Question:

Step 1: Express $b_{ij}$ in terms of $a_{ij}$. Since $b_{ij}$ is the sum of all elements in the $i$-th row of $A$ except $a_{ij}$: $$b_{ij} = \sum_{k=1}^{3} a_{ik} - a_{ij} = R_i - a_{ij}$$ where $R_i$ is the sum of all elements in the $i$-th row. Step 2: Write $B$ in matrix form. Let $S$ be the matrix where each element in row $i$ equals $R_i$ (the row sum). Then: $$B = S - A$$ where $S_{ij} = R_i$ for all $j$, meaning $S = A \cdot J$ where $J$ is the $3\times 3$ all-ones matrix. Step 3: Express $S$ using $A$. Note $S = A \cdot \mathbf{1}\mathbf{1}^T$ where $\mathbf{1}$ is the all-ones column vector... More precisely, $S = A \cdot E$ where $E$ is the $3\times3$ matrix of all ones. So: $$B = AE - A = A(E - I)$$ where $I$ is the identity matrix. Step 4: Since $AC = B = A(E - I)$ and $A$ is non-singular: $$C = A^{-1} \cdot A(E - I) = E - I$$ Step 5: Compute $E - I$ explicitly: $$E - I = \begin{pmatrix} 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \end{pmatrix}$$ This matrix is symmetric, so $C$ is symmetric. Option (a) is correct. Step 6: Check if $C$ is diagonal. $C = E - I$ has off-diagonal entries equal to 1, so it is NOT diagonal. Option (b) is incorrect. Step 7: Compute $|B|$. Since $B = A(E-I)$: $$|B| = |A| \cdot |E - I|$$ Compute $|E - I|$ where $E - I = \begin{pmatrix} 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \end{pmatrix}$: $$|E-I| = 0(0-1) - 1(0-1) + 1(1-0) = 0 + 1 + 1 = 2$$ Step 8: Therefore $|B| = 2|A|$. Option (d) is correct.