SOLVING SYSTEMS OF LINEAR EQUATIONS An equation is said to be linear if every variable has degree...

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SOLVING SYSTEMS OF LINEAR EQUATIONS

• An equation is said to be linear if every variable has degree equal to one (or zero)

• is a linear equation

• is NOT a linear equation

8254 zyx

832 13 zyx

Review these familiar techniques for solving 2 equations in 2 variables. The same techniques will be

extended to accommodate larger systems.

13

52

yx

yx

13

1536

yx

yx

13

2

yx

x

Times 3

Add

Substitute to solve: y=1

147 x

L1 represents line one L2 represents line two

13

52

yx

yx13

1536

yx

yx

13

2

yx

x

L1 is replaced by 3L1

L1 is replaced by L1 + L2

L1 is replaced by (1/7)L1

13

147

yx

x

13

52

yx

yx

1 3

14 7

y x

x

13

1536

yx

yx

13

2

yx

x

These systems are said to be EQUIVALENT because they have the SAME SOLUTION.

PERFORM ANY OF THESE OPERATIONS ON A SYSTEM OF LINEAR EQUATIONS TO

PRODUCE AN EQUIVALENT SYSTEM:

• INTERCHANGE two equations (or lines)

• REPLACE Ln with k Ln , k is NOT ZERO

• REPLACE Ln with Ln + cLm

• note: Ln is always part of what replaces it.

EXAMPLES:

is equivalent to

L1

L2

L3

L1

L3

L2

L1

L2

L3

L1

4L2

L3

is equivalent to

is equivalent to

L1

L2

L3 + 2L1

L1

L2

L3

253

343

0842

zyx

zyx

zyx

1 z

343 zyx

042 zyxReplace L1 with (1/2) L1

Replace L3 with L3 + L2

1

343

042

z

zyx

zyx

Replace L2 with L2 + L1

1z

3y

042 zyx

1

3

042

z

y

zyx

1

3

z

y

6x +4zReplace L1 withL1 + 2 L2

1

3

64

z

y

zxReplace L1 with L1 + - 4 L3

1

3

z

y

2x

253

343

0842

zyx

zyx

zyx

is EQUIVALENT to

1121

0111

2233

zyx

zyx

zyx

To solve the following system, we look for an equivalent systemwhose solution is more obvious. In the process, we manipulateonly the numerical coefficients, and it is not necessary to rewritevariable symbols and equal signs:

1121

0111

2233

zyx

zyx

zyx

This rectangular arrangement of numbers is called a MATRIX

1121

0111

2233

1121

0111

2011

2

Replace L1 with L1 + 2 L2

1121

0111

2233

1010

0111

2011

1

Replace L3 with L3 + L2

1121

0111

2233

1121

0111

2011

1010

2100

2011

1

Replace L2 with L2 + 1L1

1121

0111

2233

1121

0111

2011

1010

0111

2011

1010

2100

2011

1121

0111

2233

1121

0111

2011

1010

0111

2011

1121

0111

2233

1121

0111

2011

1010

2100

2011

1010

0111

2011

1121

0111

2233

1121

0111

2011

1010

2100

2011

1010

0111

2011

2100

1010

2011

Interchange L2 and L3

1121

0111

2233

1121

0111

2011

1010

2100

2011

1010

0111

2011

2100

1010

2011

Replace L1 with L1 + -1 L2

Replace L3 with -1 L3

Replace L2 with -1 L2

1121

0111

2233

1121

0111

2011

1010

2100

2011

1010

0111

2011

2100

1010

2011

1001

2100

1010

1121

0111

2233

1001

2100

1010

The original matrix representsa system that is equivalent to this final matrix whose solution is obvious

1121

0111

2233

1001 x

2100 z

1010 y

The original matrix representsa system that is equivalent to this final matrix whose solution is obvious

The zeros

The diagonal of ones

1001 x

2100 z

1010 y

Note the format of the matrix that yields this obvious solution:

2100

1010

1001

Whenever possible, aim for this format.