Given A = [ * 20 c 1&3 2&2 ] ; I = [ * 20 c 1&0 0&1 ] . If A - I … - Vidyadip

MCQ

Given $A =$$\left[ {\begin{array}{*{20}{c}}1&3\\2&2\end{array}} \right]$ ; $I =$$\left[ {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right]$ . $If A - \lambda I$ is a singular matrix then

A
$\lambda \in \phi$
✓
$\lambda ^2 - 3\lambda - 4 = 0$
C
$\lambda ^2 + 3\lambda + 4 = 0$
D
$\lambda ^2 - 3\lambda - 6 = 0$

✓

Answer

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 $\lambda ^2 - 3\lambda - 4 = 0$

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