7.4 The Matrix Method

A matrix is a rectangular array of numbers. A matrix can be used to represent a system of linear equations.

Consider the system of equations

S1
| x +   y + z =  0
|     2 y - z = -1
| x       + z =  1
This system can be represented by the coefficient matrix 

1   1   1   0
0   2  -1  -1
1   0   1   1
This matrix has three rows and four columns. The equivalent system corresponding to the solution will look like
S
|   x      =  a
|     y    =  b
|       z  =  c
which will have a corresponding coefficient matrix 

1   0   0   a
0   1   0   b
0   0   1   c
The zeros below the main diagonal will be achieved using the operations discussed in Section 7.2 which result in equivalent linear systems. The zeros above the main diagonal will be achieved by back substitution. 

1   1   1   0
0   2  -1  -1
1   0   1   1 E3 -> - E1 + E3

1   1   1   0
0   2  -1  -1
0  -1   0   1 E2 -> E2 + E3

1   1   1   0
0   1  -1   0
0  -1   0   1 E3 -> E2 + E3

1   1   1   0
0   1  -1   0
0   0  -1   1 E3 -> - E3


1   1   1   0
0   1  -1   0
0   0   1  -1 

Now that the system is triangularized, we begin the back substitution phase: 
E2 -> E3 + E2

1   1   1   0
0   1   0  -1
0   0   1  -1 E1 -> - E3 + E1

1   1   0   1
0   1   0  -1
0   0   1  -1 E1 -> - E2 + E1

1   0   0   2
0   1   0  -1
0   0   1  -1 Which corresponds to the system

S
| x       =   2
|   y     =  -1
|     z   =  -1
Thus the fourth column of the matrix represents the solution.

Exercise 7.4.1

Use the matrix method to solve the system
S1
|  x  +    y + z  =   0
|- x  +  2 y      =  -1
|  x         + z  =   1
Solution

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