chapter 8.1 matrices and systems of equations. matrix- a streamlined technique for solving systems...
TRANSCRIPT
CHAPTER 8.1
Matrices and Systems of Equations
Matrix- a streamlined technique for solving systems of linear equations that involves the use of a rectangular array of numbers
11 12 13
21 22 23
31 32 33
a a a
a a a
a a a
M rows
N columnsM x N
2
11 3 0
2
2 4
3 6
6
0
5
1 1
2 2
3 1
1 4
Order of Matrices
system
augmented matrix
coefficient matrix
2 4 6 10
6 2 6
4 2 2
x y z
x y z
x z
4 2 3
2 5 1
0 3 2
2 3 3 4
1 2 4 3
0 3 1 5
4 8
2 3
5 0
x y
y z
x z
1 4 0 8
0 1 2 3
1 0 5 0
Writing an Augmented Matrix1. Begin by writing the linear system and aligning the variables.
2. Next, use the coefficients and constant terms as the matrix entries. Include zeroes for each missing coefficients.
Elementary Row Operations
1. Interchange two rows.
2. Multiply a row by a nonzero constant.
3. Add a multiple of a row to another row.
Associated Augmented Matrix
1 2 3 9
1 3 0 4
2 5 5 17
1 2 3 9
0 1 3 5
2 5 5 17
1 2 3 9
0 1 3 5
0 1 1 1
1 2R R
1 32R R
1 2 3 9
0 1 3 5
0 0 2 4
2 3R R
3
1
2R
1 2 3 9
0 1 3 5
0 0 1 2
Use back-substitution to find the solution
1
1
2
x
y
z
Row-Echelon Form and Reduced Row-Echelon Form
1. All rows consisting of zeroes occur at the bottom of the matrix.
2. For each row that does not consist of zeroes, the first nonzero entry is 1(called a leading 1).
3. For two successive (nonzero) rows, the leading 1 in the higher row is farther to the left than the leading 1 in the lower row.
Reduced row-echelon form- if every column that has a leading 1 has zeroes in every position above and below its leading 1.
For example:
Row-Echelon Form
1 2 1 4
0 1 0 3
0 0 1 2
1 5 2 1 3
0 0 1 3 2
0 0 0 1 4
0 0 0 0 1
Reduced row-echelon form
1 0 0 1
0 1 0 2
0 0 1 3
0 0 0 0
0 1 0 5
0 0 1 3
0 0 0 0
Solve the system
2 3
2 2
2 4 3 2
4 7 19
y z w
x y z
x y z w
x y z w
Switch row 1 with row 2
1 2 1 0 2
0 1 1 2 3
2 4 1 3 2
1 4 7 1 19
1 3
1 4
2R R
R R
2 46R R
1 2 1 0 2
0 1 1 2 3
0 0 3 3 6
0 0 0 13 39
1 2 1 0 2
0 1 1 2 3
0 0 3 3 6
0 6 6 1 21
3
1
3R
1 2 1 0 2
0 1 1 2 3
0 0 1 1 2
0 0 0 13 39
4
1
13R
1 2 1 0 2
0 1 1 2 3
0 0 1 1 2
0 0 0 1 3
2 2
2 3
2
3
x y z
y z w
z w
w
Use back-substitution1
2
1
3
x
y
z
w
Gaussian Elimination with Back-Substitution
1. Write the augmented matrix of the system of linear equations.
2. Use elementary row operations to rewrite the augmented matrix in row-echelon form.
3. Write the system of linear equations corresponding to the matrix in row-echelon form and use back-substitution to find the solution.
Solve the system
1 1 2 4
1 0 1 6
2 3 5 4
3 2 1 1
1 2R R
1 1 2 4
0 1 1 2
2 3 5 4
3 2 1 1
1 32R R
1 1 2 4
0 1 1 2
0 1 1 4
3 2 1 1
1 4
3R R
2 3R R
1 1 2 4
0 1 1 2
0 0 0 2
0 5 7 11
inconsistent, no solution
1 1 2 4
0 1 1 2
0 1 1 4
0 5 7 11
In row 3, 0 2
1 2 3 9
0 1 3 5
0 0 1 2
Apply additional elementary row operations until you obtain a matrix in reduced row-echelon form.
2 12R R
1 0 9 19
0 1 3 5
0 0 1 2
3 1
3 2
9
3
R R
R R
1 0 0 1
0 1 0 1
0 0 1 2
1
1
2
x
y
z
Solve the system
2 4 2 0
3 5 0 1
1 2 1 0
3 5 0 1
1 2 1 0
0 1 3 1
1 2 1 0
0 1 3 1
1 0 5 2
0 1 3 1
5 2
3 1
x z
y z
5 2
1 3
x z
y z
Let z a where a is
A real number, then the solution set has the form
( 5 2, 3 1, )a a a =infinite number of solutions
1
1
2R
1 23R R
2R
2 12R R