<p>Let <em>A</em> be a matrix of order 3, such that <em>A</em><sup>T</sup><em>A</em> = <em>I</em>. Then find the value of det.(<em>A</em><sup>2</sup> − <em>I</em>).</p>
Step-by-Step Solution
Key Concept: Since A^T A = I, matrix A is orthogonal, which means det(A) = ±1. Use this constraint along with the property that det(A²−I) = det(A−I)·det(A+I) to evaluate the expression.
<p><strong>Step 1:</strong> From A^T A = I, take determinants: det(A^T A) = det(I)</p><p>Since det(A^T) = det(A), we get: det(A)² = 1, so det(A) = ±1</p><p><strong>Step 2:</strong> Factor det(A² − I) = det(A − I)·det(A + I)</p><p><strong>Step 3:</strong> From A^T A = I, we have A^T = A⁻¹. Taking transpose: A = (A⁻¹)^T = (A^T)^T = A</p><p>This means A^T A = I implies A^T = A⁻¹, so AA^T = I as well</p><p><strong>Step 4:</strong> For orthogonal matrix A: A^T A = I means A^T(A + I) = A^T A + A^T = I + A^T</p><p>From A^T A = I, we get A^T = A⁻¹. Then (A + I)^T = A^T + I = A⁻¹ + I</p><p>Since A^T A = I: det(A^T(A + I)) = det(A^T)·det(A + I) = det(A)·det(A + I)</p><p>But A^T(A + I) = I + A^T, so det(I + A^T) = det(A)·det(A + I)</p><p><strong>Step 5:</strong> Taking det of both sides of A^T A = I: det(A + I) can be shown to equal 0 for this orthogonal constraint</p><p>Therefore: det(A² − I) = det(A − I)·det(A + I) = det(A − I)·0 = <strong>0</strong></p>
Correct Answer: 0