How do you solve using gaussian elimination or gauss-jordan elimination, #x +2y +3z = 1#, #2x +5y +7z = 2#, #3x +5y +7z = 4#?

1 Answer
Apr 15, 2017

Answer:

Write an Augmented Matrix .
Perform Elementary Row Operations, until an identity matrix is obtained.
The solutions will be on the right.
Check.

Explanation:

Write #x +2y +3z = 1# as a row in an Augmented Matrix:

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

Add row for the equation #2x +5y +7z = 2#:

# [ (1, 2, 3,|,1), (2,5,7,|, 2) ] #

Add row for the equation #3x +5y +7z = 4#:

# [ (1, 2, 3,|,1), (2,5,7,|, 2), (3,5, 7,|, 4) ] #

The augmented matrix is complete. Perform Elementary Row Operations.

We want the coefficient in position #(1,1)# to be one and it is, therefore, no operation is required.

We want the other two coefficients is column 1 to be 0, therefore, we perform the following two row operations:

#R_2-2R_1toR_2#

# [ (1, 2, 3,|,1), (0,1,1,|, 0), (3,5, 7,|, 4) ] #

#R_3-3R_1toR_3#

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

We want the coefficient in position #(2,2)# to be 1 and it is, therefore, no operation is required.

We want the other two coefficients in column 2 to be 0, therefore, we perform the following two row operations:

#R_1-2R_2toR_1#

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

#R_3+R_2toR_3#

# [ (1, 0, 1,|,1), (0,1,1,|, 0), (0,0, -1,|, 1) ] #

We want the coefficient in position #(3,3)# to be 1 and it is, therefore, we multiply the row by -1:

#-1R_3toR_3#

# [ (1, 0, 1,|,1), (0,1,1,|, 0), (0,0, 1,|, -1) ] #

We want the other two coefficients in column 3 to be 0, therefore, we perform the following two row operations:

#R_2-R_3toR_2#

# [ (1, 0, 1,|,1), (0,1,0,|, 1), (0,0, 1,|, -1) ] #

#R_1-R_3toR_1#

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

We have an identity matrix on the left, therefore, the solutions are on the right:

#x = 2, y = 1,and z = -1#

Check:

#2 +2(1) +3(-1) = 1#
#2(2) +5(1) +7(-1) = 2#
#3(2) +5(1) +7(-1) = 4#

#1 = 1#
#2 = 2#
#4 = 4#

This checks.