Question

If $A=\left[\begin{array}{cc}4& 2\\ -1& 1\end{array}\right]$, then $(A-2I)(A-3I)=$

• A. $0$
• B. $A$
• C. $I$
• D. $5I$

Answer 1

Answer: (A)

$0$

$A=\left[\begin{array}{cc}4& 2\\ -1& 1\end{array}\right]$
$\Rightarrow$ $(A-2I)(A-3I)=A^2-5AI+6I^2$
$A^2-5A+6I$
by Hemilton rule $\left|\begin{array}{cc}4-\lambda & 2\\ -1& 1-\lambda \end{array}\right|=0$

after expanding determinant we got;
$\Rightarrow 4-5\lambda +{\lambda }^2+2=0$
$\Rightarrow {\lambda }^2-5\lambda +6=0$
$\Rightarrow A^2-5A+6I=0$
Ans: 0

Or

It can be done using scalar multiplication and followed by matrix multiplication.