The minimum value of $\begin{vmatrix}2 & 2 & 2 \\ 2 & 2+x & 2 \\ 2 & 2 & 2+x\end{vmatrix}$, $x \in R$ is

Calculate the determinant $\begin{vmatrix}2 & 2 & 2 \\ 2 & 2+x & 2 \\ 2 & 2 & 2+x\end{vmatrix}$ where $x \in R$.

Using cofactor expansion along Row 1:

$= 2\begin{vmatrix}2+x & 2 \\ 2 & 2+x\end{vmatrix} - 2\begin{vmatrix}2 & 2 \\ 2 & 2+x\end{vmatrix} + 2\begin{vmatrix}2 & 2+x \\ 2 & 2\end{vmatrix}$

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Calculate each 2×2 determinant:

$\begin{vmatrix}2+x & 2 \\ 2 & 2+x\end{vmatrix}$

$= (2+x)(2+x) - (2)(2)$

$= (2+x)^2 - 4$

$\begin{vmatrix}2 & 2 \\ 2 & 2+x\end{vmatrix}$

$= 2(2+x) - (2)(2)$

$= 4 + 2x - 4$

$= 2x$

$\begin{vmatrix}2 & 2+x \\ 2 & 2\end{vmatrix}$

$= (2)(2) - (2)(2+x)$

$= 4 - 4 - 2x$

$= -2x$

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Substituting back:

$= 2[(2+x)^2 - 4] - 2[2x] + 2[-2x]$

$= 2(2+x)^2 - 8 - 4x - 4x$

$= 2(2+x)^2 - 8x - 8$

Expanding $(2+x)^2 = 4 + 4x + x^2$:

$= 2(4 + 4x + x^2) - 8x - 8$

$= 8 + 8x + 2x^2 - 8x - 8$

$= 2x^2$

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To minimize $f(x) = 2x^2$:

Since $x^2 \geq 0$ for all real numbers, the minimum value of $x^2$ is $0$ when $x = 0$.

Therefore, the minimum value of $2x^2 = 2(0) = 0$.

The minimum value is $0$.