The number of solutions of the following system of linear homogeneous equations $x-y+z=0, x+2 y-z=0$ and $2…
The number of solutions of the following system of linear homogeneous equations $x-y+z=0, x+2 y-z=0$ and $2 x+y+3 z=0$ is
1
8
Countable infinite
Uncountable
Solution
To find the number od solutions of the following system of linear homogeneous equations.
$\begin{aligned} x-y+z & =0 ...(i)\\ x+2 y-z & =0 ...(ii)\\ 2 x+y+3 z & =0 ...(iii)\end{aligned}$
We can write the above equations in matrix form i.e.
$
\left[\begin{array}{ccc}
1 & -1 & 1 \\
1 & 2 & -1 \\
2 & 1 & 3
\end{array}\right]\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]=0
$
or $A X=0$, where $A=\left[\begin{array}{ccc}1 & -1 & 1 \\ 1 & 2 & -1 \\ 2 & 1 & 3\end{array}\right]$
Since, we know that, if $|A|=0$, then only a non-trivial solution exist.
Consider, $|A|=\left[\begin{array}{ccc}1 & -1 & 1 \\ 1 & 2 & -1 \\ 2 & 1 & 3\end{array}\right]$
$
\begin{aligned}
& \therefore|A|=1[2 \times 3-(-1)]+1[3+2]+1[1-4] \\
& \quad=7+5-3=9 \neq 0 \\
& \because|A| \neq 0
\end{aligned}
$
Hence, there is no non-trivial solutions. $\Rightarrow$ There will be only 1 trivial solution