<p>Let \(A = \begin{bmatrix}1 & -1 & 1\\ 2 & 1 & -3\\ 1 & 1 & 1\end{bmatrix}\) and \(10B = \begin{bmatrix}4 & 2 & 2\\ -5 & 0 & \alpha\\ 1 & -2 & 3\end{bmatrix}\). If \(B\) is the inverse of \(A\), then find the value of \(\alpha\).</p>
Step-by-Step Solution
Key Concept: If B is the inverse of A, then AB = I. Use the matrix multiplication condition to find α by computing the product and equating to identity matrix.
<p><strong>Step 1:</strong> Since B is the inverse of A, we have AB = I.</p><p><strong>Step 2:</strong> Given 10B = [4, 2, 2; -5, 0, α; 1, -2, 3], so B = [0.4, 0.2, 0.2; -0.5, 0, α/10; 0.1, -0.2, 0.3]</p><p><strong>Step 3:</strong> Compute AB and equate to I. Focus on the (1,3) entry of AB:</p><p>AB₁₃ = 1(0.2) + (-1)(α/10) + 1(0.3) = 0.2 - α/10 + 0.3 = 0.5 - α/10</p><p>Since AB = I, we need AB₁₃ = 0, so: 0.5 - α/10 = 0</p><p><strong>Step 4:</strong> Solving: α/10 = 0.5 → α = 5</p><p><strong>Verification:</strong> Check (2,3) entry: 2(0.2) + 1(α/10) + (-3)(0.3) = 0.4 + 0.5 - 0.9 = 0 ✓</p><p>∴ Answer: <strong>5</strong></p>
Correct Answer: 5