# How do you solve using gaussian elimination or gauss-jordan elimination, x-2y+z=1, 2x-3y+z=5, -x-2y+3z=-13?

Oct 11, 2017

$x = 7 + t$, $y = 3 + t$, $z = t$

#### Explanation:

Begin by creating the augmented matrix, or a matrix with the $x$, $y$, and $z$ coefficients on the left side of a vertical bar, and the constant values on the right side:

$\left[\begin{matrix}1 & - 2 & 1 & | & 1 \\ 2 & - 3 & 1 & | & 5 \\ - 1 & - 2 & 3 & | & - 13\end{matrix}\right]$

We now perform row operations on the 3 rows in order to reduce the left portion of the augmented matrix to the identity form:

$\left[\begin{matrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{matrix}\right]$

We can do any of the following:

• Swap any two rows
• Multiply a single row by a non-zero constant value
• Add/Subtract a constant multiple of a row to another row

Using the notation ${R}_{n}$ to refer to row $n$, here's the steps I'd use:

$\left[\begin{matrix}1 & - 2 & 1 & | & 1 \\ 2 & - 3 & 1 & | & 5 \\ - 1 & - 2 & 3 & | & - 13\end{matrix}\right] \left.\begin{matrix}\null \\ - 2 {R}_{1} \rightarrow \\ + {R}_{1}\end{matrix}\right.$

$\left[\begin{matrix}1 & - 2 & 1 & | & 1 \\ 0 & 1 & - 1 & | & 3 \\ 0 & - 4 & 4 & | & - 12\end{matrix}\right] \left.\begin{matrix}\null \\ \rightarrow \\ + 4 {R}_{2}\end{matrix}\right.$

$\left[\begin{matrix}1 & - 2 & 1 & | & 1 \\ 0 & 1 & - 1 & | & 3 \\ 0 & 0 & 0 & | & 0\end{matrix}\right] \left.\begin{matrix}+ 2 {R}_{2} \\ \rightarrow \\ \null\end{matrix}\right.$

$\left[\begin{matrix}1 & 0 & - 1 & | & 7 \\ 0 & 1 & - 1 & | & 3 \\ 0 & 0 & 0 & | & 0\end{matrix}\right]$

We can do nothing further [useful]. Notice the bottom line has all zero values. This indicates that this system of equations does not result in a unique $z$ value. In fact, we can translate this reduced row matrix back to a system of three unknowns in two equations:

$\left\{\begin{matrix}x & \null & - z & = & 7 \\ \null & y & - z & = & 3\end{matrix}\right.$

Any value of $z$ chosen (out of infinitely many) will result in a unique solution. For instance, if $z = 1$, we can find $x = 8$ and $y = 4$.

It's commonly the case for problems like this with infinite solutions that we define everything in terms of a template variable $t$ like so:

$x = 7 + t$, $y = 3 + t$, $z = t$