Question

Using the factor theorem it is found that b+c, c+a and a+b are three factors of the determinant $\left| \begin{matrix}

• 2a & a + b & a + c \ b + a & - 2b & b + c \ c + a & c + b & - 2c \end{matrix} \right|$.

The other factor in the value of the determinant is

• A. 4
• B. 2
• C. a+b+c
• D. None of these

Answer 1

Answer: (A)

4

Answer 2

Let$\left| \begin{matrix}

• 2a & a + b & a + c \ b + a & - 2b & b + c \ c + a & c + b & - 2c \end{matrix} \right| = k(b + c)(c + a)(a + b)$Putting

a = b = c =$\frac { \alpha \beta \gamma ( - 1 ) ( \beta - \alpha ) ( \gamma - \alpha ) } { ( 1 - \alpha ) ( 1 - \beta ) ( 1 - \gamma ) }$

then $\left| \begin{matrix}

• 2\lambda & 2\lambda & 2\lambda \ 2\lambda & - 2\lambda & 2\lambda \ 2\lambda & 2\lambda & - 2\lambda \end{matrix} \right| = k(2\lambda)(2\lambda)(2\lambda)$

∴ k = $\left| \begin{matrix}

• 1 & 1 & 1 \ 1 & - 1 & 1 \ 1 & 1 & - 1 \end{matrix} \right| = - 1(0) - 1( - 2) + 1(2) = 4$