Correct option: B.$\lambda ^2 - 3\lambda - 4 = 0$
b
$A - \lambda I$
$=$ $\left[ {\begin{array}{*{20}{c}}1&3\\2&2\end{array}} \right]$ - $\left[ {\begin{array}{*{20}{c}}\lambda &0\\0&\lambda\end{array}} \right]$ $=$ $\left[ {\begin{array}{*{20}{c}}{1 - \lambda }&3\\2&{2 - \lambda }\end{array}} \right]$
$= (1 - \lambda ) (2 - \lambda ) = \lambda ^2 - 3\lambda + 2 = 0$
i.e. for $A - \lambda I$ to be singular $\lambda ^ 2 - 3\lambda + 2 = 0$
since $A - \lambda I$ is singular ==> det. $(A - \lambda I)$ $= 0$
.hence $\left[ {\begin{array}{*{20}{c}}{1 - \lambda }&3\\2&{2 - \lambda }\end{array}} \right]$ $= 0$
$==> 2 - \lambda - 2\lambda + \lambda ^2 - 6 = 0$ or $\lambda ^2 - 3\lambda - 4 = 0$