Matrices & Determinants
Matrix Equations
Grade 12
Question:
<p>If \(\begin{bmatrix} 2 & 1 \\ 3 & 2 \end{bmatrix} A \begin{bmatrix} -3 & 2 \\ 5 & -3 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\), then \(A =\)</p>
<p>\(\begin{bmatrix} 1 & 1 \\ 1 & 0 \end{bmatrix}\)</p>
<p>\(\begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}\)</p>
<p>\(\begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix}\)</p>
<p>\(-\begin{bmatrix} 1 & 1 \\ 1 & 0 \end{bmatrix}\)</p>
Step-by-Step Solution
Key Concept: Recognize that the matrices on the left and right of A are inverses of each other. If P·A·Q = I, then A = P⁻¹·Q⁻¹, which can be computed by finding inverses of 2×2 matrices using the formula A⁻¹ = (1/det(A))·adj(A).
<p><strong>Step 1:</strong> Let P = ⌊2 1; 3 2⌋ and Q = ⌊-3 2; 5 -3⌋. We have P·A·Q = I.</p><p><strong>Step 2:</strong> Multiply both sides by P⁻¹ on the left: A·Q = P⁻¹.</p><p><strong>Step 3:</strong> Multiply both sides by Q⁻¹ on the right: A = P⁻¹·Q⁻¹.</p><p><strong>Step 4:</strong> Find P⁻¹: det(P) = 2(2) - 1(3) = 4 - 3 = 1. So P⁻¹ = ⌊2 -1; -3 2⌋.</p><p><strong>Step 5:</strong> Find Q⁻¹: det(Q) = (-3)(-3) - 2(5) = 9 - 10 = -1. So Q⁻¹ = ⌊-3 -2; -5 -3⌋ = ⌊3 2; 5 3⌋.</p><p><strong>Step 6:</strong> Compute A = P⁻¹·Q⁻¹ = ⌊2 -1; -3 2⌋ · ⌊3 2; 5 3⌋ = ⌊6-5 4-3; -9+10 -6+6⌋ = ⌊1 1; 1 0⌋.</p><p>∴ Answer: A = ⌊1 1; 1 0⌋</p>
Correct Answer: A