<p><strong>For Problems 7 and 8</strong><br>Consider an arbitrary \(3 \times 3\) non-singular matrix \(A = [a_{ij}]\). A matrix \(B = [b_{ij}]\) is formed such that \(b_{ij}\) is the sum of all the elements except \(a_{ij}\) in the \(i\)th row of \(A\).<br><br>The value of \(|B|\) is equal to</p>
Step-by-Step Solution
Key Concept: Each element b_ij equals the sum of all elements in row i minus a_ij. This linear relationship means each row of B can be expressed as a multiple of the row sum vector, making B a rank-deficient matrix with determinant 0.
<p><strong>Step 1:</strong> Define the relationship. Let S_i = sum of all elements in row i of A. Then b_ij = S_i - a_ij.</p><p><strong>Step 2:</strong> Write row i of B as: [b_i1, b_i2, b_i3] = [S_i - a_i1, S_i - a_i2, S_i - a_i3]</p><p><strong>Step 3:</strong> Notice that each row can be written as: Row_i(B) = S_i[1, 1, 1] - Row_i(A)</p><p><strong>Step 4:</strong> Consider the vector v = [1, 1, 1]^T. For each row i: b_i1 + b_i2 + b_i3 = 3S_i - (a_i1 + a_i2 + a_i3) = 3S_i - S_i = 2S_i</p><p><strong>Step 5:</strong> This means each row of B satisfies: b_i1 + b_i2 + b_i3 = 2S_i (constant for that row). The columns of B are linearly dependent because the column sum relationships force rank(B) < 3.</p><p><strong>Step 6:</strong> Verify: Perform column operation C_1 + C_2 + C_3 on B. Each entry becomes 2S_i, giving a column of constants. This means the three columns of B are linearly dependent.</p><p><strong>Step 7:</strong> Since rank(B) < 3, the matrix B is singular.</p><p>∴ Answer: <strong>|B| = 0</strong></p>
Correct Answer: C