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(C) matrix is singular if its determinant is equal to $0$. We calculate the determinant of $\Delta(x)$:$\det(\Delta(x)) = \begin{vmatrix} -x & x & 2 \

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$0$

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(C) matrix is singular if its determinant is equal to $0$. We calculate the determinant of $\Delta(x)$:$\det(\Delta(x)) = \begin{vmatrix} -x & x & 2 \ 2 & x & -x \ x & -2 & -x \end{vmatrix} = 0$Expanding along the first row:$-x(x(-x) - (-x)(-2)) - x(2(-x) - (-x)(x)) + 2(2(-2) - x(x)) = 0$$-x(-x^2 - 2x) - x(-2x + x^2) + 2(-4 - x^2) = 0$$x^3 + 2x^2 + 2x^2 - x^3 - 8 - 2x^2 = 0$$2x^2 - 8 = 0$$2x^2 = 8 \Rightarrow x^2 = 4 \Rightarrow x = \pm 2$Thus,there are $n = 2$ values of $x$ for which the matrix is singular.Now,we need to find $\det(\Delta(n)) = \det(\Delta(2))$.Substituting $x = 2$ into the matrix $\Delta(x)$:$\Delta(2) = \begin{bmatrix} -2 & 2 & 2 \ 2 & 2 & -2 \ 2 & -2 & -2 \end{bmatrix}$$\det(\Delta(2)) = -2(2(-2) - (-2)(-2)) - 2(2(-2) - (-2)(2)) + 2(2(-2) - 2(2))$$= -2(-4 - 4) - 2(-4 + 4) + 2(-4 - 4)$$= -2(-8) - 2(0) + 2(-8) = 16 - 0 - 16 = 0$.

If $n$ is the number of values of $x$ for which the matrix $\Delta(x) = \begin{bmatrix} -x & x & 2 \\ 2 & x & -x \\ x & -2 & -x \end{bmatrix}$ is singular,then find the value of $\det(\Delta(n))$.

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