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Y+ Z = 3,
-7Z = -14.
The process of backward substitution in Gaussian elimination consists in
finding the values of the unknowns, starting from the last equation and
working upwards. Thus, we solve for Z first:
Next, we substitute Z=2 into equation 2 (E2), and solve E2 for Y:
Next, we substitute Z=2 and Y = 1 into E1, and solve E1 for X:
The solution is, therefore, X = -1, Y = 1, Z = 2.
Example of Gaussian elimination using matrices
The system of equations used in the example above can be written as a matrix
equation
A
⋅
x
=
b
, if we use:
.
4
3
14
,
,
1
2
4
1
2
3
6
4
2
−
−
=
=
−
−
=
b
x
A
Z
Y
X
To obtain a solution to the system matrix equation using Gaussian elimination,
we first create what is known as the
augmented matrix
corresponding to
A
,
i.e.,
