<p><strong>66.</strong> 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 ______.</p>
Step-by-Step Solution
Key Concept: Since B is the inverse of A, we have det(B) · det(A) = 1, so det(A) = 1/det(B). The condition det(A) + 1 = 0 means det(A) = -1, which gives det(B) = -1. Calculate det(B) and solve for α.
<p><strong>Step 1:</strong> Use the relationship between determinants of inverse matrices.</p><p>Since B = A⁻¹, we have: det(B) · det(A) = 1</p><p>Therefore: det(A) = 1/det(B)</p><p><strong>Step 2:</strong> Apply the given condition det(A) + 1 = 0.</p><p>det(A) = -1</p><p>So: 1/det(B) = -1, which means det(B) = -1</p><p><strong>Step 3:</strong> Calculate det(B) using cofactor expansion along Row 1.</p><p>det(B) = 5(2·(-1) - 1·3) - 2α(0·(-1) - 1·α) + 1(0·3 - 2·α)</p><p>= 5(-2 - 3) - 2α(0 - α) + 1(0 - 2α)</p><p>= 5(-5) - 2α(-α) + (-2α)</p><p>= -25 + 2α² - 2α</p><p><strong>Step 4:</strong> Set det(B) = -1 and solve for α.</p><p>-25 + 2α² - 2α = -1</p><p>2α² - 2α - 24 = 0</p><p>α² - α - 12 = 0</p><p>(α - 4)(α + 3) = 0</p><p>α = 4 or α = -3</p><p><strong>Step 5:</strong> Find the sum of all values of α.</p><p>Sum = 4 + (-3) = 1</p><p>∴ Answer: <strong>1</strong></p>
Correct Answer: 1