Question

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 \(\det(A+B)=\)

• A. \(3\)
• B. \(1\)
• C. \(-1\)
• D. \(0\)

Hint:

While evaluating determinants, always look for a row or column containing maximum zeros. Expanding along that row or column simplifies the calculation significantly.

Answer 1

The Correct Option is A

Step 1: Find the matrix \(A+B\).
Given, \[ A= \begin{bmatrix} 1 & b & c\\ b & 2 & 3\\ c & 3 & 4 \end{bmatrix} \] and \[ B= \begin{bmatrix} 0 & b & c\\ -b & 0 & 2\\ -c & -2 & 0 \end{bmatrix} \] Adding corresponding elements, \[ A+B= \begin{bmatrix} 1+0 & b+b & c+c\\ b-b & 2+0 & 3+2\\ c-c & 3-2 & 4+0 \end{bmatrix} \] Thus, \[ A+B= \begin{bmatrix} 1 & 2b & 2c\\ 0 & 2 & 5\\ 0 & 1 & 4 \end{bmatrix} \]

Step 2: Evaluate the determinant.
Now, \[ \det(A+B)= \begin{vmatrix} 1 & 2b & 2c\\ 0 & 2 & 5\\ 0 & 1 & 4 \end{vmatrix} \] Expand along the first column because it contains two zeros.
So, \[ \det(A+B) =1\cdot \begin{vmatrix} 2 & 5\\ 1 & 4 \end{vmatrix} \] \[ =1\cdot (2\times 4-5\times 1) \] \[ =1\cdot (8-5) \] \[ =3 \]

Step 3: Final conclusion.
Therefore, \[ \boxed{3} \]