Matrices & Determinants
Orthogonal Matrices
Grade 12
Question:
<p>If both \(A - \dfrac{I}{2}\) and \(A + \dfrac{I}{2}\) are orthogonal matrix, then which of the following statements are <strong>incorrect</strong>? (where \(I\) is an identity matrix order same as that of \(A\).)</p>
<p>\(A\) is skew-symmetric matrix of odd order.</p>
<p>\(A^2 = \dfrac{3}{4}I\)</p>
<p>\(A\) is skew-symmetric matrix of even order.</p>
<p>\(A\) is orthogonal</p>
Step-by-Step Solution
Key Concept: For a matrix to be orthogonal, M^T M = I. Apply this condition to both (A - I/2) and (A + I/2) simultaneously, then subtract the resulting equations to derive constraints on A.
<p><strong>Step 1:</strong> Since (A - I/2) is orthogonal: (A - I/2)^T(A - I/2) = I</p><p>Expanding: A^T A - (1/2)A^T - (1/2)A + I/4 = I</p><p>Simplifying: A^T A - (1/2)(A^T + A) = 3I/4 ... (1)</p><p><strong>Step 2:</strong> Since (A + I/2) is orthogonal: (A + I/2)^T(A + I/2) = I</p><p>Expanding: A^T A + (1/2)A^T + (1/2)A + I/4 = I</p><p>Simplifying: A^T A + (1/2)(A^T + A) = 3I/4 ... (2)</p><p><strong>Step 3:</strong> Subtract equation (1) from equation (2):</p><p>(A^T + A) = 0, so A^T = -A (A is skew-symmetric)</p><p><strong>Step 4:</strong> Substitute A^T = -A into equation (1):</p><p>A^T A - 0 = 3I/4</p><p>Since A^T A = AA^T for any matrix and A^T = -A: (-A)A = 3I/4</p><p>This gives -A² = 3I/4, or A² = -3I/4</p><p><strong>Step 5:</strong> This is impossible for real matrices since A² must be positive semi-definite for real A, but -3I/4 is negative definite. Therefore, no such real matrix A exists that satisfies both conditions simultaneously.</p><p>∴ Any statement claiming such a matrix exists or has specific properties is <strong>incorrect</strong>.</p>
Correct Answer: A