<p>Let \(A\) be a \(3 \times 3\) matrix such that \(A\begin{bmatrix} 1 & 2 & 3 \\ 0 & 2 & 3 \\ 0 & 1 & 1 \end{bmatrix} = \begin{bmatrix} 0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{bmatrix}\). Then \(A^{-1}\) is:</p>
<p>\(\begin{bmatrix} 3 & 1 & 2 \\ 3 & 0 & 2 \\ 1 & 0 & 1 \end{bmatrix}\)</p>
<p>\(\begin{bmatrix} 3 & 2 & 1 \\ 3 & 2 & 0 \\ 1 & 1 & 0 \end{bmatrix}\)</p>
<p>\(\begin{bmatrix} 0 & 1 & 3 \\ 0 & 2 & 3 \\ 1 & 1 & 1 \end{bmatrix}\)</p>
<p>\(\begin{bmatrix} 1 & 2 & 3 \\ 0 & 1 & 1 \\ 0 & 2 & 3 \end{bmatrix}\)</p>
Step-by-Step Solution
Key Concept: To find A⁻¹, recognize that A · B = C means A = C · B⁻¹, so A⁻¹ = B · C⁻¹. The key is computing B⁻¹ efficiently and understanding that C is a permutation matrix (C⁻¹ = Cᵀ).
<p><strong>Step 1:</strong> Given: A·B = C where B = [1,2,3; 0,2,3; 0,1,1] and C = [0,0,1; 1,0,0; 0,1,0]</p><p><strong>Step 2:</strong> From A·B = C, multiply both sides by B⁻¹ on the right: A = C·B⁻¹</p><p><strong>Step 3:</strong> Therefore A⁻¹ = (C·B⁻¹)⁻¹ = B·C⁻¹</p><p><strong>Step 4:</strong> Note that C is a permutation matrix (cyclic shift), so C⁻¹ = Cᵀ = [0,1,0; 0,0,1; 1,0,0]</p><p><strong>Step 5:</strong> Compute B·C⁻¹ by matrix multiplication:</p><p>Row 1 of B × C⁻¹ columns: (1·0+2·0+3·1, 1·1+2·0+3·0, 1·0+2·1+3·0) = (3, 1, 2)</p><p>Row 2 of B × C⁻¹ columns: (0·0+2·0+3·1, 0·1+2·0+3·0, 0·0+2·1+3·0) = (3, 0, 2)</p><p>Row 3 of B × C⁻¹ columns: (0·0+1·0+1·1, 0·1+1·0+1·0, 0·0+1·1+1·0) = (1, 0, 1)</p><p><strong>∴ Answer: B</strong> where A⁻¹ = [3,1,2; 3,0,2; 1,0,1]</p>
Correct Answer: B