<p>If \(k \in R_0\), then \(\det(\text{adj}(kI_n))\) is equal to</p>
Step-by-Step Solution
Key Concept: The adjugate of a scalar multiple of identity matrix kIₙ is (k^(n-1))Iₙ, and its determinant follows the property det(adj(A)) = (det(A))^(n-1) for invertible matrices.
<p><strong>Step 1:</strong> Find adj(kIₙ). Since kIₙ is a scalar multiple of identity, all diagonal entries equal k and all off-diagonal entries are 0.</p><p><strong>Step 2:</strong> Use the property: adj(kA) = k^(n-1)·adj(A) for any n×n matrix A. Therefore, adj(kIₙ) = k^(n-1)·adj(Iₙ) = k^(n-1)·Iₙ.</p><p><strong>Step 3:</strong> Calculate the determinant: det(adj(kIₙ)) = det(k^(n-1)·Iₙ) = (k^(n-1))^n = k^(n(n-1)).</p><p><strong>Alternatively (direct formula):</strong> Use det(adj(A)) = (det(A))^(n-1). Here det(kIₙ) = k^n, so det(adj(kIₙ)) = (k^n)^(n-1) = k^(n(n-1)).</p><p>∴ Answer: <strong>k^(n(n-1))</strong> or <strong>k^(n² - n)</strong></p>
Correct Answer: B