1.
Systems of Linear Equations
http://www.lahc.edu/math/precalculus/math_260a.html
2.
Systems of Linear Equations
We need one piece of information to solve for one
unknown quantity.
3.
Systems of Linear Equations
We need one piece of information to solve for one
unknown quantity. If 2 hamburgers cost 4 dollars and
x is the cost of a hamburger, then 2x = 4 or x = $2.
4.
Systems of Linear Equations
We need one piece of information to solve for one
unknown quantity. If 2 hamburgers cost 4 dollars and
x is the cost of a hamburger, then 2x = 4 or x = $2.
If there are two unknowns x and y, we need two
pieces of information to solve them.
5.
Systems of Linear Equations
We need one piece of information to solve for one
unknown quantity. If 2 hamburgers cost 4 dollars and
x is the cost of a hamburger, then 2x = 4 or x = $2.
If there are two unknowns x and y, we need two
pieces of information to solve them.
Example A:
Let x = cost of a hamburger, y = cost of a fry. If two
hamburgers and a fry cost $7, and one hamburger
and one fry cost $5. what is the price of each?
6.
Systems of Linear Equations
We need one piece of information to solve for one
unknown quantity. If 2 hamburgers cost 4 dollars and
x is the cost of a hamburger, then 2x = 4 or x = $2.
If there are two unknowns x and y, we need two
pieces of information to solve them.
Example A:
Let x = cost of a hamburger, y = cost of a fry. If two
hamburgers and a fry cost $7, and one hamburger
and one fry cost $5. what is the price of each?
We translate these information in to two equations:
7.
Systems of Linear Equations
We need one piece of information to solve for one
unknown quantity. If 2 hamburgers cost 4 dollars and
x is the cost of a hamburger, then 2x = 4 or x = $2.
If there are two unknowns x and y, we need two
pieces of information to solve them.
Example A:
Let x = cost of a hamburger, y = cost of a fry. If two
hamburgers and a fry cost $7, and one hamburger
and one fry cost $5. what is the price of each?
2x + y = 7
We translate these information in to two equations:
Eq. 1
8.
Systems of Linear Equations
We need one piece of information to solve for one
unknown quantity. If 2 hamburgers cost 4 dollars and
x is the cost of a hamburger, then 2x = 4 or x = $2.
If there are two unknowns x and y, we need two
pieces of information to solve them.
Example A:
Let x = cost of a hamburger, y = cost of a fry. If two
hamburgers and a fry cost $7, and one hamburger
and one fry cost $5. what is the price of each?
2x + y = 7
x + y = 5{ Eq. 1
Eq. 2
We translate these information in to two equations:
9.
Systems of Linear Equations
We need one piece of information to solve for one
unknown quantity. If 2 hamburgers cost 4 dollars and
x is the cost of a hamburger, then 2x = 4 or x = $2.
If there are two unknowns x and y, we need two
pieces of information to solve them.
Example A:
Let x = cost of a hamburger, y = cost of a fry. If two
hamburgers and a fry cost $7, and one hamburger
and one fry cost $5. what is the price of each?
2x + y = 7
x + y = 5{
A group of equations such as this is called a system
of equations.
Eq. 1
Eq. 2
We translate these information in to two equations:
10.
Systems of Linear Equations
We are to find values for x and for y that will satisfy all
equations in the system.
11.
Systems of Linear Equations
We are to find values for x and for y that will satisfy all
equations in the system. Such a set of values is
called a solution for the system.
12.
Systems of Linear Equations
We are to find values for x and for y that will satisfy all
equations in the system. Such a set of values is
called a solution for the system. To solve the system,
we note that the difference in the two orders is one
hamburger and the difference in price is $2.
13.
Systems of Linear Equations
We are to find values for x and for y that will satisfy all
equations in the system. Such a set of values is
called a solution for the system. To solve the system,
we note that the difference in the two orders is one
hamburger and the difference in price is $2. So the
price of the hamburger is $2.
14.
Systems of Linear Equations
We are to find values for x and for y that will satisfy all
equations in the system. Such a set of values is
called a solution for the system. To solve the system,
we note that the difference in the two orders is one
hamburger and the difference in price is $2. So the
price of the hamburger is $2. In algebra, we subtract
the equations in the system:
2x + y = 7
x + y = 5)
Eq. 1
Eq. 2
15.
Systems of Linear Equations
We are to find values for x and for y that will satisfy all
equations in the system. Such a set of values is
called a solution for the system. To solve the system,
we note that the difference in the two orders is one
hamburger and the difference in price is $2. So the
price of the hamburger is $2. In algebra, we subtract
the equations in the system:
2x + y = 7
x + y = 5)
x = 2
Eq. 1
Eq. 2
16.
Systems of Linear Equations
We are to find values for x and for y that will satisfy all
equations in the system. Such a set of values is
called a solution for the system. To solve the system,
we note that the difference in the two orders is one
hamburger and the difference in price is $2. So the
price of the hamburger is $2. In algebra, we subtract
the equations in the system:
2x + y = 7
x + y = 5)
x = 2
Eq. 1
Eq. 2
Put x = 2 back into Eq. 2, we get y = 3.
17.
Systems of Linear Equations
We are to find values for x and for y that will satisfy all
equations in the system. Such a set of values is
called a solution for the system. To solve the system,
we note that the difference in the two orders is one
hamburger and the difference in price is $2. So the
price of the hamburger is $2. In algebra, we subtract
the equations in the system:
2x + y = 7
x + y = 5)
x = 2
Eq. 1
Eq. 2
Put x = 2 back into Eq. 2, we get y = 3.
Hence a hamburger is $2 and a fry is $3.
18.
Systems of Linear Equations
We are to find values for x and for y that will satisfy all
equations in the system. Such a set of values is
called a solution for the system. To solve the system,
we note that the difference in the two orders is one
hamburger and the difference in price is $2. So the
price of the hamburger is $2. In algebra, we subtract
the equations in the system:
2x + y = 7
x + y = 5)
x = 2
Eq. 1
Eq. 2
Put x = 2 back into Eq. 2, we get y = 3.
Hence a hamburger is $2 and a fry is $3.
This is called the elimination method - we eliminate
variables by adding or subtracting two equations.
19.
Elimination method reduces the number of
variables in the system one at a time.
Systems of Linear Equations
20.
Elimination method reduces the number of
variables in the system one at a time. The method
changes a given system into a smaller system.
Systems of Linear Equations
21.
Elimination method reduces the number of
variables in the system one at a time. The method
changes a given system into a smaller system. We
do this successively until the solution is clear.
Systems of Linear Equations
22.
Elimination method reduces the number of
variables in the system one at a time. The method
changes a given system into a smaller system. We
do this successively until the solution is clear.
Systems of Linear Equations
Example B:
Let x = cost of a hamburger, y = cost of a fry,
z = cost of a soda.
2 hamburgers, 3 fries and 3 soda cost $13.
1 hamburger, 2 fries and 2 soda cost $8.
3 hamburgers, 2 fries, 3 soda cost $13.
Find x, y, and z.
23.
Elimination method reduces the number of
variables in the system one at a time. The method
changes a given system into a smaller system. We
do this successively until the solution is clear.
Systems of Linear Equations
We translate the information into a system of three
equations with three unknowns
Example B:
Let x = cost of a hamburger, y = cost of a fry,
z = cost of a soda.
2 hamburgers, 3 fries and 3 soda cost $13.
1 hamburger, 2 fries and 2 soda cost $8.
3 hamburgers, 2 fries, 3 soda cost $13.
Find x, y, and z.
24.
2x + 3y + 3z = 13
x + 2y + 2z = 8
3x + 2y + 3z = 13
{
E 1
E 2
E 3
Systems of Linear Equations
25.
2x + 3y + 3z = 13
x + 2y + 2z = 8
3x + 2y + 3z = 13
{
E 1
E 2
E 3
Select x to eliminate since there is 1x in E2.
Systems of Linear Equations
26.
2x + 3y + 3z = 13
x + 2y + 2z = 8
3x + 2y + 3z = 13
{
E 1
E 2
E 3
Select x to eliminate since there is 1x in E2.
-2*E 2 + E1:
Systems of Linear Equations
27.
2x + 3y + 3z = 13
x + 2y + 2z = 8
3x + 2y + 3z = 13
{
E 1
E 2
E 3
Select x to eliminate since there is 1x in E2.
-2*E 2 + E1:
-2x – 4y – 4z = -16
+) 2x + 3y + 3z = 13
Systems of Linear Equations
28.
2x + 3y + 3z = 13
x + 2y + 2z = 8
3x + 2y + 3z = 13
{
E 1
E 2
E 3
Select x to eliminate since there is 1x in E2.
-2*E 2 + E1:
-2x – 4y – 4z = -16
+) 2x + 3y + 3z = 13
0 – y – z = - 3
Systems of Linear Equations
29.
2x + 3y + 3z = 13
x + 2y + 2z = 8
3x + 2y + 3z = 13
{
E 1
E 2
E 3
Select x to eliminate since there is 1x in E2.
-2*E 2 + E1:
-2x – 4y – 4z = -16
+) 2x + 3y + 3z = 13
0 – y – z = - 3
-3*E 2 + E3:
Systems of Linear Equations
30.
2x + 3y + 3z = 13
x + 2y + 2z = 8
3x + 2y + 3z = 13
{
E 1
E 2
E 3
Select x to eliminate since there is 1x in E2.
-2*E 2 + E1:
-2x – 4y – 4z = -16
+) 2x + 3y + 3z = 13
0 – y – z = - 3
-3*E 2 + E3:
-3x – 6y – 6z = -24
+) 3x + 2y + 3z = 13
Systems of Linear Equations
31.
2x + 3y + 3z = 13
x + 2y + 2z = 8
3x + 2y + 3z = 13
{
E 1
E 2
E 3
Select x to eliminate since there is 1x in E2.
-2*E 2 + E1:
-2x – 4y – 4z = -16
+) 2x + 3y + 3z = 13
0 – y – z = - 3
-3*E 2 + E3:
-3x – 6y – 6z = -24
+) 3x + 2y + 3z = 13
0 – 4y – 3z = -11
Systems of Linear Equations
32.
2x + 3y + 3z = 13
x + 2y + 2z = 8
3x + 2y + 3z = 13
{
E 1
E 2
E 3
Select x to eliminate since there is 1x in E2.
-2*E 2 + E1:
-2x – 4y – 4z = -16
+) 2x + 3y + 3z = 13
0 – y – z = - 3
-3*E 2 + E3:
-3x – 6y – 6z = -24
+) 3x + 2y + 3z = 13
0 – 4y – 3z = -11
Group these into a system of two equations with
two unknowns.
Systems of Linear Equations
33.
2x + 3y + 3z = 13
x + 2y + 2z = 8
3x + 2y + 3z = 13
{
E 1
E 2
E 3
Select x to eliminate since there is 1x in E2.
-2*E 2 + E1:
-2x – 4y – 4z = -16
+) 2x + 3y + 3z = 13
0 – y – z = - 3
-3*E 2 + E3:
-3x – 6y – 6z = -24
+) 3x + 2y + 3z = 13
0 – 4y – 3z = -11
Group these into a system of two equations with
two unknowns.
Systems of Linear Equations
Hence we reduced the system to a simpler one.
34.
– y – z = - 3
– 4y – 3z = -11{
Systems of Linear Equations
35.
– y – z = - 3
– 4y – 3z = -11{ 4y + 3z = 11
{ y + z = 3 Eq.4
Eq. 5
Systems of Linear Equations
*(-1)
36.
– y – z = - 3
– 4y – 3z = -11{ 4y + 3z = 11
{ y + z = 3 Eq.4
Eq. 5
Systems of Linear Equations
*(-1)
Let's eliminate y.
37.
– y – z = - 3
– 4y – 3z = -11{ 4y + 3z = 11
{ y + z = 3 Eq.4
Eq. 5
-3*E 4 + E5:
Systems of Linear Equations
*(-1)
Let's eliminate y.
38.
– y – z = - 3
– 4y – 3z = -11{ 4y + 3z = 11
{ y + z = 3 Eq.4
Eq. 5
-3*E 4 + E5:
+) 4y + 3z = 11
-3y – 3z = -9
Systems of Linear Equations
*(-1)
Let's eliminate y.
39.
– y – z = - 3
– 4y – 3z = -11{ 4y + 3z = 11
{ y + z = 3 Eq.4
Eq. 5
-3*E 4 + E5:
+) 4y + 3z = 11
-3y – 3z = -9
y + 0 = 2
y = 2
Systems of Linear Equations
*(-1)
Let's eliminate y.
40.
– y – z = - 3
– 4y – 3z = -11{ 4y + 3z = 11
{ y + z = 3 Eq.4
Eq. 5
-3*E 4 + E5:
+) 4y + 3z = 11
-3y – 3z = -9
y + 0 = 2
y = 2
To get z, set 2 for y in E4:
Systems of Linear Equations
*(-1)
Let's eliminate y.
41.
– y – z = - 3
– 4y – 3z = -11{ 4y + 3z = 11
{ y + z = 3 Eq.4
Eq. 5
-3*E 4 + E5:
+) 4y + 3z = 11
-3y – 3z = -9
y + 0 = 2
y = 2
To get z, set 2 for y in E4: 2 + z = 3 z = 1
Systems of Linear Equations
*(-1)
Let's eliminate y.
42.
– y – z = - 3
– 4y – 3z = -11{ 4y + 3z = 11
{ y + z = 3 Eq.4
Eq. 5
-3*E 4 + E5:
+) 4y + 3z = 11
-3y – 3z = -9
y + 0 = 2
y = 2
To get z, set 2 for y in E4: 2 + z = 3 z = 1
For x, set 2 for y , set 1 for z in E2:
Systems of Linear Equations
*(-1)
Let's eliminate y.
43.
– y – z = - 3
– 4y – 3z = -11{ 4y + 3z = 11
{ y + z = 3 Eq.4
Eq. 5
-3*E 4 + E5:
+) 4y + 3z = 11
-3y – 3z = -9
y + 0 = 2
y = 2
To get z, set 2 for y in E4: 2 + z = 3 z = 1
For x, set 2 for y , set 1 for z in E2:
x + 2*2 + 2*1 = 8
Systems of Linear Equations
*(-1)
Let's eliminate y.
44.
– y – z = - 3
– 4y – 3z = -11{ 4y + 3z = 11
{ y + z = 3 Eq.4
Eq. 5
-3*E 4 + E5:
+) 4y + 3z = 11
-3y – 3z = -9
y + 0 = 2
y = 2
To get z, set 2 for y in E4: 2 + z = 3 z = 1
For x, set 2 for y , set 1 for z in E2:
x + 2*2 + 2*1 = 8 x + 6 = 8 x = 2
Systems of Linear Equations
*(-1)
Let's eliminate y.
45.
– y – z = - 3
– 4y – 3z = -11{ 4y + 3z = 11
{ y + z = 3 Eq.4
Eq. 5
-3*E 4 + E5:
+) 4y + 3z = 11
-3y – 3z = -9
y + 0 = 2
y = 2
To get z, set 2 for y in E4: 2 + z = 3 z = 1
For x, set 2 for y , set 1 for z in E2:
x + 2*2 + 2*1 = 8 x + 6 = 8 x = 2
Systems of Linear Equations
*(-1)
Hence the hamburger cost $2, the fry cost $2 and
the drink cost $1.
Let's eliminate y.
46.
A matrix is a rectangular table of numbers.
Matrix Notation
47.
A matrix is a rectangular table of numbers.
For example
5 2 -3
4 -1 0
-1 2 -1
6 -2 3
11 9 -4
Matrix Notation
are matrices.
48.
A matrix is a rectangular table of numbers.
For example
5 2 -3
4 -1 0
-1 2 -1
6 -2 3
11 9 -4
Matrix Notation
are matrices.
Systems of linear equations can be put into matrices
and solved using matrix notation.
49.
A matrix is a rectangular table of numbers.
For example
5 2 -3
4 -1 0
-1 2 -1
6 -2 3
11 9 -4
Matrix Notation
are matrices.
Systems of linear equations can be put into matrices
and solved using matrix notation.
x + 4y = -7
2x – 3y = 8{
For example, the system
1 4 -7
2 -3 8
may be written as the matrix:
x y #
50.
A matrix is a rectangular table of numbers.
For example
5 2 -3
4 -1 0
-1 2 -1
6 -2 3
11 9 -4
Matrix Notation
are matrices.
Systems of linear equations can be put into matrices
and solved using matrix notation.
x + 4y = -7
2x – 3y = 8{
For example, the system
1 4 -7
2 -3 8
may be written as the matrix:
x y #
This is called the augmented matrix for the system.
51.
A matrix is a rectangular table of numbers.
For example
5 2 -3
4 -1 0
-1 2 -1
6 -2 3
11 9 -4
Matrix Notation
are matrices.
Systems of linear equations can be put into matrices
and solved using matrix notation.
x + 4y = -7
2x – 3y = 8{
For example, the system
1 4 -7
2 -3 8
may be written as the matrix:
x y #
This is called the augmented matrix for the system.
Each row of the matrix corresponds to an equation,
each column corresponds to a variable and the last
column corresponds to numbers.
52.
Matrix Notation
An augmented matrix can be easily converted back to
a system .
53.
Matrix Notation
An augmented matrix can be easily converted back to
a system .
For example, the matrix
2 -3 8
1 4 -7
is the system x + 4y = -7{2x – 3y = 8
54.
Operations of the equations correspond to operations
of rows in the matrices.
Matrix Notation
An augmented matrix can be easily converted back to
a system .
For example, the matrix
2 -3 8
1 4 -7
is the system x + 4y = -7{2x – 3y = 8
55.
Operations of the equations correspond to operations
of rows in the matrices. Three such operations are
important. These are the elementary row operations:
Matrix Notation
An augmented matrix can be easily converted back to
a system .
For example, the matrix
2 -3 8
1 4 -7
is the system x + 4y = -7{2x – 3y = 8
56.
Matrix Notation
I. Switch two rows.
An augmented matrix can be easily converted back to
a system .
For example, the matrix
2 -3 8
1 4 -7
is the system x + 4y = -7{2x – 3y = 8
Operations of the equations correspond to operations
of rows in the matrices. Three such operations are
important. These are the elementary row operations:
57.
Matrix Notation
I. Switch two rows. Switching row i with row j is
notated as Ri Rj.
An augmented matrix can be easily converted back to
a system .
For example, the matrix
2 -3 8
1 4 -7
is the system x + 4y = -7{2x – 3y = 8
Operations of the equations correspond to operations
of rows in the matrices. Three such operations are
important. These are the elementary row operations:
58.
Matrix Notation
1 4 -7
2 -3 8
For example,
R1 R2
I. Switch two rows. Switching row i with row j is
notated as Ri Rj.
An augmented matrix can be easily converted back to
a system .
For example, the matrix
2 -3 8
1 4 -7
is the system x + 4y = -7{2x – 3y = 8
Operations of the equations correspond to operations
of rows in the matrices. Three such operations are
important. These are the elementary row operations:
59.
Matrix Notation
1 4 -7
2 -3 8
For example,
R1 R2 2 -3 8
1 4 -7
I. Switch two rows. Switching row i with row j is
notated as Ri Rj.
An augmented matrix can be easily converted back to
a system .
For example, the matrix
2 -3 8
1 4 -7
is the system x + 4y = -7{2x – 3y = 8
Operations of the equations correspond to operations
of rows in the matrices. Three such operations are
important. These are the elementary row operations:
60.
Matrix Notation
II. Multiply a row by a constant k = 0.
61.
Matrix Notation
II. Multiply a row by a constant k = 0. Multiply row i by
k is notated as k*Ri.
62.
Matrix Notation
II. Multiply a row by a constant k = 0. Multiply row i by
k is notated as k*Ri.
1 4 -7
2 -3 8
-3*R2
For example,
63.
Matrix Notation
II. Multiply a row by a constant k = 0. Multiply row i by
k is notated as k*Ri.
1 4 -7
2 -3 8
-3*R2 1 4 -7
-6 9 -24
For example,
64.
III. Add the multiple of a row to another row.
Matrix Notation
II. Multiply a row by a constant k = 0. Multiply row i by
k is notated as k*Ri.
1 4 -7
2 -3 8
-3*R2 1 4 -7
-6 9 -24
For example,
65.
III. Add the multiple of a row to another row. k times
row i added to row j is notated as “k*Ri add Rj”.
Matrix Notation
II. Multiply a row by a constant k = 0. Multiply row i by
k is notated as k*Ri.
1 4 -7
2 -3 8
-3*R2 1 4 -7
-6 9 -24
For example,
66.
III. Add the multiple of a row to another row. k times
row i added to row j is notated as “k*Ri add Rj”.
1 4 -7
2 -3 8
For example,
-2*R1 add R2
Matrix Notation
II. Multiply a row by a constant k = 0. Multiply row i by
k is notated as k*Ri.
1 4 -7
2 -3 8
-3*R2 1 4 -7
-6 9 -24
For example,
67.
III. Add the multiple of a row to another row. k times
row i added to row j is notated as “k*Ri add Rj”.
1 4 -7
2 -3 8
For example,
-2*R1 add R2
-2 -8 14
write a copy of -2*R1
Matrix Notation
II. Multiply a row by a constant k = 0. Multiply row i by
k is notated as k*Ri.
1 4 -7
2 -3 8
-3*R2 1 4 -7
-6 9 -24
For example,
68.
III. Add the multiple of a row to another row. k times
row i added to row j is notated as “k*Ri add Rj”.
1 4 -7
2 -3 8
For example,
-2*R1 add R2 1 4 -7
0 -11 22
-2 -8 14
write a copy of -2*R1
Matrix Notation
II. Multiply a row by a constant k = 0. Multiply row i by
k is notated as k*Ri.
1 4 -7
2 -3 8
-3*R2 1 4 -7
-6 9 -24
For example,
69.
III. Add the multiple of a row to another row. k times
row i added to row j is notated as “k*Ri add Rj”.
1 4 -7
2 -3 8
For example,
-2*R1 add R2 1 4 -7
0 -11 22
-2 -8 14
write a copy of -2*R1
Fact: Performing elementary row operations on a
matrix does not change the solution of the system.
Matrix Notation
II. Multiply a row by a constant k = 0. Multiply row i by
k is notated as k*Ri.
1 4 -7
2 -3 8
-3*R2 1 4 -7
-6 9 -24
For example,
70.
Matrix Notation
The elimination method may be carried out with
matrix notation.
71.
Elimination Method in Matrix Notation:
Matrix Notation
The elimination method may be carried out with
matrix notation.
72.
Matrix Notation
The elimination method may be carried out with
matrix notation.
Elimination Method in Matrix Notation:
1. Apply row operations to transform the matrix to the
upper diagonal form where all the entries below the
main diagonal (the lower left triangular region) are 0 .
73.
* * * *
* * * *
* * * *
Row operations
Matrix Notation
The elimination method may be carried out with
matrix notation.
Elimination Method in Matrix Notation:
1. Apply row operations to transform the matrix to the
upper diagonal form where all the entries below the
main diagonal (the lower left triangular region) are 0 .
74.
* * * *
* * * *
* * * *
Row operations * * * *
* * *
* *
0
0 0
Matrix Notation
The elimination method may be carried out with
matrix notation.
Elimination Method in Matrix Notation:
1. Apply row operations to transform the matrix to the
upper diagonal form where all the entries below the
main diagonal (the lower left triangular region) are 0 .
75.
* * * *
* * * *
* * * *
Row operations * * * *
* * *
* *
0
0 0
2. Starting from the bottom row, get the answer
for one of the variable.
Matrix Notation
The elimination method may be carried out with
matrix notation.
Elimination Method in Matrix Notation:
1. Apply row operations to transform the matrix to the
upper diagonal form where all the entries below the
main diagonal (the lower left triangular region) are 0 .
76.
* * * *
* * * *
* * * *
Row operations * * * *
* * *
* *
0
0 0
2. Starting from the bottom row, get the answer
for one of the variable. Then go up one row, use
the solution already obtained to get another answer of
another variable.
Matrix Notation
The elimination method may be carried out with
matrix notation.
Elimination Method in Matrix Notation:
1. Apply row operations to transform the matrix to the
upper diagonal form where all the entries below the
main diagonal (the lower left triangular region) are 0 .
77.
Elimination Method in Matrix Notation:
1. Apply row operations to transform the matrix to the
upper diagonal form where all the entries below the
main diagonal (the lower left triangular region) are 0 .
* * * *
* * * *
* * * *
Row operations * * * *
* * *
* *
0
0 0
2. Starting from the bottom row, get the answer
for one of the variable. Then go up one row, use
the solution already obtained to get another answer of
another variable. Repeat the process, working from
the bottom row to the top row to extract all solutions.
Matrix Notation
The elimination method may be carried out with
matrix notation.
78.
x + 4y = -7
2x – 3y = 8{ Eq. 1
Eq. 2
Example C: Solve using matrix notation:
Matrix Notation
79.
1 4 -7
2 -3 8
x + 4y = -7
2x – 3y = 8{ Eq. 1
Eq. 2
Example C: Solve using matrix notation:
Put the system into a matrix:
Matrix Notation
80.
1 4 -7
2 -3 8
-2*R1 add R2
x + 4y = -7
2x – 3y = 8{ Eq. 1
Eq. 2
Example C: Solve using matrix notation:
Put the system into a matrix:
Matrix Notation
81.
1 4 -7
2 -3 8
-2*R1 add R2
-2 -8 14
x + 4y = -7
2x – 3y = 8{ Eq. 1
Eq. 2
Example C: Solve using matrix notation:
Put the system into a matrix:
Matrix Notation
82.
1 4 -7
2 -3 8
-2*R1 add R2 1 4 -7
0 -11 22
-2 -8 14
x + 4y = -7
2x – 3y = 8{ Eq. 1
Eq. 2
Example C: Solve using matrix notation:
Put the system into a matrix:
Matrix Notation
83.
1 4 -7
2 -3 8
-2*R1 add R2 1 4 -7
0 -11 22
-2 -8 14
x + 4y = -7
2x – 3y = 8{ Eq. 1
Eq. 2
Example C: Solve using matrix notation:
Put the system into a matrix:
Start from the bottom row: -11y = 22 y = -2
Matrix Notation
84.
1 4 -7
2 -3 8
-2*R1 add R2 1 4 -7
0 -11 22
-2 -8 14
x + 4y = -7
2x – 3y = 8{ Eq. 1
Eq. 2
Example C: Solve using matrix notation:
Put the system into a matrix:
Start from the bottom row: -11y = 22 y = -2
Go up one row to R1 and set y = -2:
Matrix Notation
85.
1 4 -7
2 -3 8
-2*R1 add R2 1 4 -7
0 -11 22
-2 -8 14
x + 4y = -7
2x – 3y = 8{ Eq. 1
Eq. 2
Example C: Solve using matrix notation:
Put the system into a matrix:
Start from the bottom row: -11y = 22 y = -2
Go up one row to R1 and set y = -2:
x + 4y = -7
x + 4(-2) = -7
Matrix Notation
86.
1 4 -7
2 -3 8
-2*R1 add R2 1 4 -7
0 -11 22
-2 -8 14
x + 4y = -7
2x – 3y = 8{ Eq. 1
Eq. 2
Example C: Solve using matrix notation:
Put the system into a matrix:
Start from the bottom row: -11y = 22 y = -2
Go up one row to R1 and set y = -2:
x + 4y = -7
x + 4(-2) = -7 x = 1
Matrix Notation
Hence the solution is {x = 1, y = -2}.
87.
1 4 -7
2 -3 8
-2*R1 add R2 1 4 -7
0 -11 22
-2 -8 14
x + 4y = -7
2x – 3y = 8{ Eq. 1
Eq. 2
Example C: Solve using matrix notation:
Put the system into a matrix:
Start from the bottom row: -11y = 22 y = -2
Go up one row to R1 and set y = -2:
x + 4y = -7
x + 4(-2) = -7 x = 1
Matrix Notation
Hence the solution is {x = 1, y = -2}.
It may also be written as the ordered pair (1, -2).
88.
Example D: Solve using matrix notation:
2x + 3y + 3z = 13
x + 2y + 2z = 8
3x + 2y + 3z = 13
{
E 1
E 2
E 3
Matrix Notation
89.
Example D: Solve using matrix notation:
2x + 3y + 3z = 13
x + 2y + 2z = 8
3x + 2y + 3z = 13
{
E 1
E 2
E 3
2 3 3 13
Put the system into a matrix:
1 2 2 8
3 2 3 13
Matrix Notation
90.
Example D: Solve using matrix notation:
2x + 3y + 3z = 13
x + 2y + 2z = 8
3x + 2y + 3z = 13
{
E 1
E 2
E 3
2 3 3 13 R1 R2
Put the system into a matrix:
1 2 2 8
3 2 3 13
2 3 3 13
1 2 2 8
3 2 3 13
Matrix Notation
91.
Example D: Solve using matrix notation:
2x + 3y + 3z = 13
x + 2y + 2z = 8
3x + 2y + 3z = 13
{
E 1
E 2
E 3
2 3 3 13 R1 R2
-2 -4 -4 -16Put the system into a matrix:
1 2 2 8
3 2 3 13
2 3 3 13
1 2 2 8
3 2 3 13
-2*R1 add R2
Matrix Notation
92.
Example D: Solve using matrix notation:
2x + 3y + 3z = 13
x + 2y + 2z = 8
3x + 2y + 3z = 13
{
E 1
E 2
E 3
2 3 3 13 R1 R2
-2 -4 -4 -16Put the system into a matrix:
1 2 2 8
3 2 3 13
2 3 3 13
1 2 2 8
3 2 3 13
-2*R1 add R2
0 -1 -1 -3
1 2 2 8
3 2 3 13
Matrix Notation
93.
Example D: Solve using matrix notation:
2x + 3y + 3z = 13
x + 2y + 2z = 8
3x + 2y + 3z = 13
{
E 1
E 2
E 3
2 3 3 13 R1 R2
-2 -4 -4 -16Put the system into a matrix:
1 2 2 8
3 2 3 13
2 3 3 13
1 2 2 8
3 2 3 13
-2*R1 add R2
0 -1 -1 -3
1 2 2 8
3 2 3 13
-3* R1 add R3
-3 -6 -6 -24
Matrix Notation
98.
0 -1 -1 -3
1 2 2 8
0 0 1 1
Matrix Notation
Extract the solution starting
from the bottom row.
99.
0 -1 -1 -3
1 2 2 8
0 0 1 1
From R3, we get z = 1.
Matrix Notation
Extract the solution starting
from the bottom row.
100.
0 -1 -1 -3
1 2 2 8
0 0 1 1
From R3, we get z = 1.
Go up one row to R2, we get
-y – z = -3
-y – (1) = -3
Matrix Notation
Extract the solution starting
from the bottom row.
101.
0 -1 -1 -3
1 2 2 8
0 0 1 1
From R3, we get z = 1.
Go up one row to R2, we get
-y – z = -3
-y – (1) = -3
3 – 1 = y
2 = y
Matrix Notation
Extract the solution starting
from the bottom row.
102.
0 -1 -1 -3
1 2 2 8
0 0 1 1
From R3, we get z = 1.
Go up one row to R2, we get
-y – z = -3
-y – (1) = -3
3 – 1 = y
2 = y
Up to R1, we get x + 2y + 2z = 8
x + 2(2) + 2(1) = 8
Matrix Notation
Extract the solution starting
from the bottom row.
103.
0 -1 -1 -3
1 2 2 8
0 0 1 1
From R3, we get z = 1.
Go up one row to R2, we get
-y – z = -3
-y – (1) = -3
3 – 1 = y
2 = y
Up to R1, we get x + 2y + 2z = 8
x + 2(2) + 2(1) = 8
x + 6 = 8
x = 2
Matrix Notation
Extract the solution starting
from the bottom row.
104.
0 -1 -1 -3
1 2 2 8
0 0 1 1
From R3, we get z = 1.
Go up one row to R2, we get
-y – z = -3
-y – (1) = -3
3 – 1 = y
2 = y
Up to R1, we get x + 2y + 2z = 8
x + 2(2) + 2(1) = 8
x + 6 = 8
x = 2
So the solution is (2, 2, 1).
Matrix Notation
Extract the solution starting
from the bottom row.
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