If | cases x - 4 & 2x & 2x 2x & x - 4 & 2x 2x & 2x & x - 4 cases |...

\left|\begin{cases}x - 4 & 2x & 2x \\ 2x & x - 4 & 2x \\ 2x & 2x & x - 4\end{cases}\right| = \left(A + Bx\right){\left(x - A\right)}^{2}

Put x = 0 ⇒ \left|\begin{cases}- 4 & 0 & 0 \\ 0 & - 4 & 0 \\ 0 & 0 & - 4\end{cases}\right| = {A}^{3} ⇒ A = - 4

\left|$\begin{cases}x - 4 & 2x & 2x \\ 2x & x - 4 & 2x \\ 2x & 2x & x - 4\end{cases}$\right| = \left(Bx - 4\right){\left(x + 4\right)}^{2}<br /><br />\left|\begin{cases}1 - \dfrac{4}{x} & 2 & 2 \\ 2 & 1 - \dfrac{4}{x} & 2 \\ 2 & 2 & 1 - \dfrac{4}{x}\end{cases}\right| = \left(B - \dfrac{4}{x}\right){\left(1 + \dfrac{4}{x}\right)}^{2}

Put x \to \infty ⇒ \left|\begin{cases}1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1\end{cases}\right| = B

On expanding the determinant along the first row, we get $B = 5$.