If B = begin{bmatrix} 5 & 2alpha & 1 0 & 2 & | vedclass

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(C) Given that $B = A^{-1}$,we know that $\det(B) = \frac{1}{\det(A)}$.First,calculate the determinant of matrix $B$:$\det(B) = 5(2(-1) - 3(1)) - 2\al

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(C) Given that $B = A^{-1}$,we know that $\det(B) = \frac{1}{\det(A)}$.First,calculate the determinant of matrix $B$:$\det(B) = 5(2(-1) - 3(1)) - 2\alpha(0(-1) - \alpha(1)) + 1(0(3) - \alpha(2))$$\det(B) = 5(-2 - 3) - 2\alpha(-\alpha) + 1(-2\alpha)$$\det(B) = 5(-5) + 2\alpha^2 - 2\alpha = 2\alpha^2 - 2\alpha - 25$.Given the condition $\det(A) + 1 = 0$,we have $\det(A) = -1$.Since $\det(B) = \frac{1}{\det(A)}$,we have $\det(B) = \frac{1}{-1} = -1$.Equating the two expressions for $\det(B)$:$2\alpha^2 - 2\alpha - 25 = -1$$2\alpha^2 - 2\alpha - 24 = 0$Dividing by $2$:$\alpha^2 - \alpha - 12 = 0$$(\alpha - 4)(\alpha + 3) = 0$.The values of $\alpha$ are $4$ and $-3$.The sum of these values is $4 + (-3) = 1$.

If $B = \begin{bmatrix} 5 & 2\alpha & 1 \\ 0 & 2 & 1 \\ \alpha & 3 & -1 \end{bmatrix}$ is the inverse of a $3 \times 3$ matrix $A$,then the sum of all values of $\alpha$ for which $\det(A) + 1 = 0$ is:

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