<p>If <em>n</em>th-order square matrix <em>A</em> is an orthogonal, then \(|\text{adj}(\text{adj } A)|\) is</p>
Step-by-Step Solution
Key Concept: For an orthogonal matrix A of order n: A·A^T = I, so |A| = ±1. The adjugate satisfies |adj(A)| = |A|^(n-1), and applying this recursively: |adj(adj A)| = |adj A|^(n-1) = (|A|^(n-1))^(n-1) = (±1)^(n-1)² = 1.
<p><strong>Step 1:</strong> Recall that for an orthogonal matrix A of order n: AA<sup>T</sup> = I, which implies |A| = ±1.</p><p><strong>Step 2:</strong> Use the determinant formula for adjugate matrices: |adj(B)| = |B|<sup>n-1</sup> for any n×n matrix B.</p><p><strong>Step 3:</strong> Apply this formula to A: |adj A| = |A|<sup>n-1</sup> = (±1)<sup>n-1</sup> = ±1.</p><p><strong>Step 4:</strong> Now apply the formula again to adj A (which is also an n×n matrix): |adj(adj A)| = |adj A|<sup>n-1</sup> = (±1)<sup>n-1</sup>.</p><p><strong>Step 5:</strong> Since (±1)<sup>n-1</sup> = ±1 when n-1 is any positive integer, but more precisely: |adj(adj A)| = [(±1)<sup>n-1</sup>]<sup>n-1</sup> = (±1)<sup>(n-1)²</sup> = 1 (as (n-1)² is always even when n ≠ 1).</p><p><strong>Note:</strong> For n ≥ 2, |adj(adj A)| = <strong>1</strong>.</p><p>∴ Answer: D</p>
Correct Answer: D