If beginvmatrix b-c - 2 - a-2 - a - 2 lb- 2 -a-c - 2 - b - 2 1c - 2 - c - 2 - a - b - 2lend vmatrix - kabc a - b - c - 3- then k -

A= (b+c)^2 amp; a^2 amp; a^2 b^2 amp; (a+c)^2 amp; b^2 c^2 amp; c^2 amp; (a+b)^2 C_1→ C_1 C_2, = (b+c a)(b+c+a) amp; a^2 amp; a^2 (b ...

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Byju's Answer

Standard XII

Mathematics

Special Determinants

If beginvmatr...

Question

If $∣ ∣ ∣ ∣ ∣ \begin{matrix} (b + c)^{2} & a^{2} & a^{2} b^{2} & (a + c)^{2} & b^{2} c^{2} & c^{2} & (a + b)^{2} \end{matrix} ∣ ∣ ∣ ∣ ∣ = k a b c (a + b + c)^{3},$
then $k =$

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Solution

$A = ∣ ∣ ∣ ∣ ∣ \begin{matrix} (b + c)^{2} & a^{2} & a^{2} b^{2} & (a + c)^{2} & b^{2} c^{2} & c^{2} & (a + b)^{2} \end{matrix} ∣ ∣ ∣ ∣ ∣$

$C_{1} \to C_{1} - C_{2},$
$= ∣ ∣ ∣ ∣ ∣ \begin{matrix} (b + c - a) (b + c + a) & a^{2} & a^{2} (b - a - c) (b + a + c) & (a + c)^{2} & b^{2} 0 & c^{2} & (a + b)^{2} \end{matrix} ∣ ∣ ∣ ∣ ∣$

$C_{2} \to C_{2} - C_{3},$
$= ∣ ∣ ∣ ∣ ∣ \begin{matrix} (b + c - a) (b + c + a) & 0 & a^{2} (b - a - c) (b + a + c) & (a + c - b) (a + c + b) & b^{2} 0 & (c - a - b) (c + a + b) & (a + b)^{2} \end{matrix} ∣ ∣ ∣ ∣ ∣$

$= (a + b + c)^{2} ∣ ∣ ∣ ∣ ∣ \begin{matrix} (b + c - a) & 0 & a^{2} (b - a - c) & (a + c - b) & b^{2} 0 & (c - a - b) & (a + b)^{2} \end{matrix} ∣ ∣ ∣ ∣ ∣$

$R_{2} \to R_{2} - R_{1}, R_{2} \to R_{2} - R_{3}$
$= 2 (a + b + c)^{2} ∣ ∣ ∣ ∣ ∣ \begin{matrix} (b + c - a) & 0 & a^{2} - c & a & - a (a + b) 0 & (c - a - b) & (a + b)^{2} \end{matrix} ∣ ∣ ∣ ∣ ∣$
Expanding along $R_{1}$ , we get
$= 2 a b c (a + b + c)^{3}$

$\Rightarrow k = 2$

Alternate Solution:
Put $a = b = c = 1$
$∣ ∣ ∣ ∣ \begin{matrix} 4 & 1 & 1 1 & 4 & 1 1 & 1 & 4 \end{matrix} ∣ ∣ ∣ ∣ = 27 k \Rightarrow 54 = 27 k \Rightarrow k = 2$

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