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If the system of linear equations $2 x+2 y+3 z=a$
$3 x-y+5 z=b$
$x-3 y+2 z=c$
where, $a, b,$ care non-zero real numbers, has more than onc solution, then
Options:
$3 x-y+5 z=b$
$x-3 y+2 z=c$
where, $a, b,$ care non-zero real numbers, has more than onc solution, then
Solution:
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Verified Answer
The correct answer is:
$b-c-a=0$
$\because$ System of equations has more than one solution
$\therefore \Delta=\Delta_{1}=\Delta_{2}=\Delta_{3}=0$ for infinite solution
$\begin{aligned}
\Delta_{1} &=\left|\begin{array}{ccc}
a & 2 & 3 \\
b & -1 & 5 \\
c & -3 & 2
\end{array}\right|=a(13)+2(5 c-2 b)+3(-3 b+c) \\
&=13 a-13 b+13 c=0
\end{aligned}$
i.c, $a-b+c=0$
or $\quad b-c-a=0$
$\therefore \Delta=\Delta_{1}=\Delta_{2}=\Delta_{3}=0$ for infinite solution
$\begin{aligned}
\Delta_{1} &=\left|\begin{array}{ccc}
a & 2 & 3 \\
b & -1 & 5 \\
c & -3 & 2
\end{array}\right|=a(13)+2(5 c-2 b)+3(-3 b+c) \\
&=13 a-13 b+13 c=0
\end{aligned}$
i.c, $a-b+c=0$
or $\quad b-c-a=0$
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