If left[begin{array}{rrr}1 & 2 & x 4 & -1 & | vedclass

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(C) matrix $A$ is singular if its determinant is zero,i.e.,$|A| = 0$.Given $A = \left[\begin{array}{rrr}1 & 2 & x \ 4 & -1 & 7 \ 2 & 4 & -6\end{arra

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$-3$

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(C) matrix $A$ is singular if its determinant is zero,i.e.,$|A| = 0$.Given $A = \left[\begin{array}{rrr}1 & 2 & x \ 4 & -1 & 7 \ 2 & 4 & -6\end{array}ight]$.Calculating the determinant along the first row:$|A| = 1((-1)(-6) - (7)(4)) - 2((4)(-6) - (7)(2)) + x((4)(4) - (-1)(2)) = 0$$|A| = 1(6 - 28) - 2(-24 - 14) + x(16 + 2) = 0$$|A| = 1(-22) - 2(-38) + x(18) = 0$$-22 + 76 + 18x = 0$$54 + 18x = 0$$18x = -54$$x = -3$

If $\left[\begin{array}{rrr}1 & 2 & x \\ 4 & -1 & 7 \\ 2 & 4 & -6\end{array}\right]$ is a singular matrix,then $x$ is equal to

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