Question

Given $A=\left[\begin{array}{cc}1& 3\\ 2& 2\end{array}\right]$, $I=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$ . If $A-\lambda I$ is a singular matrix then

• A. $\lambda =4$
• B. ${\lambda }^2-3\lambda -4=0$
• C. $\lambda =-1$
• D. ${\lambda }^2-3\lambda -6=0$

Answer 1

Answer: (A), (B), (C)

The correct options are
A ${\lambda }^2-3\lambda -4=0$
B $\lambda =-1$
C $\lambda =4$

$A-I\lambda$ is a singular matrix

$\Rightarrow A-I\lambda =0$

$\Rightarrow \left[\begin{array}{cc}1& 3\\ 2& 2\end{array}\right]-\lambda \left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]=0$

$\Rightarrow \left[\begin{array}{cc}1& 3\\ 2& 2\end{array}\right]-\left[\begin{array}{cc}\lambda & 0\\ 0& \lambda \end{array}\right]=0$

$\Rightarrow \left|\begin{array}{cc}1-\lambda & 3\\ 2& 2-\lambda \end{array}\right|=0$

$\Rightarrow (1-\lambda )(2-\lambda )-6=0$

$\Rightarrow 2-\lambda -2\lambda +{\lambda }^2-6=0$

$\Rightarrow {\lambda }^2-3\lambda -4=0$

$\Rightarrow {\lambda }^2-4\lambda +\lambda -4=0$

$\Rightarrow \lambda (\lambda -4)+1(\lambda -4)=0$

$\Rightarrow (\lambda -4)(\lambda +1)=0$

$\therefore \lambda =4,-1$