Use gauss Jordan method to solve the systems of simultaneous below #4x-y+3z=11#, #2x+z=5#, #2z-3x+y=5#?

1 Answer
Douglas K.
Sep 24, 2017

The Gauss-Jordan Method uses an Augmented matrix and Elementary Row Operations until an identity matrix is obtained on the left, then the solution is the column vector on the right.

Explanation:

We start the process by writing each equation as a row in an Augmented Matrix

Write, #4x-y+3z=11# as a row in an augmented matrix:

#[ (4,-1,3,|,11) ]#

Add a row for the #2x+z=5#:

#[ (4,-1,3,|,11), (2,0,1,|,5) ]#

Before we add the final row for,

#2z-3x+y=5#

, we must write it as,

#-3x+y+2z=5#

, then add it to the matrix:

#[ (4,-1,3,|,11), (2,0,1,|,5), (-3,1,2,|,5) ]#

Now that we have the augmented matrix, we perform Elementary Row Operations.

We want a 1 in position #(1,1)#, therefore, we add row 3 to row 1 and put the results is row 1; this is written as #R_1 + R_3 to R_1#

#[ (1,0,5,|,16), (2,0,1,|,5), (-3,1,2,|,5) ]#

We want a 0 in position #(2,1)#, therefore, we multiply row 1 by -2, add it to row 2 and put the results in row 2, #-2R_1 + R_2 to R_2#:

#[ (1,0,5,|,16), (0,0,-9,|,-27), (-3,1,2,|,5) ]#

We want a 0 in position #(3,1)#, therefore, we #3R_1+R_3toR_3#:

#[ (1,0,5,|,16), (0,0,-9,|,-27), (0,1,17,|,53) ]#

We want a 1 in position #(2,2)#, therefore, we #R_2 harr R_3#:

#[ (1,0,5,|,16), (0,1,17,|,53), (0,0,-9,|,-27) ]#

We want a 1 in position #(3,3)#, therefore, we #-1/9R_3toR_3#:

#[ (1,0,5,|,16), (0,1,17,|,53), (0,0,1,|,3) ]#

We want a 0 in position #(2,3)#, therefore, we #-17R_3+R_2toR_2#

#[ (1,0,5,|,16), (0,1,0,|,2), (0,0,1,|,3) ]#

We want a 0 in position #(1,3)#, therefore, we #-5R_3+R_1toR_1#:

#[ (1,0,0,|,1), (0,1,0,|,2), (0,0,1,|,3) ]#

We have an Identity Matrix on the left, therefore, the values of x, y, and z can be read from the column vector on the right:

#x = 1, y = 2, z = 3#

I used WolframAlpha to check my work.

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