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# Systems of equations - PowerPoint PPT Presentation

Systems of equations . With Gaussian elimination. System of equations. Find all pairs of x and y values that make the equations true. System of equations. Swap the order of the rows R1 <-> R2. System of equations. Multiply a row by a number -4 * R1  R1. System of equations.

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### Systems of equations

With Gaussian elimination

Find all pairs of x and y values that make the equations true.

Swap the order of the rows

R1 <-> R2

Multiply a row by a number

-4*R1  R1

Add a row to another row

R1 + R2  R2

Multiply a row by a number

-¼*R1  R1

Multiply a row by a number and add it to another row

-4*R1 + R2  R2

• Swap rows R1<-> R2

• Multiply a row by a number k*R1  R1

• Add rows together R1 + R2  R2

• Multiply-add shortcut k*R1 + R2  R2

• A method that you can use to solve ANY system of equations (no matter how big), using only two rules.

• Multiply a row by a number k*R1  R1

• Multiply-add shortcut k*R1 + R2  R2

Method: Gaussian Elimination

• Today’s fun irrelevant fact: Gauß is my great-great-great-great-great-great-great-grand-advisor

• Gauß Gerling Plucker Klein  Bocher Ford  Engen Steffe Thompson  Castillo-Garsow

• Write equations in standard form

• Use multiply to get 1x in the top equation

• Use multiply-add to get 0x in all other equations.

• Use multiply to get 1y in the second equation

• Use multiply-add to get 0y in all other equations.

• Repeat for all variables.

• Get your system in standard form

(All the variables on one side, all the constants on the other)

4x + 8y - 4z = 8

2x + 3y + 4z = 4

5x + 8y + 1z = 7

• Use multiply to get 1x in the top equation

4x + 8y - 4z = 8 (1/4) * R1 --> R1

2x + 3y + 4z = 4

5x + 8y + 1z = 7

1x + 2y - 1z = 2

2x + 3y + 4z = 4

5x + 8y + 1z = 7

• Use multiply-add to get 0xs everywhere else

1x + 2y - 1z = 2

2x + 3y + 4z = 4 -2 * R1 + R2 --> R2

5x + 8y + 1z = 7 -5 * R1 + R3 --> R3

1x + 2y - 1z = 2

0x - 1y + 6z = 0

0x - 2y + 6z = -3

• Use multiply to get 1y in the second equation

1x + 2y - 1z = 2

0x - 1y + 6z = 0 -1 * R2 --> R2

0x - 2y + 6z = -3

1x + 2y - 1z = 2

0x + 1y - 6z = 0

0x - 2y + 6z = -3

• Use multiply-add to get 0ys in all other equations

• You can do all of these now, but I’m going to put one off for later.

1x + 2y - 1z = 2

0x + 1y - 6z = 0

0x - 2y + 6z = -3 2 * R2 + R3 --> R3

1x + 2y - 1z = 2

0x + 1y - 6z = 0

0x + 0y - 6z = -3

• Use multiply to get 1z in the third equation

1x + 2y - 1z = 2

0x + 1y - 6z = 0

0x + 0y - 6z = -3 (-1/6) * R3 --> R3

1x + 2y - 1z = 2

0x + 1y - 6z = 0

0x + 0y + 1z = 0.5

• Get 0z in all other equations

1x + 2y - 1z = 2 1 * R3 + R1 --> R1

0x + 1y - 6z = 0 6 * R3 + R2 --> R2

0x + 0y + 1z = 0.5

1x + 2y + 0z = 2.5

0x + 1y + 0z = 3

0x + 0y + 1z = 0.5

• Finish my incomplete step

• Get 0y in all other equations

1x + 2y + 0z = 2.5 -2 * R2 + R1 --> R1

0x + 1y + 0z = 3

0x + 0y + 1z = 0.5

1x + 0y + 0z = -3.5

0x + 1y + 0z = 3

0x + 0y + 1z = 0.5

-3x − 9y = -6-3x − 13y = -8

• x = -2, y = 0

• x = 0, y = 8/13

• x = 1/2, y = 1/2

• x = -1/2, y = -1/2

• None of the above

-3x − 9y = -6 (-1/3)*R1 ->R1

-3x − 13y = -8

1x+ 3y = 2

-3x − 13y = -8 3R1 + R2 -> R2

1x + 3y = 2

0x − 4y = -2 (-1/4)R2 -> R2

1x + 3y = 2 (-3)R2 + R1 -> R1

0x+ 1y = ½

1x + 0y = 1/2

0x + 1y = 1/2

C

-3x − 9y = -6 (-1/3)*R1 ->R1

-3x − 13y = -8

1x+ 3y = 2

-3x − 13y = -8 3R1 + R2 -> R2

1x + 3y = 2

0x − 4y = -2 (-1/4)R2 -> R2

1x + 3y = 2 (-3)R2 + R1 -> R1

0x+ 1y = ½

1x + 0y = 1/2

0x + 1y = 1/2

What is the system of equations corresponding to the augmented matrix below?

• 2x+3y = 4, x + 2y = 3

• 3x+2y = 4, 2x + y = 3

• 2x+y = 4, 3x + 2y = 3

• x+y = 4, x + 2y = 3

• None of the above

What is the system of equations corresponding to the augmented matrix below?

• 2x+3y = 4, x + 2y = 3

In my calculator, I set the matrix [A]

• Solve

4x + 8y - 4z = 8

2x + 3y + 4z = 4

5x + 8y + 1z = 7

Then I used the command rref([A])

The calculator output was

So the answer is

x=-3.5

y=3

z=0.5

Special situations showing work)

• If, at the end you wind up with something impossible, then there are NO SOLUTIONS

• Example:

The last row:

0x + 0y = 1 is impossible,

So there are NO SOLUTIONS.

Special situations showing work)

• If, at the end you wind up with something that is always true, then there are INFINITELY MANY SOLUTIONS

• Example:

The last row:

0x + 0y = 0 is always true,

So there are INFINITELY MANY SOLUTIONS.

Solve the following system. showing work)

• x = 0, y = 3, z = 2

• x = 5, y = 3, z = 2

• x = 1, y = 3, z = 2

• x = -2, y = 3, z = 2

• None of the above

x showing work)=-2

y=3

z=2 D