81 5
. 21
6 5
8 5
. 5
. 61
6 2
8 6
5 .
3 27
6 8
2 5
. 10
− =
− +
+ −
= −
+ −
− =
− −
+
8.8 a Pivoting is necessary, so switch the first and third rows,
38 6
2 34
7 3
20 2
8
3 2
1 3
2 1
3 2
1
− =
− −
− =
+ −
− −
= −
+ −
x x
x x
x x
x x
x
Multiply the first equation by –3–8 and subtract the result from the second equation to eliminate the a
21
term from the second equation. Then, multiply the first equation by 2–8 and subtract the result from the third equation to eliminate the a
31
term from the third equation.
43 5
. 1
75 .
5 5
. 26
75 .
7 375
. 1
20 2
8
3 2
3 2
3 2
1
− =
− −
− =
+ −
− =
− +
−
x x
x x
x x
x
Pivoting is necessary so switch the second and third row,
5 .
26 75
. 7
375 .
1 43
5 .
1 75
. 5
20 2
8
3 2
3 2
3 2
1
− =
+ −
− =
− −
− =
− +
−
x x
x x
x x
x
Multiply pivot row 2 by –1.375–5.75 and subtract the result from the third row to eliminate the a
32
term.
21739 .
16 8.108696
43 5
. 1
75 .
5 20
2 8
3 3
2 3
2 1
− =
− =
− −
− =
− +
−
x x
x x
x x
The solution can then be obtained by back substitution
2 108696
. 8
21739 .
16
3
− =
− =
x
8 75
. 5
2 5
. 1
43
2
= −
− +
− =
x
82 4
8 8
1 2
2 20
1
= −
− −
+ −
= x
b Check:
20 2
2 8
4 8
34 2
7 8
4 3
38 2
8 6
4 2
− =
− −
+ −
− =
− +
− −
− =
− −
−
8.9 Multiply the first equation by –0.40.8 and subtract the result from the second equation to
eliminate the x
1
term from the second equation.
⎪⎭ ⎪
⎬ ⎫
⎪⎩ ⎪
⎨ ⎧
= ⎪⎭
⎪ ⎬
⎫ ⎪⎩
⎪ ⎨
⎧ ⎥
⎥ ⎦
⎤ ⎢
⎢ ⎣
⎡ −
− −
105 5
. 45
41 8
. 4
. 4
. 6
. 4
. 8
.
3 2
1
x x
x
Multiply pivot row 2 by –0.40.6 and subtract the result from the third row to eliminate the x
2
term.
⎪⎭ ⎪
⎬ ⎫
⎪⎩ ⎪
⎨ ⎧
= ⎪⎭
⎪ ⎬
⎫ ⎪⎩
⎪ ⎨
⎧ ⎥
⎥ ⎦
⎤ ⎢
⎢ ⎣
⎡ −
− 3333
. 135
5 .
45 41
533333 .
4 .
6 .
4 .
8 .
3 2
1
x x
x
The solution can then be obtained by back substitution
75 .
253 533333
. 3333
. 135
3
= =
x
245 6
. 75
. 253
4 .
5 .
45
2
= −
− =
x
75 .
173 8
. 245
4 .
41
1
= −
− =
x
b Check:
105 75
. 253
8 .
245 4
. 25
75 .
253 4
. 245
8 .
75 .
173 4
. 41
245 4
. 75
. 173
8 .
= +
− =
− +
− =
−
8.10 The mass balances can be written as
83 200
400
3 33
2 23
1 13
2 23
2 21
1 12
1 13
1 12
2 21
+ =
+ +
= +
= +
c Q
c Q
c Q
c Q
c Q
c Q
c Q
c Q
c Q
or collecting terms
200 400
3 33
2 23
1 13
2 23
21 1
12 2
21 1
13 12
= +
− −
= +
+ −
= −
+
c Q
c Q
c Q
c Q
Q c
Q c
Q c
Q Q
Substituting the values for the flows and expressing in matrix form
⎪⎭ ⎪
⎬ ⎫
⎪⎩ ⎪
⎨ ⎧
= ⎪⎭
⎪ ⎬
⎫ ⎪⎩
⎪ ⎨
⎧ ⎥
⎥ ⎦
⎤ ⎢
⎢ ⎣
⎡ −
− −
− 200
400 120
60 40
80 80
20 120
3 2
1
c c
c
A solution can be obtained with MATLAB as A = [120 -20 0;-80 80 0;-40 -60 120];
b = [400 0 200]; c = a\b
c = 4.0000
4.0000 5.0000
8.11 Equations for the amount of sand, fine gravel and coarse gravel can be written as
