Matrices and Determinants 2 Question 7
8. If
$ \left|\begin{matrix} a-b-c & 2 a & 2 a \\ 2 b & b-c-a & 2 b \\ 2 c & 2 c & c-a-b \end{matrix}\right| $
$=(a+b+c)(x+a+b+c)^{2}, x \neq 0$ and $a+b+c \neq 0$, then $x$ is equal to
(2019 Main, 11 Jan II)
(a) $-(a+b+c)$
(b) $-2(a+b+c)$
(c) $2(a+b+c)$
(d) $a b c$
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Answer:
Correct Answer: 8. (b)
Solution:
- Let $\Delta=$ $\left|\begin{matrix}a-b-c & 2 a & 2 a \\ 2 b & b-c-a & 2 b \\ 2 c & 2 c & c-a-b\end{matrix}\right|$
Applying $R_{1} \rightarrow R_{1}+R_{2}+R_{3}$, we get
$ \begin{aligned} \Delta & =\left|\begin{matrix} a+b+c & a+b+c & a+b+c \\ 2 b & b-c-a & 2 b \\ 2 c & 2 c & c-a-b \end{matrix}\right| \\ & =(a+b+c)\left|\begin{matrix} 1 & 1 & 1 \\ 2 b & b-c-a & 2 b \\ 2 c & 2 c & c-a-b \end{matrix}\right| \end{aligned} $
(taking common $(a+b+c)$ from $\left.R_{1}\right)$
Applying $C_{2} \rightarrow C_{2}-C_{1}$ and $C_{3} \rightarrow C_{3}-C_{1}$, we get
$ \Delta=(a+b+c)\left|\begin{matrix} 1 & 0 & 0 \\ 2 b & -(a+b+c) & 0 \\ 2 c & 0 & -(a+b+c) \end{matrix}\right| $
Now, expanding along $R_{1}$, we get
$ \Delta=(a+b+c) 1 .(a+b+c)^{2}-0 $
$ =(a+b+c)^{3}=(a+b+c)(x+a+b+c)^{2} \text { (given) } $
$\Rightarrow(x+a+b+c)^{2}=(a+b+c)^{2}$
$\Rightarrow \quad x+a+b+c= \pm(a+b+c)$
$\Rightarrow \quad x=-2(a+b+c)$
