Question

Evaluate the determinant:
\[ \left| \begin{array}{ccc} b+c & a-b & a\\ c+a & b-c & b\\ a+b & c-a & c\\ \end{array} \right| \]

• A. \(3abc-a^3-b^3-c^3\)
• B. \(3abc+a^3+b^3+c^3\)
• C. \(3abc-a^3+b^3-c^3\)
• D. \(3abc+a^3-b^3-c^3\)

Hint:

For a \(3\times 3\) determinant, expand carefully along any row or column and simplify term by term. Always keep track of signs \(+,-,+\).

Answer 1

The Correct Option is A

Concept:
To evaluate a determinant of order \(3\), we can expand it along any row or column. Here, we expand along the first row.

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

Step 2: Expand along the first row
\[ \Delta=(b+c) \begin{vmatrix} b-c & b\\ c-a & c \end{vmatrix} -(a-b) \begin{vmatrix} c+a & b\\ a+b & c \end{vmatrix} +a \begin{vmatrix} c+a & b-c\\ a+b & c-a \end{vmatrix} \]

Step 3: Evaluate the minor determinants
\[ \begin{vmatrix} b-c & b\\ c-a & c \end{vmatrix} = (b-c)c - b(c-a) = bc - c^2 - bc + ab = ab - c^2 \]

Similarly, after simplifying the complete expansion, we get:
\[ \Delta = 3abc - a^3 - b^3 - c^3 \]

Step 4: Final Answer
\[ \therefore \Delta = 3abc - a^3 - b^3 - c^3 \]
\[ \therefore \text{Correct Answer is (A)} \]