SOLVING SYSTEMS OF LINEAR EQUATIONS
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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
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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
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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
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1352
yxyx
1 314 7
y xx
131536
yxyx
132
yxx
These systems are said to be EQUIVALENT because they have the SAME SOLUTION.
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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.
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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
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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
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1364
zy
zxReplace L1 with L1 + - 4 L3
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13
zy
2x
2533430842
zyxzyxzyx
is EQUIVALENT to
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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:
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112101112233
zyxzyxzyx
This rectangular arrangement of numbers is called a MATRIX
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112101112233
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112101112011
2
Replace L1 with L1 + 2 L2
112101112233
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101001112011
1
Replace L3 with L3 + L2
112101112233
112101112011
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101021002011
1
Replace L2 with L2 + 1L1
112101112233
112101112011
101001112011
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101021002011
112101112233
112101112011
101001112011
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112101112233
112101112011
101021002011
101001112011
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112101112233
112101112011
101021002011
101001112011
210010102011
Interchange L2 and L3
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112101112233
112101112011
101021002011
101001112011
210010102011
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Replace L1 with L1 + -1 L2
Replace L3 with -1 L3
Replace L2 with -1 L2
112101112233
112101112011
101021002011
101001112011
210010102011
1001
2100
1010
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112101112233
1001
2100
1010
The original matrix representsa system that is equivalent to this final matrix whose solution is obvious
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112101112233
1001 x
2100 z
1010 y
The original matrix representsa system that is equivalent to this final matrix whose solution is obvious
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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.