Question

If$\left| \begin{matrix} (b + c)^{2} & a^{2} & a^{2} \\ b^{2} & (c + a)^{2} & b^{2} \\ c^{2} & c^{2} & (a + b)^{2} \end{matrix} \right| = kabc(a + b + c)^{3}$, then the value of k is.

• A. – 1
• B. 1
• C. 2
• D. –2

Answer 1

Answer: (C)

2

Answer 2

Operate $C_{2} \rightarrow C_{2} - C_{1},C_{3} \rightarrow C_{3} - C_{1}$ and take out $a + b + c$ from $C_{2}$ as well as from $C_{3}$ to get

$\Delta = (a + b + c)^{2}$ $\left| \begin{matrix} (b + c)^{2} & a - b - c & a - b - c \\ b^{2} & c + a - b & 0 \\ c^{2} & 0 & a + b - c \end{matrix} \right|$

(Operate$R_{1} \rightarrow R_{1} - R_{2} - R_{3}$)

= $(a + b + c)^{2}\left| \begin{matrix} 2bc & - 2c & - 2b \\ b^{2} & c + a - b & 0 \\ c^{2} & 0 & a + b - c \end{matrix} \right|$

(Operate $C_{2} \rightarrow C_{2} + \frac{1}{b}C_{1}$ and $C_{3} \rightarrow C_{3} + \frac{1}{c}C_{1})$

= $(a + b + c)^{2}\left| \begin{matrix} 2bc & 0 & 0 \\ b^{2} & c + a & \frac{b^{2}}{c} \\ c^{2} & \frac{c^{2}}{b} & a + b \end{matrix} \right|$

$= (a + b + c)^{2}\lbrack 2bc\{(a + b)(c + a) - bc\}\rbrack = 2abc(a + b + c)^{3}$.