<p>Let \(P = [a_{ij}]\) be a \(3 \times 3\) matrix and let \(Q = [b_{ij}]\), where \(b_{ij} = 2^{i+j} a_{ij}\) for \(1 \leq i, j \leq 3\). If the determinant of \(P\) is 2, then the determinant of the matrix \(Q\) is</p>
Step-by-Step Solution
Key Concept: When each element of a matrix is multiplied by a scaling factor that depends on its position, the determinant scales by the product of all scaling factors. Specifically, if row i is multiplied by factor kᵢ, det(Q) = k₁·k₂·k₃·det(P).
<p><strong>Step 1:</strong> Express Q in terms of P using the scaling relationship.</p><p>Since b_ij = 2^(i+j)·a_ij, we can write:</p><p>Q = [2^(i+j)·a_ij]</p><p><strong>Step 2:</strong> Factor out powers of 2 from rows and columns.</p><p>• From row 1: factor out 2¹</p><p>• From row 2: factor out 2²</p><p>• From row 3: factor out 2³</p><p>• From column 1: factor out 2¹</p><p>• From column 2: factor out 2²</p><p>• From column 3: factor out 2³</p><p><strong>Step 3:</strong> Calculate the total scaling factor.</p><p>det(Q) = (2¹·2²·2³)·(2¹·2²·2³)·det(P)</p><p>= 2^(1+2+3)·2^(1+2+3)·det(P)</p><p>= 2^6·2^6·det(P)</p><p>= 2^12·2</p><p>= 2^13</p><p><strong>Alternative approach:</strong> Each element a_ij is multiplied by 2^(i+j). Summing exponents: Σ(i+j) for all 9 elements = 3(1+2+3) + 3(1+2+3) = 18+18 = ... Actually: for each i,j pair: (1+1),(1+2),(1+3),(2+1),...,(3+3), sum = 3·(1+2+3) + 3·(1+2+3) = 36, but we factor by rows/columns to get 2^(1+2+3)·2^(1+2+3) = 2^12.</p><p>∴ det(Q) = 2^12·2 = <strong>2^13 or 8192</strong></p>
Correct Answer: D