If $\begin{bmatrix}x&8\\4&x\end{bmatrix}=\begin{bmatrix}6&2\\18&6\end{bmatrix}$, then x is/are equal to

$±4\sqrt{2}$

The correct answer is Option (1) → $±4\sqrt{2}$

Left determinant:

$\det\begin{bmatrix} x & 8 \\ 4 & x \end{bmatrix} = x^2 - 32$

Right determinant:

$\det\begin{bmatrix} 6 & 2 \\ 18 & 6 \end{bmatrix} = 36 - 36 = 0$

Equate:

$x^2 - 32 = 0$

$x^2 = 32$

$x = \pm 4\sqrt{2}$

$x = \pm 4\sqrt{2}$