Question

Evaluate \( \begin{vmatrix} 1 & a & b+c 1 & b & c+a 1 & c & a+b \end{vmatrix} \)

• A. \( 1 \)
• B. \( 0 \)
• C. \( (1-a)(1-b)(1-c) \)
• D. \( a+b+c \)
• E. \( 2(a+b+c) \)

Hint:

If after row operations two rows become multiples, determinant is zero.

Answer 1

The Correct Option is B

Concept: Use row operations to simplify determinant.

Step 1: Write determinant: \[ \begin{vmatrix} 1 & a & b+c 1 & b & c+a 1 & c & a+b \end{vmatrix} \]

Step 2: Apply row operations: \[ R_2 \to R_2 - R_1,\quad R_3 \to R_3 - R_1 \]

Step 3: Perform subtraction carefully: \[ R_2: (1-1, b-a, (c+a)-(b+c)) = (0, b-a, a-b) \] \[ R_3: (1-1, c-a, (a+b)-(b+c)) = (0, c-a, a-c) \]

Step 4: New determinant becomes: \[ \begin{vmatrix} 1 & a & b+c 0 & b-a & a-b 0 & c-a & a-c \end{vmatrix} \]

Step 5: Notice pattern: \[ (b-a) = -(a-b),\quad (c-a)=-(a-c) \] So rows are proportional.

Step 6: When two rows are proportional → determinant = 0.

Step 7: Final answer: \[ \boxed{0} \]