Gaussian Elimination
Gaussian elimination for the solution of a linear system transforms the
system Sx =finto an equivalent system U x =cwith upper triangular
matrix U(that means all entries in Ubelow the diagonal are zero).
This transformation is done by applying three types of transformations to
the augmented matrix (S|f).
Type 1: Interchange two equations; and
Type 2: Replace an equation with the sum of the same equation and a
multiple of another equation.
Once the augmented matrix (U|f)is transformed into (U|c), where U
is an upper triangular matrix, we can solve this transformed system
Ux =cusing backsubstitution.
M. Heinkenschloss - CAAM335 Matrix Analysis Gaussian Elimination and Matrix Inverse (updated September 3, 2010) – 1
Example 1
Suppose that we want to solve
2 4 −2
4 9 −3
−2−3 7
x1
x2
x3
=
2
8
10
.(1)
We apply Gaussian elimination. To keep track of the operations, we use,
e.g., R2=R2−2∗R1, which means that the new row 2 is computed by
subtracting 2 times row 1 from row 2.
2 4 −2 2
4 9 −3 8
−2−3 7 10
→
2 4 −2 2
0 1 1 4
0 1 5 12
R2=R2−2∗R1
R3=R3+R1
→
2 4 −2 2
0 1 1 4
0 0 4 8
R3=R3−R2
The original system (1) is equivalent to
2 4 −2
0 1 1
0 0 4
x1
x2
x3
=
2
4
8
.(2)
The system (2) can be solved by backsubstitution. We get the solution
x3= 8/4=2, x2= (4−1∗2)/1=2, x1= (2−4∗2−(−2)∗2)/2 = −1.
M. Heinkenschloss - CAAM335 Matrix Analysis Gaussian Elimination and Matrix Inverse (updated September 3, 2010) – 2
Example 2
Suppose that we want to solve
2 3 −2
1−2 3
4−1 4
x1
x2
x3
=
f1
f2
f3
.(3)
We apply Gaussian elimination.
2 3 −2f1
1−2 3 f2
4−1 4 f3
→
2 3 −2f1
0−7/2 4 f2−f1/2
0−7 8 f3−2f1
R2=R2−0.5∗R1
R3=R3−2∗R1
→
2 3 −2f1
0−7/2 4 f2−f1/2
0 0 0 f3−2f2−f1
R3=R3−2R2
The original system (3) is equivalent to
2 3 −2
0−7/2 4
0 0 0
x1
x2
x3
=
f1
f2−f1/2
f3−2f2−f1
.(4)
The last equation in system (4) reads 0x1+ 0x2+ 0x3=f3−2f2−f1.
This can only be satisfied of the right hand side satisfies
f3−2f2−f1= 0, for example if f1=f2=f3= 1.
M. Heinkenschloss - CAAM335 Matrix Analysis Gaussian Elimination and Matrix Inverse (updated September 3, 2010) – 3
Example 2 (cont.)
If f3−2f2−f1= 0, then x3can be chosen arbitrarily and x2, x1can be
determined by backsubstitution.
If f3−2f2−f1= 0, then
x3=any scalar, x2= (f2−f1/2−4x3)∗(−2/7), x1= (f1−3x2+2x3)/2.
For example if f1=f2=f3= 1, and if we choose x3= 0, then
x3= 0, x2=−1/7, x1= 5/7.
M. Heinkenschloss - CAAM335 Matrix Analysis Gaussian Elimination and Matrix Inverse (updated September 3, 2010) – 4
