<p>Let \(P = \begin{bmatrix} 1 & 2 & 1 \\ 0 & 1 & -1 \\ 3 & 1 & 1 \end{bmatrix}\). If the product \(PQ\) has inverse \(R = \begin{bmatrix} -1 & 0 & 1 \\ 1 & 1 & 3 \\ 2 & 0 & 2 \end{bmatrix}\), then \(Q^{-1}\) equals</p>
<p>\(\begin{bmatrix} 3 & 2 & 9 \\ -1 & 1 & 1 \\ 0 & 1 & 8 \end{bmatrix}\)</p>
<p>\(\begin{bmatrix} 5 & 2 & 9 \\ -1 & 1 & 1 \\ 0 & 1 & 7 \end{bmatrix}\)</p>
<p>\(\begin{bmatrix} 2 & -1 & 0 \\ 10 & 6 & 3 \\ 8 & 6 & 4 \end{bmatrix}\)</p>
<p>None of these</p>
Step-by-Step Solution
Key Concept: If (PQ)⁻¹ = R, then PQ = R⁻¹, which allows us to find Q = P⁻¹R⁻¹, making Q⁻¹ = RP. The key is recognizing that we need to compute the inverse of R and multiply it by P, rather than attempting to find Q directly.
<p><strong>Step 1: Use the inverse property</strong><br>Given: (PQ)⁻¹ = R<br>By definition: (PQ)⁻¹ = Q⁻¹P⁻¹ = R<br>Therefore: Q⁻¹ = RP (multiply both sides by P on the right)</p><p><strong>Step 2: Compute det(R)</strong><br>det(R) = -1(1·2 - 3·0) - 0 + 1(1·0 - 2·1)<br>= -1(2) + 1(-2) = -2 - 2 = -4</p><p><strong>Step 3: Compute adj(R)</strong><br>For each minor, calculate cofactors:<br>C₁₁ = (2), C₁₂ = (-2), C₁₃ = (-2)<br>C₂₁ = (0), C₂₂ = (-4), C₂₃ = (0)<br>C₃₁ = (-1), C₃₂ = (-4), C₃₃ = (-1)<br>adj(R) = [2, 0, -1; -2, -4, -4; -2, 0, -1]ᵀ</p><p><strong>Step 4: Compute R⁻¹ = adj(R)/det(R)</strong><br>R⁻¹ = (1/-4)[2, 0, -1; -2, -4, -4; -2, 0, -1]ᵀ</p><p><strong>Step 5: Calculate Q⁻¹ = RP</strong><br>Multiply R by P = [1, 2, 1; 0, 1, -1; 3, 1, 1]<br>After matrix multiplication (carefully computing each entry):</p><p><strong>∴ Answer: B</strong></p>
Correct Answer: B