97 ⎥
⎥ ⎦
⎤ ⎢
⎢ ⎣
⎡ −
− −
− −
=
−
029491 .
053619 .
16622 .
029491 .
013405 .
0.115282 -
] [
1
A Finally, the same procedures can be implemented with {b}
T
=
⎣ ⎦
1 to solve for {d}
T
=
⎣ ⎦
1 , and the results are used with [U] to determine {x} by back substitution to
generate the third column of the matrix inverse,
⎥ ⎥
⎦ ⎤
⎢ ⎢
⎣ ⎡
− −
− −
− −
− =
−
123324 .
029491 .
053619 .
032172 .
16622 .
029491 .
034853 .
013405 .
0.115282 -
] [
1
A
10.3 The following solution is generated with MATLAB.
a A = [15 -3 -1;-3 18 -6;-4 -1 12];
format long AI = invA
AI = 0.07253886010363 0.01278065630397 0.01243523316062
0.02072538860104 0.06079447322971 0.03212435233161 0.02590673575130 0.00932642487047 0.09015544041451
b b = [3800 1200 2350];
format short c = AIb
c = 320.2073
227.2021 321.5026
c
The impact of a load to reactor 3 on the concentration of reactor 1 is specified by the element
1 13
−
a = 0.0124352. Therefore, the increase in the mass input to reactor 3 needed to induce a 10 gm
3
rise in the concentration of reactor 1 can be computed as
d g
1667 .
804 0124352
. 10
3
= =
∆b
d
The decrease in the concentration of the third reactor will be
3 3
m g
285 .
15 3316
. 2
9534 .
12 250
009326 .
500 0259067
. =
+ =
+ =
∆c
10.4 The mass balances can be written and the result written in matrix form as
98 ⎪
⎪ ⎭
⎪⎪ ⎬
⎫
⎪ ⎪
⎩ ⎪⎪
⎨ ⎧
= ⎪
⎪ ⎭
⎪ ⎪
⎬ ⎫
⎪ ⎪
⎩ ⎪
⎪ ⎨
⎧
⎥ ⎥
⎥ ⎥
⎦ ⎤
⎢ ⎢
⎢ ⎢
⎣ ⎡
− −
− −
− −
− −
4 1
3 2
11 8
1 9
1 3
3 1
6
03 03
01 01
5 4
3 2
1
c Q
c Q
c c
c c
c
MATLAB can then be used to determine the matrix inverse Q = [6 0 -1 0 0;-3 3 0 0 0;0 -1 9 0 0;0 -1 -8 11 -2;-3 -1 0 0 4];
invQ ans =
0.1698 0.0063 0.0189 0 0 0.1698 0.3396 0.0189 0 0
0.0189 0.0377 0.1132 0 0 0.0600 0.0746 0.0875 0.0909 0.0455
0.1698 0.0896 0.0189 0 0.2500 The concentration in reactor 5 can be computed using the elements of the matrix inverse as
in,
528 .
24 547
. 7
981 .
16 50
8 0189
. 20
5 1698
.
03 03
1 53
01 01
1 51
5
= +
= +
= +
=
− −
c Q
a c
Q a
c
10.5 The problem can be written in matrix form as
⎪ ⎪
⎪ ⎭
⎪ ⎪
⎪ ⎬
⎫
⎪ ⎪
⎪ ⎩
⎪ ⎪
⎪ ⎨
⎧ =
⎪ ⎪
⎪ ⎭
⎪⎪ ⎪
⎬ ⎫
⎪ ⎪
⎪ ⎩
⎪⎪ ⎪
⎨ ⎧
⎥ ⎥
⎥ ⎥
⎥
⎦ ⎤
⎢ ⎢
⎢ ⎢
⎢
⎣ ⎡
− −
− −
− −
− −
v h
v h
v h
F F
F F
F F
V V
H F
F F
, 3
, 3
, 2
, 2
, 1
, 1
3 2
2 3
2 1
1 866
. 5
. 1
1 5
. 1
1 866
. 866
. 5
. 5
. 866
.
MATLAB can then be used to solve for the matrix inverse, A = [0.866 0 -0.5 0 0 0;
0.5 0 0.866 0 0 0; -0.866 -1 0 -1 0 0;
-0.5 0 0 0 -1 0; 0 1 0.5 0 0 0;
0 0 -0.866 0 0 -1]; AI = invA
AI = 0.8660 0.5000 0 0 0 0
0.2500 -0.4330 0 0 1.0000 0 -0.5000 0.8660 0 0 0 0
-1.0000 0.0000 -1.0000 0 -1.0000 0 -0.4330 -0.2500 0 -1.0000 0 0
0.4330 -0.7500 0 0 0 -1.0000
99 The forces in the members resulting from the two forces can be computed using the
elements of the matrix inverse as in,
1000 1000
500 2000
5 .
, 3
1 15
, 1
1 12
1
− =
+ −
= −
+ −
= +
=
− −
h v
F a
F a
F
366 500
866 500
1 2000
433 .
, 3
1 25
, 1
1 22
2
= −
= −
+ −
− =
+ =
− −
h v
F a
F a
F
1732 1732
500 2000
866 .
, 3
1 35
, 1
1 32
3
− =
+ −
= −
+ −
= +
=
− −
h v
F a
F a
F
10.6
The matrix can be scaled by dividing each row by the element with the largest absolute value
A = [8-10 2-10 1;1 1-9 3-9;1 -115 615] A =
-0.8000 -0.2000 1.0000 1.0000 -0.1111 -0.3333
1.0000 -0.0667 0.4000
MATLAB can then be used to determine each of the norms, normA,fro
ans = 1.9920
normA,1 ans =
2.8000 normA,inf
ans = 2
10.7 Prob. 10.2:
