Matrices & Determinants
Orthogonal Matrices
Grade 12
Question:
<p>If \(\mathbf{A} = \begin{bmatrix} \cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha \end{bmatrix}\), then the matrix <strong>A</strong> is</p>
<p>(a) symmetric matrix</p>
<p>(b) skew-symmetric matrix</p>
<p>(c) identity matrix</p>
<p>(d) orthogonal matrix</p>
Step-by-Step Solution
Key Concept: An orthogonal matrix satisfies A^T A = I. Verify this by computing A^T and multiplying with A using trigonometric identity sin²α + cos²α = 1.
<p><strong>Solution:</strong></p><p>Given: $\mathbf{A} = \begin{bmatrix} \cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha \end{bmatrix}$</p><p>For an orthogonal matrix, we need $\mathbf{A}^T \mathbf{A} = \mathbf{I}$.</p><p>$\mathbf{A}^T = \begin{bmatrix} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{bmatrix}$</p><p>$\mathbf{A}^T \mathbf{A} = \begin{bmatrix} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{bmatrix} \begin{bmatrix} \cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha \end{bmatrix}$</p><p>$= \begin{bmatrix} \cos^2 \alpha + \sin^2 \alpha & 0 \\ 0 & \sin^2 \alpha + \cos^2 \alpha \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \mathbf{I}$</p><p>Therefore, <strong>A</strong> is an orthogonal matrix.</p>
Correct Answer: d