Section 4-2
(26) 2 R1 ---> R1
(28) 2/3R2 + R1 ---> R1
(30) 4R1 + R2 ---> R2
(38) Solve the system:
2x1
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+
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x2
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=
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0
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x1
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-
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2x2
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=
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-5
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R1 <--->R2
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-2R1+R2--->R2
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1/5R2--->R2
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2R2+R1--->R1
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so the
solution is |
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(40) Solve the system:
2x1
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-
|
3x2
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=
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-2
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-4x1
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+
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6x2
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=
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7
|
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2R1+R2--->R2
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(46) Solve the system:
-6x1
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+
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2x2
|
=
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4
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3x1
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-
|
-1x2
|
=
|
-2
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First we put this into matrix form, below. We could start with multiplying the first row by -1/6 because that would give a 1 in the upper left. However, we would end up with fractions for the other entries in the frist row. Thus, let's start with getting a 0 under the 6.
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1/2R1+R2--->R2
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-1/6R1--->R1
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This corresponds
to the system |
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Since any pair solves the second equation, the solutions to the system will be those pairs which solve the first. Thus the solution is
x1
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=
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1/3t
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-
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2/3
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x2
|
=
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t
|