If | array ccc a-b-c & 2 a & 2 a 2 b & b-c-a & 2 b 2 c & 2 c & c-a-b array | =(a+b+c)(x+a+b+c)^ 2 , x ≠ 0 and a+b+c ≠ 0, then x is equal to

Correct answer: D. -2(a+b+c) Δ=| array ccc a-b-c & 2 a & 2 a 2 b & b-c-a & 2 b 2 c & 2 c & c-a-b array | R_ 1 arrow R_ 1 +R_ 2 +R Δ=| array ccc a+b+c & a+b+c & a+b+c 2 b & b-c-a & 2 b 2 c & 2 c & c-a-b array | =(a+b+c)| array ccc 1 & 1 & 1 2 b & b-c-a & 2 b 2 c & 2 c & c-a-b array | array l C_ 1 arrow C_ 1 -C_ y , C_ 2 arrow C_ 2 -C_ 3 Δ=(a+b+c)| array ccc 0 & 0 & 1 0 & -b-c-a & 2 b c+a+b & c+a+b & c-a-b array | array =(a+b+c)(a+b+c)^ 2 Hence, x=-2(a+b+c)

JEE Main 201911 JanuaryEvening Shift

Exam: JEE Main
Year: 2019
Date: 11 January
Shift: Evening Shift
Subject: Mathematics
Chapter: Determinants

JEE Main 2019 · 11 January · Evening Shift · Mathematics · Determinants

If | array ccc a-b-c & 2 a & 2 a 2 b & b-c-a & 2 b 2 c & 2 c &...

\left|

\begin{array}{ccc}a-b-c & 2 a & 2 a \\ 2 b & b-c-a & 2 b \\ 2 c & 2 c & c-a-b\end{array}

\right| =(a+b+c)(x+a+b+c)^{2}, x \neq 0

and

a+b+c \neq 0,

then

x

is equal to$

Options

Aabc
B $-(a+b+c)$
C $2(a+b+c)$
D $-2(a+b+c)$

Correct answer

D. $-2(a+b+c)$

Step-by-step solution

\Delta=\left|

\begin{array}{ccc}a-b-c & 2 a & 2 a \\ 2 b & b-c-a & 2 b \\ 2 c & 2 c & c-a-b\end{array}

\right|

R_{1} \rightarrow R_{1}+R_{2}+R

\Delta=\left|

\begin{array}{ccc}a+b+c & a+b+c & a+b+c \\ 2 b & b-c-a & 2 b \\ 2 c & 2 c & c-a-b\end{array}

\right|

=(a+b+c)\left|

\begin{array}{ccc}1 & 1 & 1 \\ 2 b & b-c-a & 2 b \\ 2 c & 2 c & c-a-b\end{array}

\right|

\begin{array}{l} C_{1} \rightarrow C_{1}-C_{y}, C_{2} \rightarrow C_{2}-C_{3} \\ \Delta=(a+b+c)\left|\begin{array}{ccc} 0 & 0 & 1 \\ 0 & -b-c-a & 2 b \\ c+a+b & c+a+b & c-a-b \end{array}

\right| \end{array}=(a+b+c)(a+b+c)^{2}

Hence,

x=-2(a+b+c)$

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