If left[begin{array}{rrr}1 & -1 & x 1 & x & | vedclass

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(D) matrix has no inverse if and only if its determinant is equal to $0$. Let $A = \left[\begin{array}{rrr}1 & -1 & x \ 1 & x & 1 \ x & -1 & 1\end{a

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$1$

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(D) matrix has no inverse if and only if its determinant is equal to $0$. Let $A = \left[\begin{array}{rrr}1 & -1 & x \ 1 & x & 1 \ x & -1 & 1\end{array}ight]$. We set $|A| = 0$: $|A| = 1(x - (-1)) - (-1)(1 - x) + x(-1 - x^2) = 0$ $|A| = 1(x + 1) + 1(1 - x) + x(-1 - x^2) = 0$ $|A| = x + 1 + 1 - x - x - x^3 = 0$ $|A| = -x^3 - x + 2 = 0$ $x^3 + x - 2 = 0$ By inspection,if $x = 1$,then $1^3 + 1 - 2 = 0$,which satisfies the equation. Alternatively,if $x = 1$,the matrix becomes $\left[\begin{array}{rrr}1 & -1 & 1 \ 1 & 1 & 1 \ 1 & -1 & 1\end{array}ight]$. Since the first and third columns are identical,the determinant is $0$. Thus,the real value of $x$ is $1$.

If $\left[\begin{array}{rrr}1 & -1 & x \\ 1 & x & 1 \\ x & -1 & 1\end{array}\right]$ has no inverse,then the real value of $x$ is

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