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The system of equations
$\begin{array}{l}
x+y+z=0 \\
2 x+3 y+z=0 \\
x+2 y=0
\end{array}$
has
Options:
$\begin{array}{l}
x+y+z=0 \\
2 x+3 y+z=0 \\
x+2 y=0
\end{array}$
has
Solution:
2168 Upvotes
Verified Answer
The correct answer is:
infinite solutions
$x+y+z=0$ ...(1)
$2 x+3 y+z=0$ ...(2)
$x+2 y=0$ ...(3)
From equation (3), we have $x=-2 y$
Putting this value of $x$ in equations (1), we get
$-2 \mathrm{y}+\mathrm{y}+\mathrm{z}=0 \Rightarrow \mathrm{y}=\mathrm{z}$
Hence $x=-2 z$
Thus, the solution of the given system of equations is $(-2 \mathrm{z}, \mathrm{z}, \mathrm{z})$, where $\mathrm{z}$ is a parameter $(Z \in R)$. Hence the system has infinite number of solutions including zero solution.
$2 x+3 y+z=0$ ...(2)
$x+2 y=0$ ...(3)
From equation (3), we have $x=-2 y$
Putting this value of $x$ in equations (1), we get
$-2 \mathrm{y}+\mathrm{y}+\mathrm{z}=0 \Rightarrow \mathrm{y}=\mathrm{z}$
Hence $x=-2 z$
Thus, the solution of the given system of equations is $(-2 \mathrm{z}, \mathrm{z}, \mathrm{z})$, where $\mathrm{z}$ is a parameter $(Z \in R)$. Hence the system has infinite number of solutions including zero solution.
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