If A = $\begin{bmatrix$}[r] 1 & 1 \ x & x \end{bmatrix}, find the value of x so that A² = O.

Given, A² = O.

$\text{or}, \begin{bmatrix$}[r] 1 & 1 \ x & x \end{bmatrix} \begin{bmatrix}[r] 1 & 1 \ x & x \end{bmatrix} = \begin{bmatrix}[r] 0 & 0 \ 0 & 0 \end{bmatrix} \\[1em] \text{L.H.S.} = \begin{bmatrix}[r] 1 & 1 \ x & x \end{bmatrix} \begin{bmatrix}[r] 1 & 1 \ x & x \end{bmatrix} \\[1em] = \begin{bmatrix}[r] 1 \times 1 + 1 \times x & 1 \times 1 + 1 \times x \ x \times 1 + x \times x & x \times 1 + x \times x \end{bmatrix} \\[1em] = \begin{bmatrix}[r] 1 + x & 1 + x \ x + x^2 & x + x^2 \end{bmatrix}

Comparing with R.H.S. we get,

⇒ 1 + x = 0 or x = -1      (…Eq 1)

⇒ x + x² = 0                    (…Eq 2)

Putting the value x = -1 in equation 2,

⇒ -1 + (-1)² = -1 + 1 = 0.

Since, x = -1 satisfies the equation,

Hence, the required value of x = -1.