Matrices & Determinants
Inverse of a Matrix
Grade 12
Question:
<p>If \(F(x) = \begin{bmatrix} \cos x & -\sin x & 0 \\ \sin x & \cos x & 0 \\ 0 & 0 & 1 \end{bmatrix}\) and \(G(y) = \begin{bmatrix} \cos y & 0 & \sin y \\ 0 & 1 & 0 \\ -\sin y & 0 & \cos y \end{bmatrix}\), then \([F(x)\, G(y)]^{-1}\) is equal to</p>
<p>\(F(-x)\, G(-y)\)</p>
<p>\(G(-y)\, F(-x)\)</p>
<p>\(F(x^{-1})\, G(y^{-1})\)</p>
<p>\(G(y^{-1})\, F(x^{-1})\)</p>
Step-by-Step Solution
Key Concept: The product of orthogonal matrices is orthogonal, so [F(x)G(y)]⁻¹ = G(y)ᵀF(x)ᵀ. Both F and G are rotation matrices with determinant 1, making them orthogonal, so their inverses equal their transposes.
<p><strong>Step 1:</strong> Recognize that F(x) and G(y) are rotation matrices (orthogonal matrices). For orthogonal matrices, A⁻¹ = Aᵀ.</p><p><strong>Step 2:</strong> Apply the property [AB]⁻¹ = B⁻¹A⁻¹ (reverse order matters):<br/>[F(x)G(y)]⁻¹ = G(y)⁻¹F(x)⁻¹ = G(y)ᵀF(x)ᵀ</p><p><strong>Step 3:</strong> Compute G(y)ᵀ by transposing G(y):<br/>G(y)ᵀ = <begin>bmatrix} \cos y & 0 & -\sin y \\ 0 & 1 & 0 \\ \sin y & 0 & \cos y \end{bmatrix} = G(-y)</p><p><strong>Step 4:</strong> Compute F(x)ᵀ by transposing F(x):<br/>F(x)ᵀ = <begin>bmatrix} \cos x & \sin x & 0 \\ -\sin x & \cos x & 0 \\ 0 & 0 & 1 \end{bmatrix} = F(-x)</p><p><strong>Step 5:</strong> Therefore [F(x)G(y)]⁻¹ = G(-y)F(-x)</p><p>∴ Answer: B</p>
Correct Answer: B