For the following system of homogeneous linear equations, find the solution if it is unique, otherwise, describe the infinite solution set in terms of an arbitrary parameter k. 4x-2y+6z=0 5x-y-z=0 2x-y+3z=0

For the following system of homogeneous linear equations, find the solution if it is unique, otherwise, describe the infinite solution set in terms of an arbitrary parameter k. 4x-2y+6z=0 5x-y-z=0 2x-y+3z=0
You can still ask an expert for help

• Questions are typically answered in as fast as 30 minutes

Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it

Willie
Step 1
Given linear equations are
4x−2y+6z=0
5x−y−z=0
2x−y+3z=0
Step 2
To find the solution of given linear equations.
First write down the given linear equations in augmented matrix.
((4,-2,6,0),(5,-1,-1,0),(2,-1,3,0))
Now use Gauss Jordan elimination method to convert above matrix,
Perform the operation ${R}_{1}\to \frac{{R}_{1}}{4}\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}{R}_{2}\to {R}_{2}-5{R}_{1}$
The matrix becomes,
$\left(\begin{array}{cccc}1& -\frac{1}{2}& \frac{3}{2}& 0\\ 0& 1& -\frac{17}{3}& 0\\ 0& 0& 0& 0\end{array}\right)$
Perform operation ${R}_{1}\to {R}_{1}+\frac{1}{2}{R}_{2}$
The matrix becomes,
$\left(\begin{array}{cccc}1& 0& -\frac{4}{3}& 0\\ 0& 1& -\frac{17}{3}& 0\\ 0& 0& 0& 0\end{array}\right)$
Here in the third row we get 0x+0y+0z=0,
This means for any value of z there will be solution of x and y .
Therefore the system of linear equation has infinitely many solutions.
Step 3
From $\left(\begin{array}{cccc}1& 0& -\frac{4}{3}& 0\\ 0& 1& -\frac{17}{3}& 0\\ 0& 0& 0& 0\end{array}\right)$ we get,
$x-\frac{4}{3}z=0$
$y-\frac{17}{3}z=0$
Now let z = k,
$⇒x-\frac{4}{3}k=0$
$⇒x=\frac{4}{3}k$
and
$⇒y-\frac{17}{3}k=0$
$⇒y=\frac{17}{3}k$
Therefore solutions are $\left(x,y,z\right)=\left(\frac{4}{3}k,\frac{17}{3}k,k\right)$