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One of the roots of
$\left|\begin{array}{ccc}x+a & b & c \\ a & x+b & c \\ a & b & x+c\end{array}\right|=0$ is
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$\left|\begin{array}{ccc}x+a & b & c \\ a & x+b & c \\ a & b & x+c\end{array}\right|=0$ is
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Verified Answer
The correct answer is:
$-(a+b+c)$
$\left|\begin{array}{ccc}x+a & b & c \\ a & x+b & c \\ a & b & x+c\end{array}\right|=0$
Applying, $C_{1} \rightarrow C_{1}+C_{2}+C_{3}$
$\left|\begin{array}{ccc}(a+b+c+x) & b & c \\ (a+b+c+x) & x+b & c \\ (a+b+c+x) & b & c+x\end{array}\right|=0$
$\begin{array}{l}
(a+b+c+x)\left|\begin{array}{ccc}
1 & b & c \\
1 & x+b & c \\
1 & b & x+c
\end{array}\right|=0 \\
C_{2} \rightarrow C_{2}-C_{1}, C_{3} \rightarrow c_{3}-C_{1} \\
(a+b+c+x)\left|\begin{array}{lll}
1 & b & c \\
0 & x & 0 \\
0 & 0 & x
\end{array}\right|=0 \\
(a+b+c+x) 1 \cdot x^{2}=0 \\
x=0,-(a+b+c) \quad(\because x \neq 0)
\end{array}$
Applying, $C_{1} \rightarrow C_{1}+C_{2}+C_{3}$
$\left|\begin{array}{ccc}(a+b+c+x) & b & c \\ (a+b+c+x) & x+b & c \\ (a+b+c+x) & b & c+x\end{array}\right|=0$
$\begin{array}{l}
(a+b+c+x)\left|\begin{array}{ccc}
1 & b & c \\
1 & x+b & c \\
1 & b & x+c
\end{array}\right|=0 \\
C_{2} \rightarrow C_{2}-C_{1}, C_{3} \rightarrow c_{3}-C_{1} \\
(a+b+c+x)\left|\begin{array}{lll}
1 & b & c \\
0 & x & 0 \\
0 & 0 & x
\end{array}\right|=0 \\
(a+b+c+x) 1 \cdot x^{2}=0 \\
x=0,-(a+b+c) \quad(\because x \neq 0)
\end{array}$
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