# How do you solve using gaussian elimination or gauss-jordan elimination, X + 2Y- 2Z=1, 2X + 3Y + Z=14, 4Y + 5Z=27?

Feb 28, 2017

$\implies \left(\begin{matrix}x \\ y \\ z\end{matrix}\right) = \left(\begin{matrix}1 \\ 3 \\ 3\end{matrix}\right)$

#### Explanation:

We will use Gaussian and back-substitution.

We kick off with the starting augmented matrix (see the NB at bottom):

$\left(\begin{matrix}1 & 2 & - 2 \\ 2 & 3 & 1 \\ 0 & 4 & 5\end{matrix}\right) \left(\begin{matrix}1 \\ 14 \\ 27\end{matrix}\right)$

$R 2 \to R 2 - 2 R 1$

$\left(\begin{matrix}1 & 2 & - 2 \\ 0 & - 1 & 5 \\ 0 & 4 & 5\end{matrix}\right) \left(\begin{matrix}1 \\ 12 \\ 27\end{matrix}\right)$

$R 3 \to R 3 + 4 R 2$

$\left(\begin{matrix}1 & 2 & - 2 \\ 0 & - 1 & 5 \\ 0 & 0 & 25\end{matrix}\right) \left(\begin{matrix}1 \\ 12 \\ 75\end{matrix}\right)$

That was really quick, actually; and this is now in row echelon form.

We can see that:

$\left(\begin{matrix}1 & 2 & - 2 \\ 0 & - 1 & 5 \\ 0 & 0 & 25\end{matrix}\right) \left(\begin{matrix}x \\ y \\ z\end{matrix}\right) = \left(\begin{matrix}1 \\ 12 \\ 75\end{matrix}\right)$

We back substitute from:

$25 z = 75 \implies z = 3$

So:

$- y + 5 z = 12 \implies y = 5 z - 12 = 3$

And:

$x + 2 y - 2 z = 1 \implies x = 1 - 2 \left(3\right) + 2 \left(3\right) = 1$

$\implies \left(\begin{matrix}x \\ y \\ z\end{matrix}\right) = \left(\begin{matrix}1 \\ 3 \\ 3\end{matrix}\right)$

NB

The augmented matrix will not necessarily be represented as it might be in your book as I can't find the functionality in the typesetter here. (So I've got a 3x3 and a column vector.)