Question

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

• A. $7I_2$
• B. $-7I_2$
• C. $5I_2$
• D. $-5I_2$

Answer 1

Answer: (B)

$-7I_2$

by Hemilton rule
each square matrix satisfy their characteristics equation so
$\left[\begin{array}{cc}3-\lambda & 1\\ -1& 2-\lambda \end{array}\right]=0$
$\Rightarrow 6-5\lambda +{\lambda }^2+1=0$
$\Rightarrow {\lambda }^2-5\lambda +7=0$
$\Rightarrow A^2-5A+7I=0$
$\Rightarrow A^2-5A=-7I_2$
Ans: B