Question

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

Answer: (D)

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

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

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

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

We find (d) satisfy the condition as

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

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

Thus,$A^2-5A-7I=0$ is correct