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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