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3 and 4 .Determinants and Matrices
easy
यदि $A = \left[ {\begin{array}{*{20}{c}}\lambda &1\\{ - 1}&{ - \lambda }\end{array}} \right]$, तो $\lambda $ के किस मान के लिये$\,{A^2} = 0$ है
A
$0$
B
$ \pm {\rm{ }}1$
C
$-1$
D
$1$
Solution
(b) ${A^2} = A\,.\,A = \left[ {\begin{array}{*{20}{c}}\lambda &1\\{ – 1}&{ – \lambda }\end{array}} \right]\,\left[ {\begin{array}{*{20}{c}}\lambda &1\\{ – 1}&{ – \lambda }\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{{\lambda ^2} – 1}&0\\0&{ – 1 + {\lambda ^2}}\end{array}} \right] = 0$,
(दिये गए अनुसार)
${\lambda ^2} – 1 = 0 \Rightarrow \lambda = \pm 1$.
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Standard 12
Mathematics
