### Systems of Equation

```Reviewing Systems of
Equations
• Substitution
• Elimination
• Matrices
Some Important
Points
• Several (2+) variables need to be solved for
• To find an exact solution, you need to generate the
same number of equations as you have variables
• For example, you’ll need 3 equations to work with if
you’re going to try and solve for 3 variables
Three Methods for
Solving
• Substitution – substituting variables to form equations
with single variables
• Elimination – combining equations to eliminate
variables (convenient only in certain situations)
• Matrices – use of matrices and their operations to
arrive at a solution
Example:
Substitution
(2)
4x + 2y - z = 20
x + 0y - z = 2
(3)
3x + y - 0z =15
(1)
Solve for x, y, and z
Example:
Substitution
(2)
4x + 2y - z = 20
x + 0y - z = 2
(3)
3x + y - 0z =15
(1)
x + 0y - z = 2
x = 2+z
z = x-2
Solve for x, y, and z
Transform equation (2) to have z in terms of x.
Example:
Substitution
(1)
4x + 2y - z = 20
(2)
z = x-2
(3)
3x + y - 0z =15
Solve for x, y, and z
Replace the original equation (2) with the new
equation.
Example:
Substitution
(1)
4x + 2y - z = 20
(2)
z = x-2
(3)
3x + y - 0z =15
3x + y - 0z =15
y =15- 3x
Solve for x, y, and z
Use the same technique on equation (3) to get y
in terms of x
Example:
Substitution
(1)
4x + 2y - z = 20
(2)
z = x-2
(3)
y =15- 3x
Solve for x, y, and z
Replace equation (3) with the new equation
Example:
Substitution
(1)
4x + 2y - z = 20
(2)
z = x-2
(3)
y =15- 3x
Solve for x, y, and z
Now substitute the expression for z in terms of
x from equation (2) into equation (1).
4x + 2y - z = 20
4x + 2y - (x - 2) = 20
Example:
Substitution
(1)
4x + 2y - z = 20
(2)
z = x-2
(3)
y =15- 3x
Solve for x, y, and z
Continue to modify this equation by substituting
equation (3) into equation (1).
4x + 2y - z = 20
4x + 2y - (x - 2) = 20
4x + 2(15- 3x) - (x - 2) = 20
Example:
Substitution
(1)
4x + 2y - z = 20
(2)
z = x-2
(3)
y =15- 3x
Solve for x, y, and z
You’ll notice that this new equation has only 1
variable (x) in it. We can readily solve for x and
use this result in further calculations
4x + 2y - z = 20
4x + 2y - (x - 2) = 20
4x + 2(15- 3x) - (x - 2) = 20
4x +30 - 6x - x + 2 = 20
4x - 6x - x = 20 -30 - 2
-3x = -12
x=4
Example:
Substitution
(1)
4x + 2y - z = 20
(2)
z = x-2
(3)
y =15- 3x
Solve for x, y, and z
Now, we can use the result of x=4 in equation
(3) to solve for y.
y =15- 3x
y =15- 3(4)
y=3
y=3
x=4
Example:
Substitution
(1)
4x + 2y - z = 20
(2)
z = x-2
(3)
y =15- 3x
z = x-2
z = (4) - 2
z=2
Solve for x, y, and z
Finally, we’ll use the result of x=4 in equation
(2) to determine the value for z.
z=2
y=3
x=4
Example:
Substitution
(2)
4x + 2y - z = 20
x + 0y - z = 2
(3)
3x + y - 0z =15
(1)
Solve for x, y, and z
Now we’ve successfully solved a system of
equations
z=2
y=3
x=4
Example: Elimination
(2)
4x + 2y - z = 20
x + 0y - z = 2
(3)
3x + y - 0z =15
(1)
Solve for x, y, and z
Example: Elimination
(2)
4x + 2y - z = 20
x + 0y - z = 2
(3)
3x + y - 0z =15
(1)
Solve for x, y, and z
We’ll use elimination to verify our work from
before. We can start be subtracting equation (2)
from equation (1)
Example: Elimination
(2)
4x + 2y - z = 20
x + 0y - z = 2
(3)
3x + y - 0z =15
(1)
(1)
-(2)
4x + 2y - z = 20
-x - 0y + z = -2
Solve for x, y, and z
Start by multiplying equation (2) by -1.
Example: Elimination
(2)
4x + 2y - z = 20
x + 0y - z = 2
(3)
3x + y - 0z =15
(1)
4x + 2y - z = 20
-(2) -x - 0y + z = -2
(1) - (2) 3x + 2y =18
(1)
Solve for x, y, and z
Add equation (1) and equation –(2). Notice that
the z term is eliminated.
Example: Elimination
(2)
3x + 2y =18
x + 0y - z = 2
(3)
3x + y - 0z =15
(1)
Solve for x, y, and z
Replace equation (1) with this new equation.
Example: Elimination
(2)
3x + 2y =18
x + 0y - z = 2
(3)
3x + y - 0z =15
(1)
3x + 2y =18
-3x - y = -15
(1)
-(3)
Solve for x, y, and z
Now subtract equation (3) from equation (1)
equation (3) by -1
Example: Elimination
(2)
3x + 2y =18
x + 0y - z = 2
(3)
3x + y - 0z =15
(1)
3x + 2y =18
-3x - y = -15
y=3
(1)
-(3)
(1) – (3)
Solve for x, y, and z
Add equation (1) and equation –(3). Notice that
the x terms cancel.
y=3
Example: Elimination
(2)
3x + 2y =18
x + 0y - z = 2
(3)
3x + y - 0z =15
(1)
Solve for x, y, and z
Use the result of y=3 in equation (1) to
determine the value of x.
3x + 2y =18
3x + 2(3) =18
3x =12
x=4
x=4
y=3
Example: Elimination
(2)
4x + 2y - z = 20
x + 0y - z = 2
(3)
3x + y - 0z =15
(1)
Solve for x, y, and z
Use the result of x=4 in equation (2) to find z.
x + 0y - z = 2
(4) - z = 2
4-2 = z
z=2
z=2
x=4
y=3
Example: Matrices
(2)
4x + 2y - z = 20
x + 0y - z = 2
(3)
3x + y - 0z =15
(1)
x
(1)
(2)
(3)
y
z
- - - - - - -
Solve for x, y, and z
The first step is to set up a matrix describing this
system of equations. This matrix will be
manipulated so that there is a diagonal of 1’s
and corresponding values on the other side of
the bar
RHS
-
1 0 0
0 1 0
0 0 1
?
?
?
Example: Matrices
(2)
4x + 2y - z = 20
x + 0y - z = 2
(3)
3x + y - 0z =15
(1)
1 0 0
0 1 0
0 0 1
?
?
?
Solve for x, y, and z
In this form, the top row will contain the
solution for x
x solution
Example: Matrices
(2)
4x + 2y - z = 20
x + 0y - z = 2
(3)
3x + y - 0z =15
(1)
1 0 0
0 1 0
0 0 1
?
?
?
Solve for x, y, and z
In this form, the middle row will contain the
solution for y
y solution
Example: Matrices
(2)
4x + 2y - z = 20
x + 0y - z = 2
(3)
3x + y - 0z =15
(1)
1 0 0
0 1 0
0 0 1
?
?
?
Solve for x, y, and z
In this form, the bottom row will contain the
solution for z
z solution
Example: Matrices
(2)
4x + 2y - z = 20
x + 0y - z = 2
(3)
3x + y - 0z =15
(1)
x
(1)
(2)
(3)
4
1
3
Solve for x, y, and z
The first step is to set up a matrix describing this
system of equations. To do this, each equation
will be transcribed into a row. These 3 rows will
be stacked on top of each other to form the
matrix. The first entry in a row will be the x
coefficient.
Example: Matrices
(2)
4x + 2y - z = 20
x + 0y - z = 2
(3)
3x + y - 0z =15
(1)
x
(1)
(2)
(3)
y
4 2
1 0
3 1
Solve for x, y, and z
The second entry in each row will be the y
coefficient
Example: Matrices
(2)
4x + 2y - z = 20
x + 0y - z = 2
(3)
3x + y - 0z =15
(1)
x
(1)
(2)
(3)
y
z
4 2 -1
1 0 -1
3 1 0
Solve for x, y, and z
The third entry in each row will be the z
coefficient
Example: Matrices
(2)
4x + 2y - z = 20
x + 0y - z = 2
(3)
3x + y - 0z =15
(1)
x
(1)
(2)
(3)
y
z
4 2 -1
1 0 -1
3 1 0
RHS
20
2
15
Solve for x, y, and z
The final entry is the right hand side (RHS) of
each equation.
Example: Matrices
(2)
4x + 2y - z = 20
x + 0y - z = 2
(3)
3x + y - 0z =15
(1)
(1)
(2)
(3)
4 2 -1
1 0 -1
3 1 0
20
2
15
Solve for x, y, and z
For matrix operations, we are allowed to do the
following things in succession:
1. We can swap rows at any time so long as all
entries retain their position within a row
Swap (1) and (2)
(1)
(2)
(3)
1 0 -1
4 2 -1
3 1 0
2
20
15
Example: Matrices
(2)
4x + 2y - z = 20
x + 0y - z = 2
(3)
3x + y - 0z =15
(1)
(1)
(2)
(3)
1 0 -1
4 2 -1
3 1 0
2
20
15
Solve for x, y, and z
For matrix operations, we are allowed to do the
following things in succession:
2. We can multiply any row by a constant so
long as it’s applied to each entry within the row
Multiply (3) by -1
(1)
(2)
(3)
1 0 -1
4 2 -1
-3 -1 0
2
20
-15
Example: Matrices
(2)
4x + 2y - z = 20
x + 0y - z = 2
(3)
3x + y - 0z =15
(1)
(1)
(2)
(3)
1 0 -1
4 2 -1
-3 -1 0
2
20
-15
Solve for x, y, and z
For matrix operations, we are allowed to do the
following things in succession:
3. We add equations together and replace one
of those equations with the results of the
Replace (2) with
(3)+(2)
(1)
(2)
(3)
1 0 -1
1 1 -1
-3 -1 0
2
5
-15
Example: Matrices
(2)
4x + 2y - z = 20
x + 0y - z = 2
(3)
3x + y - 0z =15
(1)
Solve for x, y, and z
1. We can swap rows at any time so long as all
entries retain their position within a row
2. We can multiply any row by a constant so
long as it’s applied to each entry within the row
3. We add equations together and replace one
of those equations with the results of the
(1)
(2)
(3)
1 0 -1
1 1 -1
-3 -1 0
2
5
-15
Replace (2) with
3(2)+(3)
(1)
1 0 -1
(2) 0
2 -3
(3) -3 -1 0
2
0
-15
Example: Matrices
(2)
4x + 2y - z = 20
x + 0y - z = 2
(3)
3x + y - 0z =15
(1)
Solve for x, y, and z
1. We can swap rows at any time so long as all
entries retain their position within a row
2. We can multiply any row by a constant so
long as it’s applied to each entry within the row
3. We add equations together and replace one
of those equations with the results of the
(1)
(2)
(3)
1 0 -1
0 2 -3
-3 -1 0
2
0
-15
Replace (3) with
3(1)+(3)
(1)
1 0 -1
(2) 0 2 -3
(3) 0 -1 -3
2
0
-9
Example: Matrices
(2)
4x + 2y - z = 20
x + 0y - z = 2
(3)
3x + y - 0z =15
(1)
Solve for x, y, and z
1. We can swap rows at any time so long as all
entries retain their position within a row
2. We can multiply any row by a constant so
long as it’s applied to each entry within the row
3. We add equations together and replace one
of those equations with the results of the
(1)
(2)
(3)
1 0 -1
0 2 -3
0 -1 -3
2
0
-9
Replace (2) with
(2)+(3)
(1)
(2)
(3)
1 0 -1
0 1 -6
0 -1 -3
2
-9
-9
Example: Matrices
(2)
4x + 2y - z = 20
x + 0y - z = 2
(3)
3x + y - 0z =15
(1)
Solve for x, y, and z
1. We can swap rows at any time so long as all
entries retain their position within a row
2. We can multiply any row by a constant so
long as it’s applied to each entry within the row
3. We add equations together and replace one
of those equations with the results of the
(1)
(2)
(3)
1 0 -1
0 1 -6
0 -1 -3
2
-9
-9
Replace (3) with
(2)+(3)
1 0 -1
(2) 0 1 -6
(3) 0 0 -9
(1)
2
-9
-18
Example: Matrices
(2)
4x + 2y - z = 20
x + 0y - z = 2
(3)
3x + y - 0z =15
(1)
Solve for x, y, and z
1. We can swap rows at any time so long as all
entries retain their position within a row
2. We can multiply any row by a constant so
long as it’s applied to each entry within the row
3. We add equations together and replace one
of those equations with the results of the
(1)
(2)
(3)
1 0 -1
0 1 -6
0 0 -9
2
-9
-18
Replace (3) with 1/9(3)
(1)
(2)
(3)
1 0 -1
0 1 -6
0 0 1
2
-9
2
Example: Matrices
(2)
4x + 2y - z = 20
x + 0y - z = 2
(3)
3x + y - 0z =15
(1)
Solve for x, y, and z
1. We can swap rows at any time so long as all
entries retain their position within a row
2. We can multiply any row by a constant so
long as it’s applied to each entry within the row
3. We add equations together and replace one
of those equations with the results of the
(1)
(2)
(3)
1 0 -1
0 1 -6
0 0 1
2
-9
2
Replace (1) with
(1)+(3)
(1)
(2)
(3)
1 0 0
0 1 -6
0 0 1
4
-9
2
Example: Matrices
(2)
4x + 2y - z = 20
x + 0y - z = 2
(3)
3x + y - 0z =15
(1)
Solve for x, y, and z
1. We can swap rows at any time so long as all
entries retain their position within a row
2. We can multiply any row by a constant so
long as it’s applied to each entry within the row
3. We add equations together and replace one
of those equations with the results of the
(1)
(2)
(3)
1 0 0
0 1 -6
0 0 1
4
-9
2
Replace (2) with
(2)+6(3)
(1)
(2)
(3)
1 0 0
0 1 0
0 0 1
4
3
2
Example: Matrices
(2)
4x + 2y - z = 20
x + 0y - z = 2
(3)
3x + y - 0z =15
(1)
Solve for x, y, and z
x=4
y=3
z=2
x
(1) 4 2 -1
(2) 1 0 -1
(3) 3 1 0
20
2
15
y
z
1 0 0
0 1 0
0 0 1
RHS
4
3
2
```