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(D) Let $A = \begin{bmatrix} \alpha & 14 & -1 \ 2 & 3 & 1 \ 6 & 2 & 3 \end{bmatrix}$.The inverse of a matrix $A$ does not exist if and only if the d

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$-2$

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(D) Let $A = \begin{bmatrix} \alpha & 14 & -1 \ 2 & 3 & 1 \ 6 & 2 & 3 \end{bmatrix}$.The inverse of a matrix $A$ does not exist if and only if the determinant of the matrix is zero,i.e.,$|A| = 0$.Calculating the determinant $|A|$:$|A| = \alpha(3 	imes 3 - 1 	imes 2) - 14(2 	imes 3 - 1 	imes 6) + (-1)(2 	imes 2 - 3 	imes 6)$$|A| = \alpha(9 - 2) - 14(6 - 6) - 1(4 - 18)$$|A| = \alpha(7) - 14(0) - 1(-14)$$|A| = 7\alpha + 14$Setting $|A| = 0$ for the inverse to not exist:$7\alpha + 14 = 0$$7\alpha = -14$$\alpha = -2$Thus,the value of $\alpha$ is $-2$.

If the inverse of the matrix $\begin{bmatrix} \alpha & 14 & -1 \\ 2 & 3 & 1 \\ 6 & 2 & 3 \end{bmatrix}$ does not exist,then the value of $\alpha$ is:

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