If A = […] and A = A^-1 , then the value of x is:

The correct answer is Option (1) → 0

Multiplying both sides of \(\mathrm{A} = \mathrm{A}^{-1}\) by A gives

    \(\mathrm{A}^2 = \mathrm{A}\mathrm{A}^{-1} = \mathrm{I}\)

Now

    \( \displaystyle \mathrm{A}^2 = \begin{bmatrix} x & 1 \\ 1 & 0 \end{bmatrix}\begin{bmatrix} x & 1 \\ 1 & 0 \end{bmatrix} = \begin{bmatrix} x^2 + 1 & x \\ x & 1 \end{bmatrix} \)

By equality of matrices with \(\mathrm{I} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\):

    \(x = 0\) (row 1, column 2)
    \(x^2 + 1 = 1\), which also gives \(x = 0\)

With x = 0, |A| = 0 − 1 = −1 ≠ 0, so A is indeed invertible.