} =\left[\begin{array}{cc}8-15+7& 5-5\\ -5+5& 3-10+7\end{array}\right] \\ =\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right] \\ =0 \end{gathered}$$

Hence, the correct option is (D).