Exercise 1
Solve, whenever possible, the following systems:
(1)
L1:
x + 3y − 2z = −7 4x + 2y − z = 3
3x + 5y = 1 (2)
L2:
3x − y + 2z = 0 x + 2y + z = 2 2x − 3y + z = 1 (3)
L3:
x + 2y − z = 2 3x − 2y + z = 2 x − 6y + 3z = −2
Exercise 1, answer
I For L1:
1 3 −2 | −7 4 2 −1 | 3
3 5 0 | 1
→
1 3 −2 | −7
7 1 0 | 13
3 5 0 | 1
→
→
1 3 −2 | −7
7 1 0 | 13
−32 0 0 | −64
whence, starting from last equation:
x =2
y =13 − 7x = 13 − 14 = −1 z = −1
2(−7 − x − 3y ) = −1
2(−7 − 2 + 3) = 3
Exercise 1, answer (cont.)
I For L2:
3 −1 2 | 0
1 2 1 | 2
2 −3 1 | 1
→
3 −1 2 | 0
−1 5 0 | 4
1 −5 0 | 2
The linear system is incompatible.
Exercise 1, answer (cont.)
I For L3:
1 2 −1 | 2
3 −2 1 | 2
1 −6 3 | −2
→
1 2 −1 | 2
0 −8 4 | −4
0 −8 4 | −4
→
→
1 2 −1 | 2 0 2 −1 | 1
0 0 0 | 0
The solutions of the linear system depend on one free variable, that can be chosen to be z. Thus
y =1
2(1 + z) = 1 2 +1
2z
x =2 − 2y + z = 2 − 1 − z + z = 1
Exercise 2
The 450 restaurants of a town are ranked in four categories I, II, III, IV (in decreasing order). The difference between the restaurants ranked in categories II or IV, and those ranked in categories I or III is 50; the number of restaurants that are ranked in IV is 120; the restaurants ranked in I or III are 180. Determine, if it is possible, the number of restaurants ranked in each category.
Exercise 2, answer
Let xi be the number of restaurants ranked in the i -th category. Then:
x1+ x2+ x3+ x4 = 450 x2+ x4− (x1+ x3) = 50
x4 = 120 x1+ x3 = 180 This is a linear system, associated with the matrix
1 1 1 1 | 450
−1 1 −1 1 | 50 0 0 0 1 | 120 1 0 1 0 | 180
Exercise 2, answer (cont.)
1 1 1 1 | 450
−1 1 −1 1 | 50 0 0 0 1 | 120 1 0 1 0 | 180
→
1 1 1 1 | 450
−1 1 −1 1 | 50 1 0 1 0 | 180 0 0 0 1 | 120
→
→
1 1 1 1 | 450 2 0 2 0 | 400 1 0 1 0 | 180 0 0 0 1 | 120
The second and third rows of this last matrix show that the data are incompatible, so there is no solution to the problem.
Exercise 3
Discuss and solve the following system of parametric linear equations with k ∈ R
Lk :
x1+ x2− 5x3 = 1 2x1− x2+ x4 = 0 x1+ x2+ 6x3+ kx4 = 1 x2+ kx3+ x4 = 2
Exercise 3, answer
1 1 −5 0 | 1
2 −1 0 1 | 0
1 1 6 k | 1
0 1 k 1 | 2
→
1 1 −5 0 | 1
0 −3 10 1 | −2
0 0 11 k | 0
0 1 k 1 | 2
→
→
1 1 −5 0 | 1
0 −3 10 1 | −2
0 0 11 k | 0
0 0 3k + 10 4 | 4
→
→
1 1 −5 0 | 1
0 −3 10 1 | −2
0 0 11 k | 0
0 0 0 −3k2+10k−4411 | 4
Thus if 3k2+ 10k − 44 = 0, that is k = −5±
√25+132
3 = −5±
√157 3 , the linear system has no solutions.
Otherwise, solving backwards starting from the last equation,
Exercise 3, answer (cont.)
x4= − 44 3k2+ 10k − 44 x3= − k
11x4= 4k 3k2+ 10k − 44 x2=2 + 10x3+ x4
3 =6k2+ 20k − 88 + 40k − 44
3(3k2+ 10k − 44) = 6k2+ 60k − 132 3(3k2+ 10k − 44) =
=2k2+ 20k − 66 3k2+ 10k − 44
x1=1 − x2+ 5x3=3k2+ 10k − 44 − 2k2− 20k + 66 + 20k
3k2+ 10k − 44 =
=k2+ 10k + 22 3k2+ 10k − 44
