<p>First row of a matrix <em>A</em> is [1 3 2]. If <br> adj. \(A = \begin{bmatrix} -2 & 4 & \alpha \\ -1 & 2 & 1 \\ 3\alpha & -5 & -2 \end{bmatrix}\), then det.(A) is</p>
Step-by-Step Solution
Key Concept: Use the property that A·(adj A) = (det A)·I to match corresponding elements. The first row of A multiplied by columns of adj(A) gives multiples of det(A), which allows you to find α and then det(A).
<p><strong>Step 1:</strong> Use the property A·(adj A) = (det A)·I</p><p>The first row of A is [1, 3, 2]. When we multiply this by the first column of adj(A), we get:</p><p>1(-2) + 3(-1) + 2(3α) = det(A)</p><p>-2 - 3 + 6α = det(A)</p><p>6α - 5 = det(A) ... (equation 1)</p><p><strong>Step 2:</strong> Multiply first row of A by second column of adj(A):</p><p>1(4) + 3(2) + 2(-5) = 0 (off-diagonal element of A·adj(A) must be 0)</p><p>4 + 6 - 10 = 0 ✓</p><p><strong>Step 3:</strong> Multiply first row of A by third column of adj(A):</p><p>1(α) + 3(1) + 2(-2) = 0</p><p>α + 3 - 4 = 0</p><p>α = 1</p><p><strong>Step 4:</strong> Substitute α = 1 into equation 1:</p><p>det(A) = 6(1) - 5 = 1</p><p>∴ Answer: det(A) = 1</p>
Correct Answer: A