Matrices & Determinants
Inverse of a Matrix
Grade 12
Question:
<p>If \(A = \begin{bmatrix} a+ib & c+id \\ -c+id & a-ib \end{bmatrix}\) and \(a^2 + b^2 + c^2 + d^2 = 1\), then \(A^{-1}\) is equal to</p>
<p>\(\begin{bmatrix} a-ib & -c-id \\ c-id & a+ib \end{bmatrix}\)</p>
<p>\(\begin{bmatrix} a+ib & -c+id \\ -c+id & a-ib \end{bmatrix}\)</p>
<p>\(\begin{bmatrix} a-ib & -c-id \\ -c-id & a+ib \end{bmatrix}\)</p>
<p>none of these</p>
Step-by-Step Solution
Key Concept: Recognize that this matrix has a special Hermitian-like structure where A·A† = I (where A† is the conjugate transpose), making A unitary and thus A⁻¹ = A†.
<p><strong>Step 1:</strong> Compute A†(conjugate transpose):<br/>A† = <span style='border: 1px solid black; padding: 2px;'><strong>[a-ib -c-id]</strong></span><br/><span style='border: 1px solid black; padding: 2px;'><strong>[c-id a+ib]</strong></span></p><p><strong>Step 2:</strong> Verify AA† by multiplying:<br/>First row, first column: (a+ib)(a-ib) + (c+id)(-c-id) = (a²+b²) + (c²+d²) = 1<br/>First row, second column: (a+ib)(-c-id) + (c+id)(a+ib) = 0<br/>By symmetry, all off-diagonal entries = 0, diagonal entries = 1</p><p><strong>Step 3:</strong> Since AA† = I, the matrix A is unitary, therefore:<br/>A⁻¹ = A† = <span style='border: 1px solid black; padding: 2px;'><strong>[a-ib -c-id]</strong></span><br/><span style='border: 1px solid black; padding: 2px;'><strong>[c-id a+ib]</strong></span></p><p>∴ Answer: A</p>
Correct Answer: A