Question

If $A=\left[\begin{array}{cc}1& 2\\ 3& 4\end{array}\right]$, then $A^2-5A$ is equal to

• A. $2I$
• B. $-2I$
• C. $3I$
• D. Null matrix

Answer 1

The correct option is A $2I$

$A=\left[\begin{array}{cc}1& 2\\ 3& 4\end{array}\right],$

Also, $A^2=\left[\begin{array}{cc}1& 2\\ 3& 4\end{array}\right]\left[\begin{array}{cc}1& 2\\ 3& 4\end{array}\right]=\left[\begin{array}{c}1\times 1+2\times 31\times 2+2\times 4\\ 3\times 1+4\times 33\times 2+4\times 4\end{array}\right]$

$\Rightarrow A^2=\left[\begin{array}{cc}7& 10\\ 15& 22\end{array}\right]$

$5A=5\times \left[\begin{array}{cc}1& 2\\ 3& 4\end{array}\right]=\left[\begin{array}{cc}5& 10\\ 15& 20\end{array}\right]$

$A^2-5A=\left[\begin{array}{cc}7& 10\\ 15& 22\end{array}\right]-\left[\begin{array}{cc}5& 10\\ 15& 20\end{array}\right]=\left[\begin{array}{cc}2& 0\\ 0& +2\end{array}\right]=2\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$

$\therefore A^2-5A=2I$