The value of b+c a a b c+a b c c a+b is

Correct answer: (C) 4abc. Let △= b+c a a b c+a b c c a+b . Applying R_1→R_1+R_2+R_3 △= b+c+b+c a+c+a+c a+b+a+b b c+a b c c a+b ⇒△= 2b+c 2a+c 2a+b b c+a b c c a+b ⇒△=2 b+c a+c a+b b c+a b c c a+b Applying R_1→R_1-R_2 ⇒△=2 b+c-b a+c-(c+a) a+b-b b c+a b c c a+b ⇒△=2 c 0 a b c+a b c c a+b Applying R_3→R_3-R_1 ⇒△=2 c 0 a b c+a b 0 c b Applying R_2→R_2-R_3 ⇒△=2 c 0 a b a 0 0 c b ⇒△=2cab-0-0+abc-0 ⇒△=2abc+abc ⇒△=4abc.

Q45 / 160 · AP EAMCET 2021 (19 Aug Shift 2) · Mathematics

$The value of |\begin{array}{ccc} b + c & a & a \\ b & c + a & b \\ c & c & a + b \end{array}| is$

A
$a b c$
B
$(a + b) (b + c) (c + a)$
Correct answer
C
$4 a b c$
D
$(a - b) (b - c) (c - a)$

Solution

Let $△ = |\begin{matrix} b + c & a & a \\ b & c + a & b \\ c & c & a + b \end{matrix}| .$

Applying $R_{1} \to R_{1} + R_{2} + R_{3}$

$△ = |\begin{matrix} b + c + b + c & a + c + a + c & a + b + a + b \\ b & c + a & b \\ c & c & a + b \end{matrix}|$

$\Rightarrow △ = |\begin{matrix} 2 (b + c) & 2 (a + c) & 2 (a + b) \\ b & c + a & b \\ c & c & a + b \end{matrix}|$

$\Rightarrow △ = 2 |\begin{matrix} b + c & a + c & a + b \\ b & c + a & b \\ c & c & a + b \end{matrix}|$

Applying $R_{1} \to R_{1} - R_{2}$

$\Rightarrow △ = 2 |\begin{matrix} b + c - b & a + c - (c + a) & a + b - b \\ b & c + a & b \\ c & c & a + b \end{matrix}|$

$\Rightarrow △ = 2 |\begin{matrix} c & 0 & a \\ b & c + a & b \\ c & c & a + b \end{matrix}|$

Applying $R_{3} \to R_{3} - R_{1}$

$\Rightarrow △ = 2 |\begin{matrix} c & 0 & a \\ b & c + a & b \\ 0 & c & b \end{matrix}|$

Applying $R_{2} \to R_{2} - R_{3}$

$\Rightarrow △ = 2 |\begin{matrix} c & 0 & a \\ b & a & 0 \\ 0 & c & b \end{matrix}|$

$\Rightarrow △ = 2 [c (a b - 0) - 0 + a (b c - 0)]$

$\Rightarrow △ = 2 [a b c + a b c]$

$\Rightarrow △ = 4 a b c .$

Hence, the correct answer is option: 'C'.

The value of b+caabc+abcca+b is

Solution

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