3 and 4 .Determinants and Matrices
easy

If $A = \left[ {\begin{array}{*{20}{c}}\lambda &1\\{ - 1}&{ - \lambda }\end{array}} \right]$, then for what value of $\lambda ,\,{A^2} = O$

A

$0$

B

$ \pm {\rm{ }}1$

C

$-1$

D

$1$

Solution

(b)  ${A^2} = A\,.\,A = \left[ {\begin{array}{*{20}{c}}\lambda &1\\{ – 1}&{ – \lambda }\end{array}} \right]\,\left[ {\begin{array}{*{20}{c}}\lambda &1\\{ – 1}&{ – \lambda }\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{{\lambda ^2} – 1}&0\\0&{ – 1 + {\lambda ^2}}\end{array}} \right] = 0$

(As given)

==> ${\lambda ^2} – 1 = 0 \Rightarrow \lambda = \pm 1$.

Standard 12
Mathematics

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