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solutions to homework 1...

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HW 1 Answer Key MATH:2550: Engineering Math III Hw 1: 4,10*,12,14*,16*,18*,20,22*,24 10)  1 −2 0 0 1 0  0 0 1 0 0 0  1 −2 0 1  0 0 0 0

3 −4 0 1 0 0 0 0 1 0 0 1

 −2 7  6 −3  7 −5  6 −3



1 0  0 0  1 0  0 0

R 7→R +4R

4 −−2−−−2−−−→

R1 7→R1 −3R4

R1 7→R1 +2R2

−−−−−−−−→

−2 1 0 0 0 1 0 0

0 0 0 0 1 0 0 1 0 0 1 0

0 0 0 1

 7 −5  6 −3  −3 −5  6 −3

Solution Set: {(−3, −5, 6, −3)} 14)  1 −3 0 5  −1 1 5 2 0 1 1 0   1 −3 0 5  0 −2 5 7 0 1 1 0   1 −3 0 5  0 1 1 0 0 −2 5 7   1 −3 0 5  0 1 1 0 0 0 7 7   1 −3 0 5  0 1 1 0 0 0 1 1   1 −3 0 5  0 1 0 −1 0 0 1 1 

 1 −−−−−−−−→ 0 0  1 R ↔R3 0 −−2−−−→ 0  1 R3 7→R3 +2R2 −− −−−−−−→ 0 0  1 R3 7→ 71 R3 0 −−−−−−→ 0  1 R 7→R2 −R3 −−2−−−− −−→ 0 0  1 R1 7→R1 +3R3 −− −−−−−−→ 0 0 R2 7→R2 +R1

−3 −2 1 −3 1 −2 −3 1 0 −3 1 0

 0 5 5 7 1 0  0 5 1 0 5 7  0 5 1 0 7 7  0 5 1 0 1 1

 0 5 0 −1  1 1  0 0 2 1 0 −1  0 1 1

−3 1 0

Solution: (2, −1, 1)

1

16) 

1 0 0 0 2 2  0 0 1 −2 3 2  1 0 0 0 2 2  0 0 1 0 3 2  1 0 0 0 2 2  0 0 1 0 0 −1

 −3 0  1 5  −2 −3 0 0  3 1 −3 −1  −2 −3 0 0  3 1 −3 −1 −2 0 3 1



1 0 R4 7→R4 +2R1  −−−−−−−−→  0 0  1 0 R4 7→R4 − 23 R2 −−−−−−−−−→  0 0  1 0 R4 7→R4 +R3 −−−−−−−−→  0 0

0 0 −2 2 2 0 0 1 3 3 2 −3

 −3 0  1 −1

 0 −2 −3 2 0 0  1 3 1 −1 −3 −1  0 0 −2 −3 2 2 0 0  0 1 3 1 0 0 0 0

0 2 0 0

System has a solution (consistent)

18) Row reduce the matrix assosiated to the equations:  1 0 1

 1 0 0

2 1 3 2 1 1

1 −1 0 1 −1 −1

 4 1 0  4 1 −4

 1 −−−−−−−−→  0 0  1 R 7→R −R2 0 −−3−−−3−−−→ 0 R3 7→R3 −R1

2 1 1 2 1 0

 4 1 −4  1 4 −1 1  0 −5

1 −1 −1

The system is inconsistent (R3 : 0 = 5), the solution set of the system is the triple intersection of the three planes and hence the planes do not have a common point 22) 

2 −3 −6 9

h 5



R2 7→R2 +3R1

−−−−−−−−→

The system is consistent only when 5 + 3h = 0.

2



2 0

−3 h 0 5 + 3h

...