Linear Algebra Exercises-n-Answers.pdf

36 Linear Algebra, by Hefferon
65 55
75
i 1
40
5
80
i 2 i 3
i 4
i5
i 7
70
i 6
We apply Kirchoff’s principle that the flow into the intersection of Willow and Shelburne must equal
the flow out to get i 1 + 25 = i 2 + 125. Doing the intersections from right to left and top to bottom
gives these equations.
i 1 − i 2 = 10
−i 1 + i 3 = 15
i 2 + i 4 = 5
−i 3 − i 4 + i 6 = −50
i 5 − i 7 = −10
−i 6 + i 7 = 30
The row operation ρ 1 + ρ 2 followed by ρ 2 + ρ 3 then ρ 3 + ρ 4 and ρ 4 + ρ 5 and finally ρ 5 + ρ 6 result in
this system.
i 1 − i 2 = 10
−i 2 + i 3 = 25
i 3 + i 4 − i 5 = 30
−i 5 + i 6 = −20
−i 6 + i 7 = −30
0 = 0
Since the free variables are i 4 and i 7 we take them as parameters.
i 6 = i 7 − 30
i 5 = i 6 + 20 = (i 7 − 30) + 20 = i 7 − 10
i 3 = −i 4 + i 5 + 30 = −i 4 + (i 7 − 10) + 30 = −i 4 + i 7 + 20
i 2 = i 3 − 25 = (−i 4 + i 7 + 20) − 25 = −i 4 + i 7 − 5
50
30
i 1 = i 2 + 10 = (−i 4 + i 7 − 5) + 10 = −i 4 + i 7 + 5
Obviously i 4 and i 7 have to be positive, and in fact the first equation shows that i 7 must be at least
30. If we start with i 7 , then the i 2 equation shows that 0 ≤ i 4 ≤ i 7 − 5.
(b) We cannot take i 7 to be zero or else i 6 will be negative (this would mean cars going the wrong
way on the one-way street Jay). We can, however, take i 7 to be as small as 30, and then there are
many suitable i 4 ’s. For instance, the solution
results from choosing i 4 = 0.
(i 1 , i 2 , i 3 , i 4 , i 5 , i 6 , i 7 ) = (35, 25, 50, 0, 20, 0, 30)
()