Download Solve Systems of Equations by Elimination Using Matrices

Solve Systems of Equations by
Elimination Using Matrices
The process of solving systems of equations using the elimination/ addition method
is determined by the coefficients of the variables - we add the equations eliminating
the terms that are exact opposites. By using matrices, we can streamline the process.
A matrix is a rectangular array of numbers. The matrix is typically enclosed with
square brackets. For example:
3
𝐴= [
1
4
]
2
𝐵= [
5 2
−4 0
−3
]
2
A is 2 × 2 matrix because it has 2 rows and 2 columns. B is a 2 × 3 matrix because
it has 2 rows and 3 columns.
A system of equations is written as an augmented matrix, whereby the coefficients
of the variables and the constants are separated by a line.
ax+by=c
dx+ey=f
is written as
[
a
d
b c
| ]
e f
The goal is to get the matrix in row echelon form to easily solve the resulting system:
[
a
d
𝑧
x y
b c
| ] in row echelon form → [
| solution for y ]
0 1
e f
The calculations used are similar to those we used previously without the augmented
matrix.
Row operations are:
1. Any two rows of the matrix can be interchanged.
2. The numbers in any row may be multiplied by any nonzero number.
3. Any row may be changed by adding the numbers of the row to the
numbers of another row.
Example 1: Solve the system: x – 3y = 10
2x + y = -1
1 −3 10
|
]
2 1 −1
−2 6 −20
[
|
]
2 1 −1
[
[
1
0
−3 10
|
]
7 −21
[
1
0
−3 10
|
]
1 −3
x – 3y = 10
y = -3
x – 3(-3) = 10
x + 9 = 10
x=1
Re-write as an augmented matrix
Multiply the first row by -2 to get exact opposite
Use the original first row
Add the rows above to get new second row
Divide the second row by 7
1
(The same as multiplying by )
7
Revert to the system with variables
The value of y is now known
Solve for x
Subtract 9 from both sides of the equation
The solution is (1, -3)
As the process becomes more familiar, we will be able to perform two row
operations simultaneously. For instance, in the above example, we could have
multiplied the first row by -2 and added it to the second row in one step.
Example 2: Solve the system: 4x – 5y = 1
2x + 3y = -5
[
4
2
−5 1
|
]
3 −5
Re-write as an augmented matrix
[
2
4
3 −5
|
]
−5 1
Interchange the rows
[
2
0
3 −5
|
]
−11 11
Multiply the first row by − 2
and then add to the second row
[
2
0
3 −5
|
]
1 −1
Divide the second row by -11
2x + 3y = -5
y = -1
Revert to the system with variables
The value of y is now known
2x + 3(-1) = -5
2x – 3 = -5
2x = -2
x = -1
Solve for x
Add 3 to both sides of the equation
Divide both sides of the equation by 2
The solution is (-1, -1)
Example 3: Solve the system: 5x – 4y = -1
-2x + 3y = 2
−4 −1
|
] Re-write as an augmented matrix
3 2
[
5
−2
[
5
−4 −1
|
] Multiply the second row by 5
−10 15 10
[
5
0
−4 −1
|
]
7 8
Multiply the first row by 2
and then add to the second row
[
5
0
−4 −1
| 8 ]
1
7
Divide the second row by 7
5x – 4y = -1
y=
Revert to the system with variables
8
The y value is now known
7
8
5x – 4( ) = -1
7
5x –
32
7
Solve for x
= -1
5x =
x=
Add
25
32
7
to both sides of the equation
Divide both sides of the equation by 5
7
5
7
5
8
7
7
The solution is ( ,
)
Recall that systems of linear equations can also be dependent systems, in which there
are an infinite number of solutions, or inconsistent systems, in which there is no
solution.
Example 4: Solve the system: 2x + 6y = 13
x + 3y = 5
2
1
1
[
2
6 13
|
]
3 5
3 5
|
]
6 13
1
0
3 5
| ]
0 3
[
[
x + 3y = 5
0=3
Re-write as an augmented matrix
Interchange the rows
Multiply the first row by − 2
and then add to the second row
Revert to the system with variables
Since the second equation is a false statement, there is no solution
1
Example 5: Solve the system: x – y = 3
2
4x – 2y = 12
1
1
[
4
3
]
12
−2
1
[
0
1
2
|
3
| ]
0 0
2
1
x– y=3
2
Re-write as an augmented matrix
Multiply the first row by − 4
and then add to the second row
Revert to the system with variables
0=0
Since the second equation is a true statement, there are infinite solutions