<p>Let \(B = \dfrac{1}{10}\begin{bmatrix} 4 & 2 & 2 \\ -5 & 0 & \alpha \\ 1 & -2 & 3 \end{bmatrix}\) be the inverse of \(A = \begin{bmatrix} 1 & -1 & 1 \\ 2 & 1 & -3 \\ 1 & 1 & 1 \end{bmatrix}\). Then \(\alpha\) equals:</p>
Step-by-Step Solution
Key Concept: If B is the inverse of A, then AB = I. Use the property that the (i,j)-th entry of AB equals the dot product of the i-th row of A with the j-th column of B, and match coefficients to find α.
<p><strong>Step 1:</strong> Since B is the inverse of A, we have AB = I.</p><p><strong>Step 2:</strong> Compute the (1,3)-entry of AB:</p><p>Row 1 of A: [1, -1, 1] and Column 3 of B: [1/5, α/10, 3/10]</p><p>(1,3)-entry = 1·(1/5) + (-1)·(α/10) + 1·(3/10) = 1/5 - α/10 + 3/10 = (2 - α + 3)/10 = (5 - α)/10</p><p><strong>Step 3:</strong> Since AB = I, the (1,3)-entry must equal 0:</p><p>(5 - α)/10 = 0</p><p>5 - α = 0</p><p>∴ α = 5</p>
Correct Answer: 5