Question

The value of determinant \(\left| {\begin{array}{*{20}{c}} {a - b - c}&{2a}&{2a}\\ {2b}&{b - c - a}&{2b}\\ {2c}&{2c}&{c - a - b} \end{array}} \right|\) is:
1. a + b + c
2. (a + b + c)²
3. (a + b + c)³
4. 0

Answer 1

Correct Answer - Option 3 : (a + b + c)³

Given:

\(\left| {\begin{array}{*{20}{c}} {a - b - c}&{2a}&{2a}\\ {2b}&{b - c - a}&{2b}\\ {2c}&{2c}&{c - a - b} \end{array}} \right|\)

Formula:

\(A = \left[ {\begin{array}{*{20}{c}} {{a_{11}}}&{{a_{12}}}&{{a_{13}}}\\ {{a_{21}}}&{{a_{22}}}&{{a_{23}}}\\ {{a_{31}}}&{{a_{32}}}&{{a_{33}}} \end{array}} \right]\)

|A| = a₁(a₂₂ × a₃₃ - a₃₂ × a₂₃) - a₁₂ (a₂₁ × a₃₃ - a₃₁ × a₂₃) + a₁₃(a₂₁ × a₃₂ - a₃₁ × a₂₂)

Calculation:

\(Let \;A = \left[ {\begin{array}{*{20}{c}} {a - b - c}&{2a}&{2a}\\ {2b}&{b - c - a}&{2b}\\ {2c}&{2c}&{c - a - b} \end{array}} \right]\)

C₁ → C₁ - C₂

\(A = \left[ {\begin{array}{*{20}{c}} {-(a + b + c)}&{2a}&{2a}\\ {a+b+c}&{b - c - a}&{2b}\\ {0}&{2c}&{c - a - b} \end{array}} \right]\)

|A| = -(a + b + c) [(b - c - a) (c - a - b) - 2b.2c)] -(a + b + c) [2a (c - a - b) - 2a. 2c] +0

= -(a + b + c) [(b - c - a) (c - a - b) - 4bc) + (2a (c - a - b) - 4ac]]

= -(a + b + c) [(b - c - a) (c - a - b) - 4bc + 2a (c - a - b) - 4ac]

= -(a + b + c) [bc - ab - b² - c² + ac + bc - ac + a² + ab - 4bc + 2ac - 2ab - 2a² - 4ac]

= -(a + b + c) [-a² - b² - c² + 2bc - 4bc + 2ac - 4ac - 2ab]

= -(a + b + c) [ -a2 - b2 - c² - 2bc - 2ac - 2ab]

= -(a + b + c) [ - (a² + b² + c² + 2ab + 2bc + 2ac)]

= (a + b + c) (a2 + b2 + c2 + 2ab + 2bc + 2ac)

= (a + b + c)³

|A| = (a + b + c)3