The equations $x-y+2 z=4$ $3 x+y+4 z=6$ $x+y+z=1$ have
The equations $x-y+2 z=4$ $3 x+y+4 z=6$ $x+y+z=1$ have
- unique solution
- infinitely many solutions
- no solution
- two solutions
Solution
Given equations, $x-y+2 z=4$
$\begin{aligned}
& 3 x+y+4 z=6 \\
& x+y+z=1
\end{aligned}$
$\begin{aligned}
& \text { Let } \Delta=\left|\begin{array}{ccc}
1 & -1 & 2 \\
3 & 1 & 4 \\
1 & 1 & 1
\end{array}\right| \\
& =1(1-4)+1(3-4)+2(3-1) \\
& =-3-1+4=0 \\
& \text { and } \Delta_1=\left|\begin{array}{ccc}
4 & -1 & 2 \\
6 & 1 & 4 \\
1 & 1 & 1
\end{array}\right| \\
& =4(1-4)+1(6-4)+2(6-1) \\
& =-12+2+10=0
\end{aligned}$
Now, $\Delta=0$ and $\Delta_1=0$
$\therefore \quad$ These equations have infinitely many solutions.
Asked in: AP EAMCET 2016
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