The values of t such that the matrix begin{bm | vedclass

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(C) matrix $A$ has no inverse if and only if its determinant is zero,i.e.,$|A| = 0$.Given $A = \begin{bmatrix} 1 & 3 & 2 \ 2 & 5 & t \ 4 & 7-t & -6

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$-3, 2$

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(C) matrix $A$ has no inverse if and only if its determinant is zero,i.e.,$|A| = 0$.Given $A = \begin{bmatrix} 1 & 3 & 2 \ 2 & 5 & t \ 4 & 7-t & -6 \end{bmatrix}$.Calculating the determinant $|A|$:$|A| = 1[5(-6) - t(7-t)] - 3[2(-6) - 4(t)] + 2[2(7-t) - 4(5)] = 0$$|A| = 1[-30 - 7t + t^2] - 3[-12 - 4t] + 2[14 - 2t - 20] = 0$$|A| = t^2 - 7t - 30 + 36 + 12t + 28 - 4t - 40 = 0$$|A| = t^2 + t - 6 = 0$Factoring the quadratic equation:$t^2 + 3t - 2t - 6 = 0$$t(t+3) - 2(t+3) = 0$$(t-2)(t+3) = 0$Thus,the values of $t$ are $t = 2$ and $t = -3$.

The values of $t$ such that the matrix $\begin{bmatrix} 1 & 3 & 2 \\ 2 & 5 & t \\ 4 & 7-t & -6 \end{bmatrix}$ has no inverse,are

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