Home | 18.022 | Chapter 15 | Section 15.4

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Example:

2 x

+3 y

+7 z

= 2

   x

-    y

-   z

= 1

   x

 

+  z

= 1

 

A =

2 

3

7

1

-1

1

1

0

1

V =

2

1

1

 

STANDARD METHOD:

GAUSSIAN ELIMINATION:

1. Solve first equation for x by dividing it by two and moving y and z terms to other side:

x + (3/2) y + (7/2) z = 1

x = 1 - (3/2) y - (7/2) z

1.Divide first row of A and V by 2:

A =

1 

3/2

7/2

1

-1

1

1

0

1

V =

1

1

1

 

2. Substitute in other equations for x:

-5/2  y  - 9/2 z = 0, for second equation.
- 3/2 y  - 5/2 z = 0, for third equation.

2. Subtract first row of A and V from each other row of A and V:

A =

1 

3/2

7/2

0

-5/2

-9/2

0

-3/2

-5/2

V =

1

0

0

 

 

3. Solve second equation for y:

y = -9/5 z

3. Divide second row by -5/2:

A =

1 

3/2

7/2

0

1

9/5

0

-3/2

-5/2

V =

1

0

0

 

4. Substitute for y in third:

1/5 z = 0

4. Add 3/2 of second row to third:

A =

1 

3/2

7/2

0

1

9/5

0

0

1/5

V =

1

0

0

 

5. Multiply third equation by 5:

z = 0

5. Multiply third row by 5:

A =

1 

3/2

7/2

0

1

9/5

0

0

1

V =

1

0

0

 

Current form of equations:

x = 1 - (3/2) y - (7/2) z
y = - 9/5 z
z = 0

Current A and V:

A =

1 

3/2

7/2

0

1

9/5

0

0

1

V =

1

0

0

 

6. Substitute for z in first two equations:

x = 1 - (3/2) y
y = 0
z = 0

6. Subtract 7/2 of third row from first and 9/5 of it from second row:

A =

1 

3/2

0

0

1

0

0

0

1

V =

1

0

0

 

7. Substitute for y in the first equation:

x = 1
y = 0
z = 0

7. Subtract 3/2 of 2nd row from 1st:

A =

1 

0

0

0

1

0

0

0

1

V =

1

0

0

 

You see, if you follow this example, that solving equations by systematically eliminating variables, and using Gaussian elimination to change a matrix into the identity matrix are essentially the same thing.