# How do you solve using gaussian elimination or gauss-jordan elimination, x₁+ 2x₂+ x₃= 2, x₁+ 3x₂- x₃ = 4, 3x₁+ 7x₂+ x₃= 8?

Feb 12, 2018

color(blue)((-2-5z , 2+2z,z)

#### Explanation:

First form an augmented matrix with the coefficients on the left and the constants in the augmented part on the right. Due to formatting limitations I will use embolden font for the augmented part.

$\left[\begin{matrix}1 & 2 & 1 & \boldsymbol{2} \\ 1 & 3 & - 1 & \boldsymbol{4} \\ 3 & 7 & 1 & \boldsymbol{8}\end{matrix}\right]$

We now get the matrix in upper diagonal form using row operations:

The notation will be:

$R 2 = R 2 + 3 R 1$

This means, row 2 is row 2 plus 3 times row 1 added to it.

$R 3 = R 3 - 3 R 1$

$\left[\begin{matrix}1 & 2 & 1 & \boldsymbol{2} \\ 1 & 3 & - 1 & \boldsymbol{4} \\ 0 & 1 & - 2 & \boldsymbol{2}\end{matrix}\right]$

$R 2 = R 2 - R 1$

$\left[\begin{matrix}1 & 2 & 1 & \boldsymbol{2} \\ 0 & 1 & - 2 & \boldsymbol{2} \\ 0 & 1 & - 2 & \boldsymbol{2}\end{matrix}\right]$

$R 3 = R 3 - R 2$

$\left[\begin{matrix}1 & 2 & 1 & \boldsymbol{2} \\ 0 & 1 & - 2 & \boldsymbol{2} \\ 0 & 0 & 0 & \boldsymbol{0}\end{matrix}\right]$

Notice we have eliminated the last row. This is due to linear dependency amongst the rows. This is because one of the equations was just a multiple of the other equations. As a result we only have two independent equations, but we have 3 variables, so we can only solve the system by assigning arbitrary values to 1 of the variables. This results in there being an infinite number of solutions.

Using back substitution:

$z = z$

$y = 2 + 2 z$

$x = - 2 - 5 z$

Solutions:

color(blue)((-2-5z , 2+2z,z)

For arbitrary $\boldsymbol{z}$