If \(b\) and \(c\) are non-zero real numbers,
\[
A=
\begin{bmatrix}
1 & b & c\\
b & 2 & 3\\
c & 3 & 4
\end{bmatrix}
\quad \text{and} \quad
B=
\begin{bmatrix}
0 & b & c\\
-b & 0 & 2\\
-c & -2 & 0
\end{bmatrix},
\]
then \(\det(A+B)=\)
Show Hint
While evaluating determinants, always look for a row or column containing maximum zeros. Expanding along that row or column simplifies the calculation significantly.
Step 1: Add the two matrices. We have $A = \begin{bmatrix} 1 & b & c \\ b & 2 & 3 \\ c & 3 & 4 \end{bmatrix}$ and $B = \begin{bmatrix} 0 & b & c \\ -b & 0 & 2 \\ -c & -2 & 0 \end{bmatrix}$. Add them element by element: \[ A + B = \begin{bmatrix} 1 & 2b & 2c \\ 0 & 2 & 5 \\ 0 & 1 & 4 \end{bmatrix} \] The first column has two zeros, making computation easy.
Step 2: Notice that $B$ is skew-symmetric. $B^T = -B$ (check: $B_{12} = b$, $B_{21} = -b$, etc.). This is helpful context but what matters is the sum.
Step 3: Expand the determinant along the first column. \[ \det(A+B) = 1 \cdot \begin{vmatrix} 2 & 5 \\ 1 & 4 \end{vmatrix} - 0 + 0 \] Only the first entry in column 1 is nonzero, so just that $2\times 2$ minor matters.
Step 5: Complete the calculation. \[ \det(A+B) = 1 \times 3 = 3 \] The $b$ and $c$ values cancelled out when adding, so the answer is independent of $b$ and $c$.
