Section 4-3
(26) Solve the system:
3x1
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+
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5x2
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-
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x3
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=
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-7
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x1
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+
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x2
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+
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x3
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=
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-1
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2x1
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+
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11x3
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=
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7
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First, we transform the system to an augmented matrix, and then reduce the matrix. In our first step, we switch two rows to get a 1 at the top left corner.
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[ |
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This reduced matrix corresponds to the following system, which gives the solution.
x1
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=
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-2
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|||
x2
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=
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0
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|||
x3
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=
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1
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(30) Solve the system
2x1
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+
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4x2
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-
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6x3
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=
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10
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3x1
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+
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3x2
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-
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3x3
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=
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6
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1/3R2 --->R2
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R1<--->R2
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-2R1+R2--->R2
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1/2R2--->R2
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-R2+ R1---> R1
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(reduced)
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Since the matrix is reduced we can now write down the solution to the original system. The reduced matrix corresponds to the system
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or |
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so the solutions are |
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(32) Solve the system
2x1
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-x2
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=
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0
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3x1
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+2x2
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=
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7
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x1
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-x2
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=
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-2
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R1 <---->R3
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-3R1+R2--->R2 -2R1+R3--->R3 |
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-5R3+R2--->R2
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Even though the last matrix is not yet reduced, we can stop, because the second row corresponds to the equation 0x1+0x2=-7, or 0=-7. Therefore it is impossible for the system to have any solutions. NO SOLUTIONS