Optimization of Chemical Processes 2nd Edition Solution Manual Complete Solutions Guide

Appendix A

Problem A.1

1 1 0 4  1 7 
(a) A B =    
 2 1 1 3  1 11

0 4  1 1 8 4 
B A=    
1 3   2 1 7 4 

A B  B A

1 2  0 4  2 10
(b) AT B     
1 1  1 3 1 7 

1 1 0 4  1 5 
(c) A B     
 2 1 1 3  3 4 

1 1 0 4  1 3
(d) A B     
 2 1 1 3  1 2

det A  (1)(1)  (1)(2)  1
(e)
det B  (0)(3)  (1)(4)  4

 1 1  3 4 
(f) adj A    adj B   
 2 1  1 0 

adj A  1 1 adj B   1 
A1   B 1  
det A  2 1 det B  1/ 4 0 
(g)


 1 1 1 1 1 0 
Verify: A1 A     
 2 1  2 1 0 1 

  1  0 4  1 0 
B 1 B     
 1/ 4 0  1 3  0 1 




1

,Problem A.2

1 2 3 1 
(a) A  3 2 1 b   2 det A  8
1 0 1 3 

1 2 3 1 
D   A b   3 2 1 2
1 0 1 1 

subtract row 1 from row 3

1 2 3 1 
D  3 2 1 2  det A  8
0 2 2 0 

multiply row 1 by 3 and subtract from row 2

1 2 3 1 
D  0  4  8  1 det A  8
0  2  2 0 

multiply row 2 by ½ and subtract from row 3


1 2 3 1

D  0  4  8  1 det A  8
0 0 2 1/2 

Thus,

1 2 3   x1   1 
0 4 8   x  =  1 
   2  
0 0 2   x3  1/ 2 

1 1
2x3   x3 
2 4
4 x2  8x3  4 x2  8(1/ 4)  1  x2  1/ 4
x1  2x2  3x3  x1  2(1/ 4)  3(1/ 4)  1



2

,  x1  3/ 4
x    1/ 4 1/ 4
T




(b) x  A1b
 2 2 2 
T

 2 2 2 
   1/ 4 1/ 4 1/ 2 
adj A  4 8 4 
1
A     1/ 4 1/ 4  1
det A 8
 1/ 4 1/ 4 1/ 2 

 1/ 4 1/ 4 1/ 2  1   3/ 4 
x   1/ 4   
1/ 4  1  2   1/ 4 
 1/ 4 1/ 4 1/ 2  1   1/ 4 



1 2 3
6
(c) x1   2 2 1  det A   3/ 4
8
1 0 1

1 1 3
x2  3 2 1  det A 
2
  1/ 4
8
1 1 1

1 2 1 
2
x3  3 2 2   det A   1/ 4
8
1 0 1 


Problem A.3

z1  3x1  x3
3 0 1
z2  x1  x2  x3 or z  Ax , A  1 1 1
0 2 1
z3  2 x2  x3

Solve x3  z1  3x1
z2  x1  x2  z1  3x1  x2  2 x1  z1


3

, 1 1 1 1 3
x2  z3  x3  z3  z1  x1
2 2 2 2 2


1 1 3
Then z2  z3  z1  x1  2 x1  z1
2 2 2
x1  z1  2z2  z3

Now x3  z1  3x1  z1  3( z1  2z2  z3 )

x3  2z1  6z2  3z3

1 1
and, x2  z3  x3
2 2
x2  z1  3z2  2z3

1 2 1
or, 
x  1 3 2  z
 2 6 3 

Also, x  A1z

 1 1 2   1 2
T
1
 adj A    
1
A     2 3 6    1 3 2 
 det A 
 1 2 3   2 6 3 
1
This checks with the result obtained using algebraic manipulation.


Problem A.4

x   xT x 
1/ 2




x1  12  12  22 
1/ 2
 6  2.45

x2  12  02  22 
1/ 2
 5  2.24


1 
x x2  1 1 2 0   5
T
1

 2



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