solving systems of linear equations

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.

1352

yxyx

131536

yxyx

132

yxx

Times 3

Add

Substitute to solve: y=1

147 x

L1 represents line one L2 represents line two

1352

yxyx

131536

yxyx

132

yxx

L1 is replaced by 3L1

L1 is replaced by L1 + L2

L1 is replaced by (1/7)L1

13147

yxx

1352

yxyx

1 314 7

y xx

131536

yxyx

132

yxx

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 toL1

L2

L3

L1

L3

L2

L1

L2

L3

L1

4L2

L3is equivalent to

is equivalent to

L1

L2

L3 + 2L1

L1

L2

L3

2533430842

zyxzyxzyx

1 z

343 zyx042 zyxReplace L1 with (1/2) L1

Replace L3 with L3 + L2

1343

042

zzyx

zyxReplace L2 with L2 + L1

1z

3y042 zyx

13042

zy

zyx

13

zy

6x +4zReplace L1 withL1 + 2 L2

1364

zy

zxReplace L1 with L1 + - 4 L3

13

zy

2x

2533430842

zyxzyxzyx

is EQUIVALENT to

112101112233

zyxzyxzyx

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:

112101112233

zyxzyxzyx

This rectangular arrangement of numbers is called a MATRIX

112101112233

112101112011

2

Replace L1 with L1 + 2 L2

112101112233

101001112011

1

Replace L3 with L3 + L2

112101112233

112101112011

101021002011

1

Replace L2 with L2 + 1L1

112101112233

112101112011

101001112011

101021002011

112101112233

112101112011

101001112011

112101112233

112101112011

101021002011

101001112011

112101112233

112101112011

101021002011

101001112011

210010102011

Interchange L2 and L3

112101112233

112101112011

101021002011

101001112011

210010102011

Replace L1 with L1 + -1 L2

Replace L3 with -1 L3

Replace L2 with -1 L2

112101112233

112101112011

101021002011

101001112011

210010102011

1001

2100

1010

112101112233

1001

2100

1010

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

112101112233

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:

210010101001

Whenever possible, aim for this format.