76 which can be plotted as
20 40
60 80
100 120
140
200 400
600 800
1000
Thus, the solution is approximately
x
1
= 400,
x
2
= 60. The solution can be checked by substituting it back into the equations to give
174 244
60 4
. 17
2400 120
160 60
10 400
1 .
1 ≈
= +
− ≈
= +
−
Therefore, the graphical solution is not very good.
b
Because the lines have very similar slopes, you would expect that the system would be ill-conditioned
c
The determinant can be computed as 86
. 20
14 .
19 2
10 2
. 17
1 .
1 4
. 17
2 10
1 .
1 =
+ −
= −
− −
= −
− This result is relatively low suggesting that the solution is ill-conditioned.
8.4 a
The determinant can be evaluated as
69 12
7 5
3 2
2 5
2 1
7 5
1 1
3 2
1 2
− =
− +
+ −
= ⎥⎦
⎤ ⎢⎣
⎡ −
+ ⎥⎦
⎤ ⎢⎣
⎡ −
− −
⎥⎦ ⎤
⎢⎣ ⎡
− −
=
D D
b
Cramer’s rule
77 9855
. 69
68 69
2 2
1 2
3 7
3 2
1
= −
− =
− −
− −
=
x
4638 .
1 69
101 69
2 5
1 3
1 7
2
2
= −
− =
− −
= x
9130 .
69 63
69 2
2 5
3 2
1 2
3
3
= −
− =
− −
− =
x
c
Pivoting is necessary, so switch the first and third rows,
2 7
3 3
2 2
2 5
3 2
3 2
1 2
1
= +
− =
− +
= −
x x
x x
x x
x Multiply pivot row 1 by 15 and subtract the result from the second row to eliminate the a
21
term.
2 7
3 6
. 2
4 .
2 2
2 5
3 2
3 2
2 1
= +
− =
− =
− x
x x
x x
x Pivoting is necessary so switch the second and third row,
6 .
2 4
. 2
2 7
3 2
2 5
3 2
3 2
2 1
= −
= +
− =
− x
x x
x x
x Multiply pivot row 2 by 2.4–3 and subtract the result from the third row to eliminate the
a
32
term.
2 .
4 .6
4 2
7 3
2 2
5
3 3
2 2
1
= =
+ −
= −
x x
x x
x The solution can then be obtained by back substitution
913043 .
6 .
4 2
. 4
3
= =
x
463768 .
1 3
913043 .
7 2
2
= −
− =
x
78 985507
. 5
463768 .
1 2
2
1
= +
= x
d
2 463768
. 1
2 985507
. 5
3 913043
. 463768
. 1
2 985507
. 2
913043 .
7 463768
. 1
3
= −
= −
+ =
+ −
8.5 Prob. 8.3:
A=[-1.1 10;-2 17.4]; detA
ans = 0.8600
Prob. 8.4: A=[0 -3 7;1 2 -1;5 -2 0];
detA ans =
-69
8.6 a The equations can be expressed in a format that is compatible with graphing x
2
versus x
1
:
4 .
9 51
. 5
. 9
5 .
1 2
1 2
+ =
+ =
x x
x x
The resulting plot indicates that the intersection of the lines is difficult to detect:
10 12
14 16
18 20
22
5 10
15 20
Only when the plot is zoomed is it at all possible to discern that solution seems to lie at about x
1
= 14.5 and x
2
= 10.
79
14.3 14.35
