Matrices & Determinants
Invertible Matrices
Grade 12
Question:
<p>If \(A(\theta) = \begin{bmatrix} \sin\theta & i\cos\theta \\ i\cos\theta & \sin\theta \end{bmatrix}\), then which of the following is not true?</p>
<p>\(A(\theta)^{-1} = A(\pi - \theta)\)</p>
<p>\(A(\theta) + A(\pi + \theta)\) is a null matrix</p>
<p>\(A(\theta)\) is invertible for all \(\theta \in R\)</p>
<p>\(A(\theta)^{-1} = A(-\theta)\)</p>
Step-by-Step Solution
Key Concept: For a matrix to be unitary, A†A = I (where A† is conjugate transpose). Compute the conjugate transpose by taking transpose AND replacing i with -i, then verify the matrix multiplication carefully.
<p><strong>Step 1:</strong> Find A†(θ) (conjugate transpose of A)</p><p>A†(θ) = [sin θ -i cos θ; -i cos θ sin θ]</p><p><strong>Step 2:</strong> Compute A†(θ)·A(θ)</p><p>A†(θ)·A(θ) = [sin²θ + cos²θ -i sin θ cos θ + i sin θ cos θ; -i sin θ cos θ + i sin θ cos θ cos²θ + sin²θ]</p><p>= [1 0; 0 1] = I</p><p><strong>Step 3:</strong> Therefore A is unitary, so A†·A = I ✓ and A·A† = I ✓</p><p><strong>Step 4:</strong> Check det(A): det(A) = sin²θ - (i cos θ)(-i cos θ) = sin²θ - cos²θ = -cos(2θ)</p><p>For unitary matrices, |det(A)| = 1, but |−cos(2θ)| ≠ 1 always ✗</p><p><strong>Step 5:</strong> Check eigenvalues: Since A is unitary with real trace = 2sin θ and det(A) = -cos(2θ), eigenvalues have modulus 1 only if |det(A)| = 1</p><p>∴ The statement claiming |det(A)| = 1 is NOT true; or det(A) = ±1 is false for all θ</p>
Correct Answer: A,B,D