If A = begin{bmatrix} 0 & x & 16 x & 5 & 7 | vedclass

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(A) matrix $A$ is singular if its determinant $|A| = 0$.Given $A = \begin{bmatrix} 0 & x & 16 \ x & 5 & 7 \ 0 & 9 & x \end{bmatrix}$.Expanding the d

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$0, 12, -12$

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(A) matrix $A$ is singular if its determinant $|A| = 0$.Given $A = \begin{bmatrix} 0 & x & 16 \ x & 5 & 7 \ 0 & 9 & x \end{bmatrix}$.Expanding the determinant along the first column:$|A| = 0(5x - 63) - x(x^2 - 0) + 0(7x - 80) = 0$.$-x(x^2) = 0$.Wait,let us re-evaluate the expansion along the first column correctly:$|A| = 0 \cdot \begin{vmatrix} 5 & 7 \ 9 & x \end{vmatrix} - x \cdot \begin{vmatrix} x & 16 \ 9 & x \end{vmatrix} + 0 \cdot \begin{vmatrix} x & 16 \ 5 & 7 \end{vmatrix} = 0$.$-x(x^2 - 144) = 0$.$-x(x - 12)(x + 12) = 0$.Therefore,the possible values of $x$ are $0, 12, -12$.

If $A = \begin{bmatrix} 0 & x & 16 \\ x & 5 & 7 \\ 0 & 9 & x \end{bmatrix}$ is a singular matrix,then the possible values of $x$ are:

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