MCQ
If $A = \left[ {\begin{array}{*{20}{c}}x&1\\1&0\end{array}} \right]$ and ${A^2}$ is the identity matrix, then $x =$
- A$1$
- B$2$
- C$3$
- ✓$0$
✓
Answer
Correct option: D.
$0$
d
(d) $A = \left[ {\begin{array}{*{20}{c}}x&1\\1&0\end{array}} \right],\therefore {A^2} = I \Rightarrow \left[ {\begin{array}{*{20}{c}}x&1\\1&0\end{array}} \right]\,\left[ {\begin{array}{*{20}{c}}x&1\\1&0\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right]$
(d) $A = \left[ {\begin{array}{*{20}{c}}x&1\\1&0\end{array}} \right],\therefore {A^2} = I \Rightarrow \left[ {\begin{array}{*{20}{c}}x&1\\1&0\end{array}} \right]\,\left[ {\begin{array}{*{20}{c}}x&1\\1&0\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right]$
==> $\left[ {\begin{array}{*{20}{c}}{{x^2} + 1}&x\\x&1\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right] \Rightarrow {x^2} + 1 = 1 \Rightarrow x = 0$.
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