Copyright © 2011 Pearson, Inc. 7.3 Multivariate Linear Systems and Row Operations.
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Transcript of Copyright © 2011 Pearson, Inc. 7.3 Multivariate Linear Systems and Row Operations.
Copyright © 2011 Pearson, Inc.
7.3Multivariate
Linear Systems and Row
Operations
Slide 7.3 - 2 Copyright © 2011 Pearson, Inc.
What you’ll learn about
Triangular Forms for Linear Systems Gaussian Elimination Elementary Row Operations and Row Echelon Form Reduced Row Echelon Form Solving Systems with Inverse Matrices Applications
… and whyMany applications in business and science are modeled by systems of linear equations in three or more variables.
Slide 7.3 - 3 Copyright © 2011 Pearson, Inc.
Equivalent Systems of Linear Equations
The following operations produce an equivalent
system of linear equations.
1. Interchange any two equations of the system.
2. Multiply (or divide) one of the equations by any nonzero real number.
3. Add a multiple of one equation to any other equation in the system.
Slide 7.3 - 4 Copyright © 2011 Pearson, Inc.
Row Echelon Form of a Matrix
A matrix is in row echelon form if the following
conditions are satisfied.
1. Rows consisting entirely of 0’s (if there are any) occur at the bottom of the matrix.
2. The first entry in any row with nonzero entries is 1.
3. The column subscript of the leading 1 entries increases as the row subscript increases.
Slide 7.3 - 5 Copyright © 2011 Pearson, Inc.
Elementary Row Operations on a Matrix
A combination of the following operations will
transform a matrix to row echelon form.
1. Interchange any two rows.
2. Multiply all elements of a row by a nonzero real number.
3. Add a multiple of one row to any other row.
Slide 7.3 - 6 Copyright © 2011 Pearson, Inc.
Example Finding a Row Echelon Form
Solve the system:
2x +3y−z=−1−x+5y+3z=−103x−y−6z=5
Slide 7.3 - 7 Copyright © 2011 Pearson, Inc.
Example Finding a Row Echelon Form
Apply elementary row operations to find a row echelon
form of the augmented matrix.
2 3 −1 −1−1 5 3 −103 −1 −6 5
⎡
⎣
⎢⎢⎢
⎤
⎦
⎥⎥⎥−R21u ruuu
1 −5 −3 102 3 −1 −13 −1 −6 5
⎡
⎣
⎢⎢⎢
⎤
⎦
⎥⎥⎥−2R1 + R2u ruuuuuuuu
1 −5 −3 100 13 5 −213 −1 −6 5
⎡
⎣
⎢⎢⎢
⎤
⎦
⎥⎥⎥−3R1 + R3u ruuuuuuuu
1 −5 −3 100 13 5 −210 14 3 −25
⎡
⎣
⎢⎢⎢
⎤
⎦
⎥⎥⎥
113
R2u ruuuu
2x +3y−z=−1−x+5y+3z=−103x−y−6z=5
Slide 7.3 - 8 Copyright © 2011 Pearson, Inc.
Example Finding a Row Echelon Form
1 −5 −3 10
0 1513
−2113
0 14 3 −25
⎡
⎣
⎢⎢⎢⎢
⎤
⎦
⎥⎥⎥⎥
−14R2 + R3u ruuuuuuuuuu
1 −5 −3 10
0 1513
−2113
0 0−3113
−3113
⎡
⎣
⎢⎢⎢⎢⎢⎢
⎤
⎦
⎥⎥⎥⎥⎥⎥
−1331
R3u ruuuuu
1 −5 −3 10
0 1513
−2113
0 0 1 1
⎡
⎣
⎢⎢⎢⎢
⎤
⎦
⎥⎥⎥⎥
Slide 7.3 - 9 Copyright © 2011 Pearson, Inc.
Example Finding a Row Echelon Form
Convert the matrix to equations and solve by substitution.
z =1;y+5 / 13=−21/ 13 so y=−2;x+10−3=10 so x=3.
The solution is 3,−2,1( ).
Solve the system:
2x +3y−z=−1−x+5y+3z=−103x−y−6z=5
Slide 7.3 - 10 Copyright © 2011 Pearson, Inc.
Reduced Row Echelon Form
If we continue to apply elementary row
operations to a row echelon form of a matrix, we
can obtain a matrix in which every column that
has a leading 1 has 0’s elsewhere. This is the
reduced echelon form.
Slide 7.3 - 11 Copyright © 2011 Pearson, Inc.
Invertible Square Linear System
Let A be the coefficient matrix of a system of n linear equations in n variables given by AX = B, where X is the n 1 matrix of variables and B is the n 1 matrix of numbers on the right-hand side of the equations. If A–1 exists, then the system of equations has the unique solution
X = A–1B.
Slide 7.3 - 12 Copyright © 2011 Pearson, Inc.
Example Solving a System Using Inverse Matrices
Solve the system
2x −3y=02x−2y=10
Slide 7.3 - 13 Copyright © 2011 Pearson, Inc.
Example Solving a System Using Inverse Matrices
Write the system as a matrix equation.
Let A =2 −32 −2
⎡
⎣⎢
⎤
⎦⎥, X =
xy
⎡
⎣⎢
⎤
⎦⎥, and B=
010
⎡
⎣⎢
⎤
⎦⎥.
Then A⋅X =2 −32 −2
⎡
⎣⎢
⎤
⎦⎥⋅
xy
⎡
⎣⎢
⎤
⎦⎥=
2x−3y2x−2y
⎡
⎣⎢
⎤
⎦⎥ so that
AX =B, where A is the coefficient matrix of the system.
A-1 exists since detA≠0. Use grapher to find
X =A-1B=1510
⎡
⎣⎢
⎤
⎦⎥. The solution of the system is (15,10).
Slide 7.3 - 14 Copyright © 2011 Pearson, Inc.
Example Fitting a Parabola to Three Points
Determine a, b, and c so that −3, 32( ), 1, 4( ) and 5, 40( )
are on the graph of f x( ) =ax2 +bx+ c.
Slide 7.3 - 15 Copyright © 2011 Pearson, Inc.
Example Fitting a Parabola to Three Points
Determine a, b, and c so that −3, 32( ), 1, 4( ) and 5, 40( )
are on the graph of f x( ) =ax2 +bx+ c.
We must have f −3( ) =32, f 1( ) =4, and f 5( ) =40
f −3( ) =9a−3b+ c=32
f 1( ) =a+b+ c=4
f 5( ) =25a+5b+ c=40
A=9 −3 11 1 125 5 1
⎡
⎣
⎢⎢⎢
⎤
⎦
⎥⎥⎥, X =
abc
⎡
⎣
⎢⎢⎢
⎤
⎦
⎥⎥⎥, and B=
32440
⎡
⎣
⎢⎢⎢
⎤
⎦
⎥⎥⎥
Slide 7.3 - 16 Copyright © 2011 Pearson, Inc.
Example Fitting a Parabola to Three Points
Determine a, b, and c so that −3, 32( ), 1, 4( ) and 5, 40( )
are on the graph of f x( ) =ax2 +bx+ c.
A grapher shows that
X =A−1B=2−35
⎡
⎣
⎢⎢⎢
⎤
⎦
⎥⎥⎥.
Thus a=2, b=−3, and c=5.
The graph of the quadratic fucntion f x( ) =ax2 +bx+ c
contains the three points (–3, 32), (1, 4) and (5, 40).
Slide 7.3 - 17 Copyright © 2011 Pearson, Inc.
Example Fitting a Parabola to Three Points
Determine a, b, and c so that −3, 32( ), 1, 4( ) and 5, 40( )
are on the graph of f x( ) =ax2 +bx+ c.
Support Graphically
The figure shows a graph
of y1=2x2 −3x+5
superimposed on a scatterplot of the three points.The points appear to lieon the curve.
Slide 7.3 - 18 Copyright © 2011 Pearson, Inc.
Quick Review
1. Find the amount of pure acid in 45L of a 58%
acid solution.
2. Find the amount of water in 30 L of a 28%
acid solution.
3. Is the point (0, −1) on the graph of the function
f (x) =x3 −4x−1?4. Solve for x in terms of the other variables:x + z + w = 2
5. Find the inverse of the matrix 2 10 3
⎡
⎣⎢
⎤
⎦⎥.
Slide 7.3 - 19 Copyright © 2011 Pearson, Inc.
Quick Review Solutions
1. Find the amount of pure acid in 45L of a 58%
acid solution. 26.1 L
2. Find the amount of water in 30 L of a 28%
acid solution. 21.6 L
3. Is the point (0, −1) on the graph of the function
f (x) =x3 −4x−1? yes
4. Solve for x in terms of the other variables:x + z + w = 2 x=2−z−w
5. Find the inverse of the matrix 2 10 3
⎡
⎣⎢
⎤
⎦⎥
1/2 −1/ 60 1/ 3
⎡
⎣⎢
⎤
⎦⎥.