<p>Given matrix \(B\) is the inverse of matrix \(A\), where \[B = \begin{bmatrix} 5 & 2\alpha & 1 \\ 0 & 2 & 1 \\ \alpha & 3 & -1 \end{bmatrix}\] and \(\det(A) + 1 = 0\). Find the sum of all values of \(\alpha\).</p>
Step-by-Step Solution
Key Concept: If B is the inverse of A, then det(A)·det(B) = 1. Since det(A) = -1, we have det(B) = -1. Expand det(B) as a function of α and solve the resulting equation.
<p><strong>Step 1:</strong> Use the given condition. Since $\det(A) + 1 = 0$, we have $\det(A) = -1$.</p><p><strong>Step 2:</strong> Apply the inverse matrix property. If $B = A^{-1}$, then $\det(A) \cdot \det(B) = 1$, so $\det(B) = \frac{1}{\det(A)} = \frac{1}{-1} = -1$.</p><p><strong>Step 3:</strong> Calculate $\det(B)$ using cofactor expansion along the first column:</p><p>$$\det(B) = 5 \cdot \det\begin{bmatrix} 2 & 1 \\ 3 & -1 \end{bmatrix} - 0 + \alpha \cdot \det\begin{bmatrix} 2\alpha & 1 \\ 2 & 1 \end{bmatrix}$$</p><p>$$= 5(-2-3) + \alpha(2\alpha - 2)$$</p><p>$$= 5(-5) + 2\alpha^2 - 2\alpha$$</p><p>$$= -25 + 2\alpha^2 - 2\alpha$$</p><p><strong>Step 4:</strong> Set $\det(B) = -1$:</p><p>$$-25 + 2\alpha^2 - 2\alpha = -1$$</p><p>$$2\alpha^2 - 2\alpha - 24 = 0$$</p><p>$$\alpha^2 - \alpha - 12 = 0$$</p><p>$$(\alpha - 4)(\alpha + 3) = 0$$</p><p>Therefore, $\alpha = 4$ or $\alpha = -3$.</p><p><strong>Step 5:</strong> Find the sum of all values: $4 + (-3) = \mathbf{1}$</p>
Correct Answer: 1