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

1 Answer
Feb 12, 2018

Answer:

#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.

#[(1,2,1,bb2),(1,3,-1,bb4),(3,7,1,bb8)]#

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

The notation will be:

#R2=R2+3R1#

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

#R3=R3-3R1#

#[(1,2,1,bb2),(1,3,-1,bb4),(0,1,-2,bb2)]#

#R2=R2-R1#

#[(1,2,1,bb2),(0,1,-2,bb2),(0,1,-2,bb2)]#

#R3=R3-R2#

#[(1,2,1,bb2),(0,1,-2,bb2),(0,0,0,bb0)]#

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+2z #

#x=-2-5z#

Solutions:

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

For arbitrary #bbz#