Matrices & Determinants
Inverse of a Matrix
Grade 12
Question:
<p>If \(A(\alpha, \beta) = \begin{bmatrix} \cos\alpha & \sin\alpha & 0 \\ -\sin\alpha & \cos\alpha & 0 \\ 0 & 0 & e^\beta \end{bmatrix}\), then \(A(\alpha, \beta)^{-1}\) is equal to</p>
<p>\(A(-\alpha, -\beta)\)</p>
<p>\(A(-\alpha, \beta)\)</p>
<p>\(A(\alpha, -\beta)\)</p>
<p>\(A(\alpha, \beta)\)</p>
Step-by-Step Solution
Key Concept: Recognize that A(α,β) is a block diagonal matrix: a 2×2 rotation matrix combined with a scalar exponential. The inverse of a block diagonal matrix is the block diagonal matrix of inverses, and the inverse of a rotation matrix is its transpose.
<p><strong>Step 1:</strong> Recognize the block diagonal structure. A(α,β) has the form:</p><p>A(α,β) = [R(α) | 0] where R(α) is the 2×2 rotation matrix and the third row/column isolates e^β</p><p><strong>Step 2:</strong> For the 2×2 rotation block, R(α)^(-1) = R(α)^T (rotation matrices are orthogonal):</p><p>R(α)^(-1) = [cos α -sin α]</p><p>[sin α cos α]</p><p><strong>Step 3:</strong> For the scalar e^β, the inverse is simply e^(-β).</p><p><strong>Step 4:</strong> Assemble the inverse matrix as a block diagonal:</p><p>A(α,β)^(-1) = <br>[cos α -sin α 0]<br>[sin α cos α 0]<br>[0 0 e^(-β)]</p><p>Which equals <strong>A(-α, -β)</strong></p><p>∴ Answer: A</p>
Correct Answer: A