3 and 4 .Determinants and Matrices
easy

यदि $A = \left[ {\begin{array}{*{20}{c}}x&1\\1&0\end{array}} \right]$और ${A^2}$ तत्समक आव्यूह हैं, तो $x =$

A

$1$

B

$2$

C

$3$

D

$0$

Solution

(d) $A = \left[ {\begin{array}{*{20}{c}}x&1\\1&0\end{array}} \right],\therefore {A^2} = I \Rightarrow \left[ {\begin{array}{*{20}{c}}x&1\\1&0\end{array}} \right]\,\left[ {\begin{array}{*{20}{c}}x&1\\1&0\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right]$

==> $\left[ {\begin{array}{*{20}{c}}{{x^2} + 1}&x\\x&1\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right] \Rightarrow {x^2} + 1 = 1 \Rightarrow x = 0$.

Standard 12
Mathematics

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