Sections 1.1 & 1.2 Intro to Systems of Linear Equations & Gauss Jordan Elimination
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Transcript of Sections 1.1 & 1.2 Intro to Systems of Linear Equations & Gauss Jordan Elimination
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Sections 1.1 & 1.2 Intro to Systems of Linear Equations
& Gauss Jordan Elimination
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Linear Algebra – Linear Equationsy = 3x – 2 2x – 3y + 4z = 9
Vector Spaces
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A solution to an equation.
A solution to multiple equations (a system of equations).
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A parametric representation of a solution.
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Geometrical meaning of a solution.
For two equations in two unknowns:One Solution No Solutions Infinite Solutions
2x + 3y = 6 2x + 3y = 6 2x + 3y = 6 4x – 2y = 12 4x + 6y = –24 4x + 6y = 12
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Geometrical meaning of a solution.
For three equations in three unknowns:
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Geometrical meaning of a solution.
For three equations in three unknowns:
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The number of solutions to a system of linear equations.
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A system is consistent if there is at least one solution to it.
A system is inconsistent if there is no solution for it.
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Equivalent systems – systems that have the same solution set.
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Hard to solve: Easy to solve:
x = 1 y = –1 z = 2
x – 2y + 3z = 9 –x + 3y = –4 →2x – 5y + 5z = 17
+ 0y + 0z0x + + 0z0x + 0y +
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Hard to solve: Easy to solve:
x = 1 y = –1 z = 2
x – 2y + 3z = 9 –x + 3y = –4 →2x – 5y + 5z = 17
12 3
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Elementary Operations:1. Interchange any two equations.2. 3.
x – 2y + 3z = 9–2x + 3y + 7z = –4 2x – 5y + 5z = 17
–2x + 3y + 7z = –4 x – 2y + 3z = 9 2x – 5y + 5z = 17
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Elementary Operations:1. 2. Multiply any equation by a nonzero scalar.3.
x – 2y + 3z = 9–2x + 3y + 7z = –420x – 50y + 50z = 170
x – 2y + 3z = 9–2x + 3y + 7z = –4 2x – 5y + 5z = 17
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Elementary Operations:1. 2. 3. Add a multiple of one equation to another equation.
x – 2y + 3z = 9–2x + 3y + 7z = –4 2x – 5y + 5z = 17
x – 2y + 3z = 9–2x + 3y + 7z = –4 2x – 5y + 5z = 17
2x – 4y +12z = 18–2x + 3y + 7z = –4 2x – 5y + 5z = 17
2x – 4y + 12z = 18 – y + 19z = 14 2x – 5y + 5z = 17 x – 2y + 3z = 9 – y + 19z = 14 2x – 5y + 5z = 17
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x = 1 y = –1 z = 2
x – 2y + 3z = 9 –x + 3y = –4 →2x – 5y + 5z = 17
+ 0y + 0z0x + + 0z0x + 0y +
Gauss Jordan Elimination:
Gaussian Elimination:
x – 2y + 3z = 9 y + 3z = 5 z = 2
x – 2y + 3z = 9 –x + 3y = –4 →2x – 5y + 5z = 17
0x + 0x + 0y +
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Ex. Use Gauss Jordan elimination to solve the system x – 2y + 3z = 9 –x + 3y = –4 2x – 5y + 5z = 17
x – 2y + 3z = 9 y + 3z = 5 2x – 5y + 5z = 17
x – 2y + 3z = 9 y + 3z = 5 –y – z = –1
x – 2y + 3z = 9 y + 3z = 5 2z = 4
x + 9z = 19 y + 3z = 5 2z = 4
(method of elimination)
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Ex. Use Gauss Jordan elimination to solve the system x – 2y + 3z = 9 –x + 3y = –4 2x – 5y + 5z = 17
x + 9z = 19 y + 3z = 5 2z = 4
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x + 9z = 19 y + 3z = 5 2z = 4
x + 9z = 19 y + 3z = 5 z = 2
x + 9z = 19 y = –1 z = 2
x = 1 y = –1 z = 2
Ex. Use Gauss Jordan elimination to solve the system x – 2y + 3z = 9 –x + 3y = –4 2x – 5y + 5z = 17
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Notice that we can use elementary row operations on a matrix to work through the Gauss Jordan method of elimination. x – 2y + 3z = 9 –x + 3y = –4 2x – 5y + 5z = 17
x – 2y + 3z = 9 y + 3z = 52x – 5y + 5z = 17
x – 2y + 3z = 9 y + 3z = 5 –y – z = –1
x – 2y + 3z = 9 y + 3z = 5 2z = 4
x + 9z = 19 y + 3z = 5 2z = 4
1 2 3 90 1 3 52 5 5 17
1 2 3 90 1 3 50 1 1 1
1 2 3 90 1 3 50 0 2 4
1 0 9 190 1 3 50 0 2 4
1 2 3 91 3 0 4
2 5 5 17
R2+R1→R2
R3–2R1→R3
R3+R2→R3
R1+2R2→R3
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x + 9z = 19 y + 3z = 5 2z = 4
1 0 9 190 1 3 50 0 2 4
Notice that we can use elementary row operations on a matrix to work through the Gauss Jordan method of elimination. x – 2y + 3z = 9 –x + 3y = –4 2x – 5y + 5z = 17
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x + 9z = 19 y + 3z = 5 2z = 4
x + 9z = 19 y + 3z = 5 z = 2
x = 1 y + 3z = 5 z = 2
x = 1 y = –1 z = 2
1 0 9 190 1 3 50 0 2 4
1 0 9 190 1 3 50 0 1 2
1 0 0 10 1 3 50 0 1 2
1 0 0 10 1 0 10 0 1 2
Notice that we can use elementary row operations on a matrix to work through the Gauss Jordan method of elimination. x – 2y + 3z = 9 –x + 3y = –4 2x – 5y + 5z = 17
½ R3→R3
R1–9R3→R1
R2–3R3→R2
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Elementary Row Operations(We are allowed to use these operations on a matrix when trying to solve a system of linear equations.)
Elementary Row Operation: Notation:1. 2. 3.
1. Interchange two rows Ri ↔ Rk
2. Multiply (or divide) a row by a nonzero constant. cRi → Ri
3. Add (or subtract) a multiple of one row to another. Ri + cRk → Ri
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We'd like to take our original augmented matrix, and through row operations put it in the following form:
where the bi's are just some constants (some numbers).
1
2
3
4
1
1 0 0 0 0 00 1 0 0 0 00 0 1 0 0 00 0 0 1 0 0
0 0 0 0 1 00 0 0 0 0 1
n
n
bbbb
bb
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1 2 3 70 0 0 00 0 0 0
1 0 5 20 1 6 3
For example, this matrix has a solution that is easy to see, (1, 3, 5), because the matrix is in the final form that we want.
1 0 0 10 1 0 30 0 1 5
This is not always possible though. The following are matrices that cannot be put into this form.
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Reduced Row Echelon FormA matrix is said to be in reduced echelon form if all of the following properties hold true:1. All rows consisting entirely of zeros are grouped at the bottom.2. The leftmost nonzero number in each row is 1 (called the leading one).3. The leading 1 of a row is to the right of the previous row's leading 1.4. All entries directly above and below a leading 1 are zeros.
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Reduced Row Echelon FormA matrix is said to be in reduced echelon form if all of the following properties hold true:1. All rows consisting entirely of zeros are grouped at the bottom.2. The leftmost nonzero number in each row is 1 (called the leading one).3. The leading 1 of a row is to the right of the previous row's leading 1.4. All entries directly above and below a leading 1 are zeros.
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x = 1 y = –1 z = 2
x – 2y + 3z = 9 –x + 3y = –4 →2x – 5y + 5z = 17
Gauss Jordan Elimination:
Gaussian Elimination:
x – 2y + 3z = 9 y + 3z = 5 z = 2
x – 2y + 3z = 9 –x + 3y = –4 →2x – 5y + 5z = 17
1 2 3 91 3 0 4
2 5 5 17
1 2 3 90 1 3 50 0 1 2
1 0 0 10 1 0 10 0 1 2
Reduced Row Echelon Form
Row Echelon Form
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Ex. Determine which of the following matrices are in reduced row echelon form.
(a) (b) (c) 1 0 0 40 1 0 50 0 1 2
1 0 0 30 1 0 40 0 0 00 0 1 5
1 0 0 20 0 1 40 1 0 1
Reduced Row Echelon Form1. All rows consisting entirely of zeros are grouped at the bottom.2. The leftmost nonzero number in each row is 1 (called the leading one).3. The leading 1 of a row is to the right of the previous row's leading 1.4. All entries directly above and below a leading 1 are zeros.
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Ex. Determine which of the following matrices are in reduced row echelon form.
(d) (e) (f) 1 0 0 0 40 0 1 0 50 0 0 1 7
1 0 3 40 1 2 10 0 0 0
1 0 30 1 20 0 0
Reduced Row Echelon Form1. All rows consisting entirely of zeros are grouped at the bottom.2. The leftmost nonzero number in each row is 1 (called the leading one).3. The leading 1 of a row is to the right of the previous row's leading 1.4. All entries directly above and below a leading 1 are zeros.
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Ex. Determine which of the following matrices are in reduced row echelon form.
(g) (h)1 0 0 20 1 0 4
2 0 0 30 1 0 50 0 1 7
Reduced Row Echelon Form1. All rows consisting entirely of zeros are grouped at the bottom.2. The leftmost nonzero number in each row is 1 (called the leading one).3. The leading 1 of a row is to the right of the previous row's leading 1.4. All entries directly above and below a leading 1 are zeros.
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Ex. Put the matrix into reduced row echelon form.1 3 113 4 62 7 17
A
1 3 113 4 62 7 17
1 3 110 13 390 13 39
1 3 110 1 30 13 39
R2 – 3R1→R2
R3 – 2R1→R3
R1 – 3R2→R1
R3 + 13R2→R3
1 0 20 1 30 0 0
–1/13 R2→R2
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Using a calculator with matrices.
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Ex. Solve the following system. x + y = 11 3x – 4y = –6 2x – 7y = –17
1 1 113 4 62 7 17
1 1 110 7 390 9 39
38
7
397
787
1 00 10 0
1 0 00 1 00 0 1
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Ex. Solve the following system of equations. x1 – 4x2 + 7x3 = –3 –2x1 + 9x2 – 4x3 = 7 x1 – 3x2 + 17x3 = –2
1 4 7 32 9 4 7
1 3 17 2
1 4 7 30 1 10 10 1 10 1
1 0 47 10 1 10 10 0 0 0
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Here's one view of the three planes: Here's a side view of the three planes:
Ex. Solve the following system of equations. x1 – 4x2 + 7x3 = –3(A graph of this system is given below.) –2x1 + 9x2 – 4x3 = 7
x1 – 3x2 + 17x3 = –2
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Ex. Solve the following system of equations. x1 – 2x2 + 3x3 = 4 –2x1 + 4x2 – 6x3 = –8 3x1 – 6x2 + 9x3 = 12
1 2 3 42 4 6 8
3 6 9 12
1 2 3 40 0 0 00 0 0 0
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Here's one view of the three planes: Here's a side view of the three planes:
Ex. Solve the following system of equations. x1 – 2x2 + 3x3 = 4(A graph of this system is given below.) –2x1 + 4x2 – 6x3 = –8
3x1 – 6x2 + 9x3 = 12
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Notation and terminology
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If we call a matrix A then we shall refer to the entries in A as follows:ai j is the number in matrix A in row i column j.
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If we call a matrix A then we shall refer to the entries in A as follows:ai j is the number in matrix A in row i column j.
If then we have the following:a1 2 =
a3 1 =
If then we have the following:b1 2 =
b2 3 =
11 13 36 5 14 0 2
A
1 3 42 2 1
B
13
–4
3
1
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Given a system like –5x + 6y – 2z = 11 we will often look at the following x + 3y = 2 two matrices. 2x – y + z = 4
Coefficient Matrix Augmented Matrix
5 6 21 3 02 1 1
5 6 2 111 3 0 22 1 1 4
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Homogenous systems
8x – 2y + 6z = 0 9x + 3y + 7z = 0 4x – 5y + 2z = 0
13x – 8y + 2z = 0 2x + 4y –10z = 0 5x – 7y + 3z = 0 x + 3y + 5z = 0