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Transcript
```8.1
MATRICES AND SYSTEMS OF EQUATIONS
What You Should Learn
• Write matrices and identify their orders.
• Perform elementary row operations on matrices.
• Use matrices and Gaussian elimination to solve
systems of linear equations.
• Use matrices and Gauss-Jordan elimination to
solve systems of linear equations.
2
Matrices
3
Matrices
A matrix having m rows and n columns is said to be of order m  n.
If m = n, the matrix is square of order m  m (or n  n). For a square
matrix, the entries a11, a22, a33, . . . are the main diagonal entries.
4
Example 1 – Order of Matrices
Determine the order of each matrix.
a.
b.
Solution:
a. This matrix has one row and one column. The order of
the matrix is 1  1.
b. This matrix has one row and four columns. The order of
the matrix is 1  4.
5
Example 1 – Order of Matrices
Determine the order of each matrix.
c.
d.
Solution:
c. This matrix has two rows and two columns. The order of
the matrix is 2  2.
d. This matrix has three rows and two columns. The order
of the matrix is 3  2.
6
Matrices
Row matrix: A matrix that has only one row
Column matrix: Amatrix that has only one column
Augmented matrix of the system: A matrix derived from a
system of linear equations (each written in standard form
with the constant term on the right)
Coefficient matrix of the system: the matrix derived from
the coefficients of the system (but not including the
constant terms)
7
Matrices
System:
x – 4y + 3z = 5
–x + 3y – z = –3
2x
– 4z = 6
Augmented Matrix:
Coefficient Matrix:
8
Elementary Row Operations
9
Elementary Row Operations
1. Interchange two equations.
Interchange two rows
2. Multiply an equation by a nonzero constant.
Multiply a row by a nonzero constant
3. Add a multiple of an equation to another equation.
Add a multiple of a row to another row
10
Elementary Row Operations
1
0

0


0

0
1
0 1
 
0 0
0 0







 1 

 0 1
1
0

0


0

0
0
1
0

0
0
0
0
1

0
0






0
0
0

1
0
0
0
0


0

1
11
Gaussian Elimination with
Back-Substitution
12
Example 6 – Gaussian Elimination with Back-Substitution
Solve the system
y + z – 2w = –3
x + 2y – z
= 2 .
2x + 4y + z – 3w = – 2
x – 4y – 7z – w = –19
Solution:
Write augmented matrix.
13
Example 6 – Solution
cont’d
Interchange R1 and R2 so
in upper left corner.
Perform operations on R3
and R4 so first column has
14
Example 6 – Solution
cont’d
Perform operations on R4
so second column has
Perform operations on R3
and R4 so third and fourth
15
Example 6 – Solution
cont’d
The matrix is now in row-echelon form, and the
corresponding system is
x + 2y – z
= 2
y + z – 2w = –3 .
z – w = –2
w= 3
Using back-substitution, you can determine that the
solution is x = –1, y = 2, z = 1, and w = 3.
16
Gaussian Elimination with Back-Substitution
17
Gauss-Jordan Elimination
18
Gauss-Jordan Elimination
With Gaussian elimination, elementary row operations are
applied to a matrix to obtain a (row-equivalent) row-echelon
form of the matrix.
A second method of elimination, called Gauss-Jordan
elimination, continues the reduction process until a
reduced row-echelon form is obtained.
19
Example 8 – Gauss-Jordan Elimination
Use Gauss-Jordan elimination to solve the system
x – 2y + 3z = 9
–x + 3y
= –4 .
2x – 5y + 5z = 17
Solution:
The row-echelon form of the linear system can be obtained
using the Gaussian elimination.
20
Example 8 – Solution
cont’d
Now, apply elementary row operations until you obtain
zeros above each of the leading 1’s, as follows.
Perform operations on R1
so second column has a
Perform operations on R1
and R2 so third column has
21
Example 8 – Solution
cont’d
The matrix is now in reduced row-echelon form. Converting
back to a system of linear equations, you have
x=1
y = –1.
z=2
Now you can simply read the solution, x = 1, y = –1, and
z = 2 which can be written as the ordered triple(1, –1, 2).
22
Gaussian Elimination with
Back substitution
1
0

0


0

0
1
0 1
 
0 0
0 0







 1 

 0 1
Gaussian Jordon
1
0

0


0

0
0
1
0

0
0
0
0
1

0
0






0
0
0

1
0
0
0
0


0

1
23
```
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