Lecture 4
Solving simple stoichiometric equations
a1FeS2  11O2  a2 Fe2O3  a3SO2
a1  2a2
a1  2a2  0a3  0
2a1  a3
2a1  0a2  a3  0
22  3a2  2a3
0a1  3a2  2a3  22
The Gauß scheme
A linear system of equations
 1  2 0  a1   0 

   
 2 0  1 a 2    0 
0 3
2  a3   22

a1a2  2a2  0a3  0
2a1  0a2 a3  a3  0
0a1  3a2  2a1a3  22
Multiplicative elements.
A non-linear system
Matrix algebra deals essentially with linear linear systems.
x  a0  a1u1  a2u2  a3u3  ... anun
Solving a linear system
 1  2 0  a1   0 

   
 2 0  1 a 2    0 
0 3
2  a3   22

 a1   0   1  2 0 
    

 a 2    0  /  2 0  1
 a   22  0 3
2 
 3   
The division through a vector or a matrix is not defined!
 a11 a12 
 b1 
; B   
A  
a
a
22 
 21
 b2 
 a11b1  a12b2 
 c1 
  C   
A  B  
a
b

a
b
 21 1 22 2 
 c2 
C  c1   b1   a11 a12 

   /   
B  c2   b2   a21 a22 
c1  a11b1  a12b2
c2  a21b1  a22b2
2 equations and four unknowns
Determinants
1 2

A  
3
4


 1 2

A  
2
4


 2  1 4  3  2
a
A   11
 a21
a12 

a22 
a
DetA  A   11
 a21
0  1 4  2  2
a

A  d
g

a12 
  a11a22  a21a12
a22 
b
e
h
c

f
i 
Det A: determinant of A
A
B
1
4
3
The determinant of linear
dependent matrices is zero.
Such matrices are called singular.
Det A
2
5
2
3
6
3
-6
Det B
A
2
4
8
3
6
9
0
B
1
4
7
Det A
1
2
7
2
8
14
0
3
6
9
1
4
6
Det B
2
5
9
0
3
6
12
Higher order determinants
det A   (1)i  j aij detSubMAij 
n
for any i =1 to n
Laplace formula
j 1
 1 2 3


A   2 4 6
7 8 9


4
SubMA11  
8
6
2
; SubMA12  
9
7
 2 3
1
; SubMA22  
SubMA21  
 8 9
7
 2 3
1
; SubMA32  
SubMA31  
 4 6
2
6
2
; SubMA13  
9
7
3
1
; SubMA23  
9
7
4

8 
2

8
3
1 2
; SubMA33  

6
2
4


det A  1(4 * 9  8 * 6)  2(2 * 9  7 * 6)  3(2 * 8  7 * 4)  2(2 * 9  8 * 3)  4(1* 9  7 * 3)  6(1* 2  7 * 2) 
7(2 * 6  4 * 4)  8(1* 6  2 * 3)  9(1* 4  2 * 2)  0
The matrix is linear dependent
The number of operations raises with the faculty of n.
For a non-singular square matrix
the inverse is defined as
A  A 1  I
A 1  A  I
A matrix is singular if it’s
determinant is zero.
a
A   11
 a21
a12 

a22 
a
DetA  A   11
 a21
Singular matrices are those where some rows or
columns can be expressed by a linear
combination of others.
Such columns or rows do not contain additional
information.
They are redundant.
 1 2 3


A   2 4 6
7 8 9


r2=2r1
a12 
  a11a22  a21a12
a22 
Det A: determinant of A
1 2 3 


A  4 5 6 
 6 9 12


r3=2r1+r2
A linear combination of vectors
V  k1V1  k2V2  k3V3  ... kn Vn
A matrix is singular if at least one
of the parameters k is not zero.
The augmented matrix

 1 2 3
 4  5 
 1 2 3 4  5








Augm A   2 4 6 ; B   3 7   C  ( A : B)   2 4 6 3 7 
7 8 9
 1 6 
7 8 9 1 6 







The trace of a square matrix is the sum of its diagonal entries.
An insect species at
three locations has the
following abundances
per season

 1 2 3  n



Trace A   2 4 6    aii  1  4  9  14
 7 8 9  i 1



The diagonal entries (trace) of
the dot product of AB’ contain
The predation rates per
the total numbers of insects
season are given by
per site kept by predators
 0.3 0.4 0.34 0.1 
 10 50 60 5 

 B   0.2 0.25 0.25 0.05


A   34 60 74 8 
 0.34 0.55 0.56 0.2 
 25 45 48 3 




 43.9 29.75 66.5 


C  AB'   60.16 40.7 87.6 
 42.12 28.4 60.73


The inverse of a 2x2 matrix
a
A   11
 a12
a21 

a22 
 a22
1

A 
a11a22  a12 a21   a12
Determinant
1
 a21 

a11 
The inverse of a square matrix only exists
if its determinant differs from zero.
Singular matrices do not have an inverse
The inverse of a diagonal matrix
 a11

 0
A
...

 0

 1

 a11

0
1
A 

 ...
 0


0 

0 
... ... 

... ann 
0 ...
a22 ...
...
0
0
1
a22
...
0

0 


... 0 

... ... 
1 
...
ann 
...
(A•B)-1 = B-1 •A-1 ≠ A-1 •B-1
The inverse can be unequivocally calculated by the Gauss-Jordan algorithm
Systems of linear equations
3 x  4 y  10
2 x  5 y  12
x
10 * 5  12 * 4
3 *12  2 *10
 0.285714 ; y 
 2.285714
3*5  2 * 4
3*5  2 * 4
Determinant
a12 
 a11 a12 
a

  Det 11
  a11a22  a21a12
 a21 a22 
 a21 a22 
 x a12 
 a11 x 


Det
Det
 y a22   xa22  ya12 ; y 
 a21 y   a11 y  a21 x ;
x
a12  a11a22  a21a12
a12  a11a22  a21a12
a
a


Det 11
Det 11
 a21 a22 
 a21 a22 
Solving a simple linear system
1
a1FeS2  11O2  a2 Fe2O3  a3SO2
1
 1  2 0   1  2 0  a1   a1   a1   1  2 0   0 

 
      
  
 2 0  1  2 0  1 a 2   I a 2    a 2    2 0  1  0 
0 3
2   0 3
2  a3   a3   a3   0 3
2   22

4FeS2  11O2  2Fe2O3  8SO2
The general solution of a linear system
 1 0 ...

0 1 ...
A 1A  I
Identity matrix I  
 ... ... ...
1

XA B
 0 0 ...

Only possible if A is not singular.
IX  XI  X
If A is singular the system has no solution.
AX  B  A 1AX  A 1B
3x  2 y  4 z  10
 3x  3 y  8 z  12
9 x  0.5 y  2.3z  1
Systems with a unique solution
The number of independent equations
equals the number of unknowns.
2
4 
 3


3
8 
3
 9  0.5 2.3 


1
0

0
...

1 
2
4 
 3


3
8 
3
 9  0.5 2.3 


X: Not singular
10  x   0.3819 
    

12   y    4.5627 
 1   z    0.0678
    

2
4 10
 3


3
8 12
3
 9  0 .5 2 .3 1 


The augmented matrix Xaug
is not singular and has the
same rank as X.
The rank of a matrix is
minimum number of
rows/columns of the largest
non-singular submatrix
A matrix is linear independent if none of the row or column vectors can be
expressed by a linear combinations of the remaining vectors
A linear combination of vectors
A  k1V1  k2V2  k3V3  ... kn Vn
A matrix of n-vectors (row or columns) is called linear dependent if it is possible to express
one of the vectors by a linear combination of the other n-1 vectors.
 1 2 3


A   2 4 6
7 8 9


r2=2r1
1 2 3 


A  4 5 6 
 6 9 12


r3=2r1+r2
The matrices are linear dependent
If a vector V of a matrix is linear dependent on the other vecors, V does not contain additional
information. It is completely defined by the other vectors. The vector V is redundant.
k1V1  k2 V2  k3V3  ...  kn Vn  k1V1  k2 V2  k3V3  ...  kn Vn  0
k1 , k2 , k3 ...kn  0
Linear independence
How to detect linear dependency
k 1 0
 2 3 1  2k1  3k 2  1k3  0


2k1  4k 2  6(2k1  3k 2 )  0  10k1  14k 2  0
190
A   2 4 6   2k1  4k 2  6k3  0 

 11k1 
k1  0  k 2  0
7
k

8
k

9
(

2
k

3
k
)

0

11
k

19
k

0
14
1
2
1
2
1
2
 7 8 9  7k  8k  9k  0
k3  0
1
2
3


k 1 2k 2
 k 2k
6k1 12k 2
k1  2k 2  7k3  0
 1 2 3
2k1  4k 2  8( 1  2 )  0

0


k

2
k

0
k
2
7
7
7
A   2 4 6   2k1  4k 2  8k3  0 
 7
 1
 k2  1
 k 2k
12k1 24k 2
2
 7 8 9  3k  6k  9k  0 3k1  6k 2  9( 1  2 )  0

 0 k1  2k 2  0
k

0
1
2
3


3
7
7
7
7
Any solution of k3=0 and k1=-2k2 satisfies the above equations. The matrix is linear
dependent.
If a matrix A is linearly independent, then any submatrix of A is also linearly independent
The rank of a matrix is the maximum number of
linearly independent row and column vectors
 1 2 3
1 2 3 




RankA   2 4 6   2 RankA   4 5 6   2
7 8 9
 6 9 12




1 2 3 4 


2 4 6 8 
RankA  
1
4 8 12 16 


 8 16 24 32


a1  2a 2  a 3  2a 4  5
2a1  3a 2  2a 3  3a 4  6
3a1  4a 2  4a 3  3a 4  7
5a1  6a 2  7a 3  8a 4  8
2x1  6x 2  5x 3  9x 4  10   2

2x1  5x 2  6x 3  7x 4  12 
 2

4x1  4x 2  7x 3  6x 4  14   4

5x1  3x 2  8x 3  5x 4  16 
 5
2x1  3x 2  4x 3  5x 4  10   2

4x1  6x 2  8x 3  10x 4  20 
 4

4x1  5x 2  6x 3  7x 4  14   4

5x1  6x 2  7x 3  8x 4  16 
 5
6
5
5
6
4
7
3
8
3
9   x1
 
7   x2

6   x3
 
5 
 x4
4

 10 



   12 

 14 




 16 

 x1

   x2
7   x3
 
8  
 x4
5 

Infinite number of 6solutions
8 10
5
6
6
7
3
No solution
6
4
8
5
6
6
7
2x1  3x 2  4x 3  5x 4  10   2

4x1  6x 2  8x 3  10x 4  12 
 4

4x1  5x 2  6x 3  7x 4  14   4

5x1  6x 2  7x 3  8x 4  16 
 5
2x1  3x 2  6x 3  9x 4  10   2

2x1  4x 2  5x 3  6x 4  12   2

4x1  5x 2  4x 3  7x 4  14 
 4
3
5   x1
 
10   x 2

7   x3
 
8  
 x4

 10 



   20 

 14 




 16 


 10 



   12 

 14 




 16 

 x1 
9 
 10 

  x2 


5 6
  12 
 x3 
 14 
4 7
 


x 

 4
3
4
5 
 10 


  x1 
12 
6
8
10  


x2 
  14 
5
6
7 

x3 



6
7
8  
 16 
x 

 4
 16 
12 14 16 



6
Infinite number of4 solutions
5
2x1  3x 2  4x 3  5x 4  10
 2

4x1  6x 2  8x 3  10x 4  12
4



4x1  5x 2  6x 3  7x 4  14
 4
 5
5x1  6x 2  7x 3  8x 4  16


10x1  12x 2  14x 3  16x 4  16 
  10
No solution
2x1  3x 2  4x 3  5x 4  10
 2

4x1  6x 2  8x 3  10x 4  12
4


4x1  5x 2  6x 3  7x 4  14
 4
 5
5x1  6x 2  7x 3  8x 4  16


10x1  12x 2  14x 3  16x 4  32 
  10
3
4
6
8
5
6
6
12
7
14
Infinite number of solutions
5 
  x1
10  
x2
7 
  x3
8  
x
 4
16 

 10 




 12 
   14 




 16 


 32 


Consistent
Rank(A) = rank(A:B) = n
Consistent
Rank(A) = rank(A:B) < n
Inconsistent
Rank(A) < rank(A:B)
Consistent
Rank(A) = rank(A:B) < n
Inconsistent
Rank(A) < rank(A:B)
Consistent
Rank(A) = rank(A:B) = n
A  X  B  A 1  A  X  A 1  B  X  A 1  B
Consistent system
Solutions extist
rank(A) = rank(A:B)
Single
solution extists
rank(A) = n
Inconsistent system
No solutions
rank(A) < rank(A:B)
Multiple
solutions extist
rank(A) < n
n1KOH  n2Cl2  n3 KClO3  n4 KCl  n5 H 2O
n1  n3  n4
n1  n3  n4  0
n1  3n3  n5
n1  3n3  n5
n1  2n5
n1  2n5
2n2  n3  n4
2n2  n3  n4  0
We have only four
equations but five
unknowns.
The system is
underdetermined.
n1
1
1
1
0
Inverse
0
-0.5
0
-1
n2
0
0
0
2
n3
-1
-3
0
-1
0
1
0
0.5
-0.33333 0.333333
0.333333 0.666667
1

1
1

0

n4
-1
0
0
-1
0
0.5
0
0
 1  1 n1   0 
   
0  3 0  n2   1 
  n5



0 0
0 n3
2
   
2  1  1 n4   0 
0
A
0
1
2
0
N*n5
2
1
0.333333
1.666667
n1
n2
n3
n4
n5
6
3
1
5
3
The missing value is found by dividing the vector through its
smallest values to find the smallest solution for natural numbers.
6KOH  3Cl2  KClO3  5KCl  3H 2O
n1Mga1Sia 2  n2 Na3 H a 4Bra5  n3Sia6 H a7  n4 Na8 H a9  n5 Mga10 Bra11
Equality of
atoms involved
n1a1  n5 a10
n1a2  n3 a6
n2 a3  n4 a8
n2 a4  n4 a9  n3 a7
n2 a5  n5 a11
Including
information on
the valences of
elements
a1  2a2
a4  a3  a5
a7  4 a 6
a8  4(a9  1)
a10  2a11
We have 16 unknows but without
experminetnal information only 11 equations.
Such a system is underdefined.
A system with n unknowns needs at least n
independent and non-contradictory equations
for a unique solution.
a1  a11
If ni and ai are unknowns we have a non-linear situation.
We either determine ni or ai or mixed variables such that no multiplications occur.
n1a1  n5 a10
n1a2  n3 a6
n2 a3  n4 a8
n2 a4  n4 a9  n3 a7
n2 a5  n5 a11
a1  2a2
a4  a3  a5
a7  4 a 6
a8  4(a9  1)
a10  2a11
a1  a11
0
 n1 0 0 0 0

 n1 0 0 0 0  n3
 0 0 n2 0 0
0

0
 0 0 0 n2 0
 0 0 0 0 n2
0

 2 1 0 0 0
0

 0 0 1 1 1 0
 0 0 0 0 0 4

0
0 0 0 0 0
0 0 0 0 0
0

0
1 0 0 0 0
0
0
0
 n5
0  a1   0 

  
0
0
0
0
0  a 2   0 
0
0
 n4
0
0  a3   0 

  
0  n3
0
 n4
0  a 4   0 
0
0
0
0
 n5  a5   0 
0
0
0
0
0  a6    0 

  
0
0
0
0
0  a7   0 
1
0
0
4
0  a8   0 

  
0
1
4
0
0  a9    4 
0
0
0
1
 2  a10  0 

  
0
0
0
0
 1  a11  0 
The matrix is singular because a1, a7, and a10 do
not contain new information
Matrix algebra helps to determine what information is needed
for an unequivocal information.
n1Mga1Sia 2  n2 Na3 H a 4Bra5  n3Sia6 H a7  n4 Na8 H a9  n5 Mga10 Bra11
From the knowledge of the salts we get n1 to n5
n1Mga1Sia 2  n2 Na3 H a 4Bra5  n3Sia6 H a7  n4 Na8 H a9  n5 Mga10 Bra11
Mg2 Si  4Na3 H a 4Bra5  SiHa7  4Na8 H a9  2MgBr2
a4  3a3  a5
a3  a8
4 a 4  a 7  4 a9
4a5  4
a8  1
a9  3
3

 1
 0

 0
 0

 0

1 1
0
0
4
0
0
1
0
0
0
0
a3
a4
a5
a7
a8
a9
Inverse
0  a3   0 
   
0  1 0  a4   0 
 1 0  4  a5   0 
    
0 0
0  a7   1 
 
0 1
0  a8   1 
0 0
1  a9   3 
0
0
We have six variables and six
equations that are not
contradictory and contain
different information.
The matrix is therefore not
singular.
a3
-3
1
0
0
0
0
a4
1
0
4
0
0
0
a5
-1
0
0
1
0
0
a7
0
0
-1
0
0
0
a8
0
-1
0
0
1
0
a9
0
0
-4
0
0
1
0
1
0
4
0
0
1
3
0
12
0
0
0
0
0
-1
0
0
0
1
1
4
0
0
1
3
0
12
1
0
0
0
0
-4
0
1
A
0
0
0
1
1
3
a3
a4
a5
a7
a8
a9
Mg2 Si  4NH 4Br  SiH4  4NH 3  2MgBr2
1
4
1
4
1
3
Linear models in biology
The logistic model of population growth
r 2
N  rN  N  c
K
t
1
2
3
4
N
1
5
15
45
We need four
measurements
r
4  1r  1  c
K
r
10  5r  25  c
K
r
30  15r  225  c
K
1 1 r 
4 1
  


 10    5 15 1 r / K 
 30 15 225 1 c 
  


K denotes the maximum possible density
under resource limitation, the carrying
capacity.
r denotes the intrinsic population growth
rate. If r > 1 the population growths, at r < 1
the population shrinks.
K  1.286 / 0.036  36
Population growth
N  1.286 N 
t
1
2
3
4
5
6
7
8
9
10
1.286 2
N  2.679
36
N
1
4.928571
13.07635
26.46055
38.15409
37.8974
38.00788
37.96091
37.98099
37.97242
N
3.928571
8.147777
13.3842
11.69354
-0.25669
0.110482
-0.04698
0.02008
-0.00856
0.003656
We have an overshot.
In the next time step the population should
decrease below the carrying capacity.
N
K
Overshot
K/2
N (t  1)  N (t )  N (t )
N (t  1)  N  1.286N 
1.286 2
N  2.679
36
Fastest
population
growth
t