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(D) matrix $A$ has no inverse if and only if its determinant is zero,i.e.,$|A| = 0$.Given $A = \left[\begin{array}{ccc}1 & -1 & x \ 1 & x & 1 \ x &

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$1$

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(D) matrix $A$ has no inverse if and only if its determinant is zero,i.e.,$|A| = 0$.Given $A = \left[\begin{array}{ccc}1 & -1 & x \ 1 & x & 1 \ x & -1 & 1\end{array}ight]$.Calculating the determinant along the first row:$|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$$-x^3 - x + 2 = 0$$x^3 + x - 2 = 0$By inspection,$x = 1$ is a root because $1^3 + 1 - 2 = 0$.Dividing $x^3 + x - 2$ by $(x - 1)$,we get $(x - 1)(x^2 + x + 2) = 0$.For $x^2 + x + 2 = 0$,the discriminant $D = b^2 - 4ac = 1^2 - 4(1)(2) = 1 - 8 = -7$.Since $D Thus,the only real value is $x = 1$.

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

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