8.6. Augmented Matrices http://www.ck12.org
Once you have your system represented as a matrix you can solve it using an augmented matrix. An augmented
matrix is two matrices that are joined together and operated on as if they were a single matrix. In the case of
solving a system, you need to augment the coefficient matrix and the constant matrix. The vertical line indicates the
separation between the coefficient matrix and the constant matrix.
1 1 1
1 2 3
2 3 4
∣∣
∣∣
∣∣
9
22
31
Reduce the matrix to reduced row echelon form and you will find the solution to the system, if one exists.
Example A
Attempt to solve the system from the guidance section.
Solution:
1 1 1
1 2 3
2 3 4
∣∣
∣∣
∣∣
9
22
31
→
→ −I →
→ − 2 I →
1 1 1
0 1 2
0 1 2
∣∣
∣∣
∣∣
9
13
13
→
→
→ −II →
1 1 1
0 1 2
0 0 0
∣∣
∣∣
∣∣
9
13
0
This system is dependent which means there are an infinite number of solutions.
Example B
Solve the following system using an augmented matrix.
x+y+z= 6
x−y−z=− 4
x+ 2 y+ 3 z= 14Solution:
1 1 1
1 − 1 − 1
1 2 3
∣∣
∣∣
∣∣
6
− 4
14
→
→ −I
→ −I →
1 1 1
0 − 2 − 2
0 1 2
∣∣
∣∣
∣∣
6
− 10
8
→ −III →
→ + 3 III →
→
1 0 − 1
0 1 4
0 1 2
∣∣
∣∣
∣∣
− 2
14
8
→
→
→ −II →
1 0 − 1
0 1 4
0 0 − 2
∣∣
∣∣
∣∣
− 2
14
− 6
→
→
→ ÷− 2 →
1 0 − 1
0 1 4
0 0 1
∣∣
∣∣
∣∣
− 2
14
3
→ +III
→ − 4 III
→
1 0 0
0 1 0
0 0 1
∣∣
∣∣
∣∣
1
2
3
Every matrix can be interpreted as its own linear system. The final augmented matrix can be interpreted as:
1 x+ 0 y+ 0 z= 1
0 x+ 1 y+ 0 z= 2
0 x+ 0 y+ 1 z= 3Which meansx= 1 ,y= 2 ,z=3.
Example C
Solve the following system using augmented Matrices.