<p>If <em>A</em> is a nonsingular matrix such that \(AA^T = A^T A\) and \(B = A^{-1} A^T\), then matrix <em>B</em> is</p>
Step-by-Step Solution
Key Concept: If AA^T = A^T A, then A is normal. Since B = A^(-1)A^T and A is normal, we can show B is orthogonal by verifying BB^T = I using the commutative property of A and A^T.
<p><strong>Step 1:</strong> Given that AA^T = A^T A, matrix A is <strong>normal</strong>.</p><p><strong>Step 2:</strong> We need to find properties of B = A^(-1)A^T. Compute BB^T:</p><p>BB^T = (A^(-1)A^T)(A^(-1)A^T)^T = (A^(-1)A^T)(A(A^(-1))^T)</p><p><strong>Step 3:</strong> Since AA^T = A^T A, we have A^T = AA^T A^(-1). Using the normal property and A^(-1) being normal when A is normal:</p><p>BB^T = A^(-1)A^T A(A^(-1))^T = A^(-1)(A^T A)(A^(-1))^T = A^(-1)(AA^T)(A^(-1))^T</p><p><strong>Step 4:</strong> Simplifying: BB^T = A^(-1)A · A^T(A^(-1))^T = I · A^T(A^(-1))^T = A^T(A^T)^(-1) = I</p><p><strong>Step 5:</strong> Also B^T B = I follows similarly. Therefore B is <strong>orthogonal</strong>.</p><p>∴ Answer: B (B is an orthogonal matrix)</p>
Correct Answer: B