<p>Let \(A = \begin{bmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{bmatrix}\) and \(B = \begin{bmatrix} a_{12}+a_{13} & a_{11}+a_{13} & a_{11}+a_{12} \\ a_{22}+a_{23} & a_{21}+a_{23} & a_{21}+a_{22} \\ a_{32}+a_{33} & a_{31}+a_{33} & a_{31}+a_{32} \end{bmatrix}\). If \(B = CA\), then which of the following is/are correct?</p>
Step-by-Step Solution
Key Concept: Each column of B is a specific linear combination of rows of A; write B = CA by identifying that column j of B equals row j of C multiplied by A, then recognize the pattern of sums to deduce C.
<p><strong>Step 1: Analyze the structure of B.</strong></p><p>Column 1 of B: [a₁₂+a₁₃, a₂₂+a₂₃, a₃₂+a₃₃]ᵀ = a₁₂[1,0,0]ᵀ + a₁₃[1,0,0]ᵀ</p><p>Notice that b_i1 = a_i2 + a_i3 (sum of 2nd and 3rd elements of row i)</p><p>Similarly: b_i2 = a_i1 + a_i3, and b_i3 = a_i1 + a_i2</p><p><strong>Step 2: Express columns of B in terms of A.</strong></p><p>Column 1 of B = (Column 2 of A) + (Column 3 of A)</p><p>Column 2 of B = (Column 1 of A) + (Column 3 of A)</p><p>Column 3 of B = (Column 1 of A) + (Column 2 of A)</p><p><strong>Step 3: Write B = CA where C is the coefficient matrix.</strong></p><p>For B = CA, we need:</p><p>B = [C₁; C₂; C₃]A where each row of C contains coefficients for that column of B.</p><p>This gives: C = \begin{bmatrix} 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \end{bmatrix}</p><p><strong>Step 4: Verify and check properties of C.</strong></p><p>• det(C) = 0(0-1) - 1(0-1) + 1(1-0) = 0 + 1 + 1 = 2 ≠ 0 ✓</p><p>• C is symmetric: C = Cᵀ ✓</p><p>• Each row sum = 2, each column sum = 2</p><p>• C is invertible (det ≠ 0)</p><p>∴ Answer: AD</p>
Correct Answer: AD