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# Solve this problem - If , then is equal to : - Matrices and Determinants - JEE Main

If $\begin{vmatrix} a-b-c & 2a & 2a\\ 2b & b-c-a & 2b\\ 2c & 2c & c-a-b \end{vmatrix}= (a+b+c) (x+a+b+c)^{2}, x\neq 0 \: and\:$

$a+b+c\neq 0$ , then $x$ is equal to :

• Option 1)

$-2(a+b+c)$

• Option 2)

abc

• Option 3)

$-(a+b+c)$

• Option 4)

$2(a+b+c)$

Answers (1)
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Value of determinants of order 3 -

-

$\begin{vmatrix} a-b-c &2a & 2a\\ 2b & b-c-a &2b \\ 2c & 2c & c-a-b \end{vmatrix}$

$R_{1}\rightarrow R_{1}+R_{2}+R_{3}$

$=\begin{vmatrix} a+b+c &a+b+c & a+b+c\\ 2b & b-c-a &2b \\ 2c & 2c & c-a-b \end{vmatrix}$

$C_{2}\rightarrow C_{2}-C_{1}$               $C_{3}\rightarrow C_{3}-C_{1}$

$=(a+b+c)\begin{vmatrix} 1 &0 & 0\\ 2b & -(a+b+c) &0 \\ 2c & 2c & c-a-b \end{vmatrix}$

$=(a+b+c)\cdot (a+b+c)^{2}$

$=> x=0 \:\: or\: \: -2(a+b+c)$

Option 1)

$-2(a+b+c)$

Option 2)

abc

Option 3)

$-(a+b+c)$

Option 4)

$2(a+b+c)$

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