Matrices & Determinants
Inverse of a Matrix
Grade 12
Question:
<p>If \(A\begin{bmatrix} 1 & 2 & 3 \\ 0 & 2 & 3 \\ 0 & 1 & 1 \end{bmatrix} = \begin{bmatrix} 0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{bmatrix}\), then \(A^{-1}\) equals</p>
<p>\(\begin{bmatrix} 3 & 1 & 2 \\ 3 & 0 & 2 \\ 1 & 0 & 1 \end{bmatrix}\)</p>
<p>\(\begin{bmatrix} 1 & 2 & 3 \\ 0 & 2 & 3 \\ 0 & 1 & 1 \end{bmatrix}\)</p>
<p>\(\begin{bmatrix} 0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{bmatrix}\)</p>
<p>\(\begin{bmatrix} 3 & 0 & 2 \\ 1 & 0 & 1 \\ 3 & 1 & 2 \end{bmatrix}\)</p>
Step-by-Step Solution
Key Concept: Instead of finding A directly, find A⁻¹ by recognizing that if AB = C, then A = CB⁻¹, so A⁻¹ = B(C)⁻¹. The key is to compute B⁻¹ and multiply it with C⁻¹ in the correct order.
<p><strong>Step 1:</strong> Given: $AB = C$ where $B = \begin{bmatrix} 1 & 2 & 3 \\ 0 & 2 & 3 \\ 0 & 1 & 1 \end{bmatrix}$ and $C = \begin{bmatrix} 0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{bmatrix}$</p><p><strong>Step 2:</strong> From $AB = C$, multiply both sides by $B^{-1}$ on the right: $A = CB^{-1}$</p><p>Taking inverse: $A^{-1} = (CB^{-1})^{-1} = BC^{-1}$</p><p><strong>Step 3:</strong> Find $C^{-1}$. Notice $C$ is a permutation matrix (cyclic shift). $C^{-1} = C^T = \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \end{bmatrix}$</p><p><strong>Step 4:</strong> Compute $A^{-1} = BC^{-1} = \begin{bmatrix} 1 & 2 & 3 \\ 0 & 2 & 3 \\ 0 & 1 & 1 \end{bmatrix} \cdot \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \end{bmatrix} = \begin{bmatrix} 3 & 1 & 2 \\ 3 & 0 & 2 \\ 1 & 0 & 1 \end{bmatrix}$</p><p>$\therefore$ Answer: (A)</p>
Correct Answer: A