Systems of linear equations, matrices, determinants




Equivalent systems of linear equations

Examples of linear systems

 
/ x+2y+3z=5
| x-y+6z=1
| 3x-2y=4
\ y+4z=8

/ x+y+3z-4t=12
\ 3x+y-2z-t=0

/ 2x+3y+4z=5
| x-y+2z=6
\ 3x-5y-z=0

Each of these systems can be expressed by the help of matrices.
 
[ 1  2  3]   [x]   [5]
[ 1 -1  6] . [y] = [1]
[ 3 -2  0]   [z]   [4]
[ 0  1  4]         [8]


                [x]
[ 1  1  3 -4]   [y]   [12]
[ 3  1 -2 -1] . [z] = [ 0]
                [t]


[ 2  3  4]   [x]   [5]
[ 1 -1  2] . [y] = [6]
[ 3 -5 -1]   [z]   [0]
The first matrix of each representation of a linear system contains the coefficients appearing in the system. This matrix is called the coefficient matrix.
The essential information of such system is this coefficient matrix A together with the column matrix B of the known terms.
We can denote such system shortly as AX=B. Then X is the column matrix of the unknowns. Therefore we can gather all essential data of the system in one matrix, by adding the column matrix of the known terms to the coefficient matrix.
For the first system we have then
 
[ 1  2  3  5]
[ 1 -1  6  1]
[ 3 -2  0  4]
[ 0  1  4  8]
This matrix is called the enlarged matrix of the system.
Conversely, when we have the enlarged matrix of a system, we know the system and we can start calculating the solutions.

equivalent systems

Two systems are equivalent if and only if they have the same set of solutions.
Now we'll define some basic actions by which a system is transformed in an equivalent system.

Action 1 :

Multiply one of the equations with a real number different from 0.
This is equivalent with :
Multiply one of the rows, of the enlarged matrix of the system, with a real number different from 0.

Action 2 :

Multiply one of the equations with a real number, and add the result to another equation, leaving the original equation unchanged.
This is equivalent with :
Multiply one of the rows, of the enlarged matrix of the system, with a real number, and add the result to another row, leaving the original row unchanged.

Action 3 :

Interchange two equations .
This is equivalent with :
Interchange two rows of the enlarged matrix of the system.

Action 4 :

Delete an equation that is equivalent with 0 = 0
This is equivalent with :
Delete a row, of the enlarged matrix of the system, that contains only zero's.

Action 5 :

If an equation of a system is equivalent with 0=k, with k not zero, then the system has no solutions.
This is equivalent with :
If a row, of the enlarged matrix of the system, is composed of zero's, except for the last number, then the system has no solutions.

Row-transformations

Each of the actions transforming a system to an equivalent system, is equivalent to a transformation of the rows of the enlarged matrix of the system. These are the so called 'row transformations'.

Now, it can be shown that, by a appropriate consequence of only these 5 actions, we can solve any system of linear equations. The method, to do this in a efficient way, is called the method of Gauss.

The Gauss method

We'll illustrate this method by means of 4 examples.

Strategy:

Only use the row transformations described above and work from top-row to bottom-row.
1. In each row make the first non-zero element (called the main element) equal to 1
2. Transform, beneath this main element, all the elements to 0
3. If a row of zeros occurs, then delete this row
4. If a row of zeros occurs, except for the last number, then the system has no solutions.

Example1

Take the system
 
/ 2x-3y+2z=21
| x+4y-z=1
\ -x+2y+z=17
or equivalently
 
[2  -3   2  21]
[1   4  -1   1]
[-1  2   1  17]
First I'll show the row transformations, then I'll state the method in words.
 
                R1 <-> R2
[1   4  -1   1]
[2  -3   2  21]
[-1  2   1  17]
                R2 + 2.R3
[1   4  -1   1]
[0   1   4  55]
[-1  2   1  17]
                R3 + R1
[1   4  -1   1]
[0   1   4  55]
[0   6   0  18]
                (1/6).R3
[1   4  -1   1]
[0   1   4  55]
[0   1   0   3]
                R3-R2
[1   4  -1   1]
[0   1   4  55]
[0   0  -4 -52]
                (-1/4).R3

[1   4  -1   1]
[0   1   4  55]
[0   0   1  13]
Now we shall return to the system.
 
/ x+4y-z=1
|   y+4z=55
\      z=13
Working from z to x, we find easily z=13 ; y=3 ; x=2.

We recall the strategy:
Only use the row transformations described above and work from top-row to bottom-row.
1. In each row make the first non-zero element (called the main element) equal to 1
2. Transform, beneath this main element, all the elements to 0
3. If a row of zeros occurs, then delete this row
4. If a row of zeros occurs, except for the last number, then the system has no solutions.

Remark: All these matrices are called row-equivalent.
The resulting matrix is called an 'echelon' matrix.
It is also possible to transform, the matrix such that each main element is the only non zero element on its column.
In that case the resulting matrix is called row-canonical.

Example2

Take the system
 
/ 3x-4y+5z-4t=12
| x-y +z -2t = 0
| 2x+y+2z+3t=52
| 2x-2y+2z-4t=0
\ 2x-3y+2z-t=4
or equivalently
 
[3  -4  5  -4  12]
[1  -1  1  -2   0]
[2   1  2   3  52]
[2  -2  2  -4   0]
[2  -3  2  -1   4]
                        R1-R4
[1  -2  3   0  12]
[1  -1  1  -2   0]
[2   1  2   3  52]
[2  -2  2  -4   0]
[2  -3  2  -1   4]
                        R2-R1;R3-2R1;R4-2R1;R5-2R1
[1  -2  3   0  12]
[0   1 -2  -2 -12]
[0   5 -4   3  28]
[0   2 -4  -4 -24]
[0   1 -4  -1 -20]
                        R3-5R2;R4-2R2;R5-R2;
[1  -2  3   0  12]
[0   1 -2  -2 -12]
[0   0  6  13  88]
[0   0  0   0   0]
[0   0 -2   1  -8]
                        delete R4;R3<->R4;
[1  -2  3   0  12]
[0   1 -2  -2 -12]
[0   0 -2   1  -8]
[0   0  6  13  88]
                        R4+3R3;(1/16)R4;R3/(-2);
[1  -2  3   0  12]
[0   1 -2  -2 -12]
[0   0  1 -1/2  4]
[0   0  0   1   4]
This is again an echelon matrix. Thus return to the system and calculate the unknowns from bottom to top.
 
/ x-2y+3z =12
| y -2z -2t =-12
| z-0.5t=4
\ t=4

x=10 ; y=8 ; z=6 ; t=4

Example3

Take the system
 
/ x+y-4z+10t=24
\ 3x-2y-2z+6t=15

[1  1  -4  10  24]
[3 -2  -2   6  15]
                        R2-3R1
[1  1  -4  10  24]
[0 -5  10 -24 -57]
                        r2/(-5)
[1  1  -4  10   24  ]
[0  1  -2  4.8  11.4]

Back to the system
/ x + y -4z +10t =24
\    y  -2z +4.8t=11.4
The fundamental difference between proceeding examples is that, in this case, it is impossible to calculate the unknowns. Therefore, we write the system as follows.
 
/ x + y = 4z -10t + 24
\    y  = 2z -4.8t+ 11.4

<=>

/ x = 2z - 5.2t + 12.6
\ y = 2z -4.8t+ 11.4

For each chosen value of z and of t, we can calculate one solution of the system.
The system has an infinitely number of solutions.
The values of z and t are arbitrary.
The set S of all solutions can be written as follows.
 
S = {2z - 5.2t + 12.6 ,2z -4.8t + 11.4 , z, t} with z, t in R

Example4

Take the system
 
/ x-3y=21
| 4x+2y=14
\ 3x+3y=7

[1  -3  21]
[4   2  14]
[3   3   7]
                (1/2).R2 ; R3-3.R1
[1  -3  21]
[2   1   7]
[0  12 -56]
                R2-2.R1
[1  -3  21]
[0   7 -35]
[0  12 -56]
                (1/7)R2 ; (1/4)R3
[1  -3  21]
[0   1  -5]
[0   3 -14]
                R3 -3.R2
[1  -3  21]
[0   1  -5]
[0   0   1]
Because of the last row, the system has no solutions.

Rank of a matrix and inverse of a matrix

Singular and regular matrices

If the determinant of a square matrix is 0, we call this matrix singular otherwise, we call the matrix regular.

Rank of a matrix

Take a fix matrix A. By crossing out, in a suitably way, some rows and some columns from A, we can construct many square matrices from A.
Doing this, search now the biggest regular square matrix.
The number of rows of that matrix is called the rank of A.

Examples

3 x 2 matrix A =

 
[1 2]
[0 0]
[4 0]
The rank is at most 2. If we delete the second row, we have the biggest regular square matrix. The rank of A is 2.

3 x 3 matrix A =

 
[4 5 7]
[1 2 3]
[2 4 6]
The rank is at most 3, but A is not a regular matrix. So, we are looking for a regular 2x2 matrix. We delete row 3 and column 3. The remaining matrix is regular. The rank of A is 2.

3 x 3 matrix A =

 
[1 2 3]
[1 2 3]
[2 4 6]
The rank is at most 3, but A is not a regular matrix. So, we are looking for a regular 2x2 matrix. But all 2x2 matrices are not regular. So, the rank of A is 1.

Application:

Take matrix A

 
[ a     a-1   a-2 ]
[3a-3   a+1   2a-4]
Determine the rank for all values of a.

Solution:
The maximum rank is 2. We select a 2 x 2 matrix from A, for example

 
[ a     a-1]
[3a-3   a+1]
The determinant of that matrix is -2a2 +7a -3. the zero's are 1/2 and 3. It follows that the rank of A is 2 if a is not in { 1/2 , 3}.

If we substitute 1/2 for a, we immediately see that the rank of A is 2.

If we substitute 3 for a, then we have for A

 
[3  2  1]
[6  4  2]
The rank of this matrix is 1.

Conclusion: If a is different from 3, the rank is 2. The rank is 1 if a = 3.

Adjoint matrix of a square matrix A

First we recall the method to calculate the cofactor of an element.

Choose in the matrix the element on the i-th row and the j-th column.
Delete this row en this column from the matrix.
Calculate the value D of the determinant of the remaining matrix.
The cofactor of the the element on the i-th row and the j-th column is (-1)i+j.D
The elements of the deleted row and column do not affect the value of this cofactor!

Now we are ready to calculate the adjoint matrix of a square matrix A.
Replace each element of A with its own cofactor and transpose the result, then you have calculated the adjoint matrix of A.

Examples

 
[ 1, 4 ]
[ 5, 8 ]

has the adjoint matrix

[ 8,  -4 ]
[ -5,  1 ]

---------------------

[ 1, 2, 4 ]
[ 3, 2, 5 ]
[ 1, 3, 4 ]

has the adjoint matrix
[ -7, 4,  2  ]
[ -7, 0,  7  ]
[ 7,  -1, -4 ]

Cofactors property

Theorem : When we multiply the elements of a row of a square matrix with the corresponding cofactors of another row, then the sum of these product is 0.

Proof:
We'll prove this property for 3x3 matrices but the method of the proof is universal.
So take P =

 
[a  b  c]
[d  e  f]
[g  h  i]
Let A,B,C,D,E,F,G,H,I be the cofactors of a,b,c,d,e,f,g,h,i.
We multiply the elements of a row, say the second, with the corresponding cofactors of another row, say the the first.
We have to prove that dA+eB+fC = 0.
Take now the matrix Q =
 
[d  e  f]
[d  e  f]
[g  h  i]
Since the matrix has two equal rows,its determinant is 0. So det(Q) = 0.
Furthermore, the cofactors of corresponding elements of the first row of P and Q are the same. These cofactors are A B and C.
Hence the calculation of det(Q) emanating from the first row gives dA+eB+fC.
Since we know that det(Q) = 0, dA+eB+fC = 0.
Q.E.D.

Inverse matrix of a regular square matrix

We say that B is an inverse matrix of A if and only if A.B=I=B.A . (I is the identity matrix.)

Uniqueness of the inverse matrix

We show that it is impossible that there are two inverse matrices for A.
Say, there are two inverse matrices B and C for A, then we have
A.B=I=B.A and A.C=I=C.A
Then, B=I.B=C.A.B=C.I=C
So B=C is the unique inverse of A.

Calculating the inverse of a regular square matrix

We'll show that A-1 = (adjoint of A) / det(A)

We'll prove this property for 3x3 matrices but the method of the proof is universal.
We show that A. (adjoint of A) / det(A) = E or equivalently that A.(adjoint of A) = det(A) . E

First we'll calculate
A.(adjoint of A) =

 
[a  b  c]   [A D G]
[d  e  f] . [B E H]  =
[g  h  i]   [C F I]

[aA+bB+cC   aD+bE+cF   aG+bH+cI]
[dA+eB+fC   dD+eE+fF   dG+eH+fI]  =
[gA+hB+iC   gD+hE+iF   gG+hH+iI]

        Because of the cofactors property,

[aA+bB+cC       0         0    ]
[  0        dD+eE+fF      0    ]  =
[  0            0      gG+hH+iI]

        The diagonal elements of this matrix are det(A)

[  det(A)      0         0    ]
[  0         det(A)      0    ]  =
[  0           0        det(A)]

        [  1     0      0]
det(A). [  0     1      0]  =
        [  0     0      1]

det(A).I
In the same way, (adjoint of A).A =det(A).I

Remark : All regular matrices have an inverse matrix and now we can calculate this inverse matrix.

Example 1

 
[ 1, 4 ]
[ 5, 8 ]

determinant =  -12

and has  adjunct matrix

[ 8,  -4 ]
[ -5,  1 ]

The inverse matrix is :
[ -2/3,  1/3  ]
[ 5/12, -1/12 ]

Example 2
 
[ 1, 2, 4 ]
[ 3, 2, 5 ]
[ 1, 3, 4 ]

determinant =  7

and has adjunct matrix

[ -7, 4,  2  ]
[ -7, 0,  7  ]
[ 7,  -1, -4 ]

The inverse matrix is

[ -1, 4/7,  2/7  ]
[ -1,  0,    1   ]
[ 1,  -1/7, -4/7 ] ]

Singular matrices and the inverse.

Assume, B is the inverse of a singular matrix A. Then, A.B = I => |A|.|B|=1 Since |A|=0, this is impossible. So, a singular matrix has no inverse.

Cramer systems

Cramer's rule

A system of n linear equations in n unknowns is called a Cramer system if and only if the matrix formed by the coefficients is regular.

There is a special method to solve such a system. This method is called Cramer's rule.

We'll prove the rule for a system of 3 equations in 3 unknowns, but the rule is universal.
Take,

 
/ ax + by  + cz  = d
| a'x+ b'y + c'z = d'         (1)
\ a"x+ b"y + c"z = d"

           |a  b  c |
with |N| = |a' b' c'|
           |a" b" c"|

Then we have
           |xa  b  c |
   x.|N| = |xa' b' c'|
           |xa" b" c"|
and using the properties of determinants
           |xa +by +cz    b  c |
   x.|N| = |xa'+b'y+c'z   b' c'|
           |xa"+b"y+c"z   b" c"|
and appealing to (1)
           |d   b  c |
   x.|N| = |d'  b' c'|
           |d"  b" c"|
Thus,
           |d   b  c |
     x   = |d'  b' c'| / |N|    (2)
           |d"  b" c"|
Similarly,
           |a   d  c |
     y   = |a'  d' c'| / |N|    (3)
           |a"  d" c"|

           |a   b  d |
     z   = |a'  b' d'| / |N|    (4)
           |a"  b" d"|
The formulas (2), (3), (4) constitute Cramer's rule. It can be proved that this solution is the only solution of (1).

Example:

Solve the system in x, y and z . This system has parameter p.

 
/ p x - y + z = 0
| 6 x + y - 2z = 2
\ px - 2y - z = 1

We find :   |N| = -4p - 18

We assume that p is not -4.5.
Men find :

 x.|N| = -5

 y.|N| =  -2p + 6

 z.|N| = 3p + 6

 So,  x = 5/(4p+18) ;  y = (p-3)/(2p+9) ;  z = (3p+6)/(-4p-18)

Homogeneous system of Cramer

If all the known terms of a Cramer system are 0, the system is homogeneous in the unknowns. It follows directly from Cramer's rule that the only solution of such system is the zero solution (all unknowns = 0). This solution is also called the obvious solution.

Example

/ x + 2 y + 4z = 0
| 3x + 2y + 5z = 0
\ x + 3y + 4z = 0

This system has only the solution x = y = z = 0

Classification of systems of linear equations

Main equations ; side equations ; main unknowns ; side unknowns

Take any system with m linear equations in n unknowns.
Let A be the matrix of coefficients of the system, and let r= rank(A).
Then there can at least one regular r x r matrix M be made from A .
Once M is chosen, we call it the main matrix.

Example:

 
/ x+2y+z+2u=1
| 4x+4y=-3
| 3x+6y=-4.5
\ 2x+4y+2z+4u=2

The matrix of coefficients is

[1  2  1  2]
[4  4  0  0]
[3  6  0  0]
[2  4  2  4]

The rank of this matrix is 3.

We choose a main matrix M from that matrix.

[1  2  1]
[4  4  0]
[3  6  0]

The equations corresponding with the rows of M are called the main equations, the other equations are the side equations.
The unknowns corresponding with the columns of M are called the main unknowns, the other columns are the side unknowns.

In previous example we have:
The main equations are the first, the second and the third one.
The main unknowns are x, y and z. The last equation is the only side equation and u is the side unknown.

Characteristic determinant

The characteristic matrix, associated with a particular side equation, is the matrix formed by adding to the main matrix :
1. a row at the bottom, with the coefficients of the main unknowns in that side equation.
2. a column at the right, with the known terms of the main equations and of the known term of that side equation.

The characteristic determinant is the determinant of the characteristic matrix.
So there are as much characteristic determinants as the number of side equations.

We continue with our example :
The characteristic determinant of the only side equation is

 
   |1  2  1  1  |
   |4  4  0 -3  |
   |3  6  0 -4.5|
   |2  4  2  2  |
With all these new concepts we can classify all the systems of linear equations.

Classification

First kind

The systems with n equations and n unknowns and with n = rank of the matrix of coefficients .
These are the Cramer systems mentioned earlier. They have exactly one solution.

Second kind

The systems with m equations and n unknowns and with m = rank of the matrix of coefficients .
In that case, there are side unknowns, but no side equations.
It can be proved that all the side unknowns can be chosen arbitrarily. With each choice of these side unknowns corresponds exactly one solution of the system.
 
Example:

/ 2x+3y+z=4
\ x+2y-z=3

Choose the main matrix

[2  3]
[1  2]
z is the side unknown. With each choice of z, corresponds exactly one solution of the system.

Third kind

The systems with m equations and n unknowns and with n = rank of the matrix of coefficients .
In that case, there are (m - n) side equations, but no side unknowns.
Thus there are (m - n) characteristic determinants. It can be proved that the system has a solution if and only if all the characteristic determinants are zero. In that case, the solution is unique! Furthermore, all the side equations are linear combinations of the main equations.
The unique solution can be found by omitting all the side equations, and solving the remaining system of the first kind.
Example:
 
/ 2x+y=1
| x+y=0
| 3x+2y=1
\ 4x+3y=1

I choose the main matrix

[2  1]
[1  1]
The characteristic determinants are
|2  1  1|
|1  1  0| = 0
|3  2  1|

and

|2  1  1|
|1  1  0| = 0
|4  3  1|

The unique solution is the solution of the system

/ 2x+y=1
\ x+y=0

x=1, y=-1

Fourth kind

The systems with m equations and n unknowns and with r = rank of the matrix of coefficients .
In that case, there are (m - r) side equations, and (n - r) side unknowns.
Thus there are (m - r) characteristic determinants. It can be proved that the system has a solution if and only if all the characteristic determinants are zero.
furthermore, all the side equations are linear combinations of the main equations and can be omitted. Then, the remaining system is a system of the second kind.
All the side unknowns can be chosen arbitrarily. With each choice of these side unknowns corresponds exactly one solution of the system.

Example 1

/ -4.5x - y + z = 0
| 6x + y - 2z = 2
\ -4.5 x - 2y - z = 1

The rang of the matrix of the system is 2.
We choose y and z as main unknowns and the third equation is a side equation. The side unknown is x. There is one characteristic matrix

[ -1 1 0 ]
[ 1 -2 2]
[-2 -1 1]

The value of de characteristic determinant is -5.
The system has no solutions.

Example 2

/ x + y - z + u = 2
| 2x - y + z + u = 3
| 3x + 2u = 5
\ 4x + y - z + 3u = 7

The rank of the matrix of coefficients is 2. We choose as main matrix M =

 
  [ 1   1]
  [ 2  -1]
The main unknowns are x and y and the main equations are the first one and the second one.
The side unknowns are z and u and the side equations are the third one and de fourth one.
There are two characteristic determinants
 
 | 1  1  2 |
 | 2 -1  3 | = 0
 | 3  0  5 |

 | 1  1  2 |
 | 2 -1  3 | = 0
 | 4  1  7 |
As the characteristic determinants are 0, the side equations may be omitted. We get a system of the second kind.
We bring the terms with side unknowns to the right hand side.

/ x + y = 2 + z - u
\ 2x - y = 3 - z - u

We can solve this remaining system as a system of the first kind.
There is exactly one solution for each value of z and u.

x = ( 5 - 2 u)/3
y = (1 + 3z - u)/3

Homogeneous equations

If all the known terms of a system are 0, the system is homogeneous in the unknowns. A homogeneous system always has the obvious solution. All properties stated in previous classification, also hold for homogeneous systems.
Because all known terms are zero, it is easy to verify that all characteristic determinants are 0.

Important conclusions

  1. A homogeneous Cramer system has exactly one solution, the obvious solution.
  2. In a homogeneous system of the second kind, we can choose all the side unknowns arbitrarily. With each choice of these side unknowns corresponds exactly one solution of the system.
  3. A homogeneous system of the third kind, has exactly one solution, the obvious solution.
  4. In a homogeneous system of the fourth kind, we can choose all the side unknowns arbitrarily and omit the side equations. With each choice of these side unknowns corresponds exactly one solution of the system.
  5. A homogeneous system of n equations in n unknowns, is either a Cramer system or a system of the fourth kind. It has respectively only the obvious solution, or an infinity number of solutions. Hence we can state the important conclusion:
    A homogeneous system of n equations in n unknowns has a solution different from the obvious one, if and only if the determinant of the coefficient matrix is zero.
  6. It is immediate that, if a solution is found for a homogeneous system, each real multiple of this solution is also a solution of the system.
Example

What are the p-values such that the system has a solution different from (0,0,0) ?

/ p x - y + z = 0
| 6 x + y - 2z = 0
\ px - 2y - z = 0

The system is homogeneous. The condition is that the matrix formed by the coefficients is singular.

The determinant is -4p - 18. So, for p = -9/2 , there are solutions different from (0,0,0).

To find these solutions, we substitute -9/2 for p.
We omit a side equation. Then we have a system of the second kind with infinitely many solutions.

Applications

Criterion for collinear points

Take three points A,B and C in a plane. The coordinates are (a,a');(b,b') and (c,c')
 
        The three points are collinear
<=>
        There is a line ux+vy+w=0 through A,B and C
<=>
        There are values for u,v,w different from 0,0,0 such that
        u.a + v.a' + w = 0
        u.b + v.b' + w = 0
        u.c + v.c' + w = 0
<=>
        The homogeneous system
        a.u + a'.v + 1.w = 0
        b.u + b'.v + 1.w = 0
        c.u + c'.v + 1.w = 0
        has a solution for u,v,w different from the obvious solution.
<=>
        |a      a'      1|
        |b      b'      1| = 0
        |c      c'      1|
The points A(a,a'), B(b,b'), C(c,c') are collinear if and only if
 
        |a      a'      1|
        |b      b'      1| = 0
        |c      c'      1|

Equation of the line AB

Take two points A(a,a') and B(b,b') in a plane. Appealing on previous application, we can write
 
        A third point P(x,y) is on the line AB
<=>
        |x      y       1|
        |a      a'      1| = 0
        |b      b'      1|
This is the equation of the line AB.
The equation of the line AB with A(a,a') and B(b,b') is
 
        |x      y       1|
        |a      a'      1| = 0
        |b      b'      1|

Criterion for concurrent lines

We start with three intersecting lines in the plane. The equations are
 
    u x + v y + w  = 0
    u'x + v'y + w' = 0
    u"x + v"y + w" = 0

  The three lines are concurrent

<=> The three lines have a common point P(xo,yo)

<=>  There is an xo and a yo such that
    u xo + v yo + w  = 0
    u'xo + v'yo + w' = 0
    u"xo + v"yo + w" = 0

<=>  There is an xo and a yo such that
    u  xo + v  yo = -w
    u' xo + v' yo = -w'
    u" xo + v" yo = -w"

<=> The system
         u  x + v  y = -w
         u' x + v' y = -w'
         u" x + v" y = -w"
    has a solution for x and y.

    The matrix of the coefficients of the system is
       [ u   v ]
       [ u'  v']
       [ u"  v"]

    Since the lines intersect each other, the rank is 2.
    We choose the first two equations as main equations

    The system has a solution for x and y
<=> The characteristic determinant is zero

<=>    | u   v  w |
       | u'  v' w'| = 0
       | u"  v" w"|
 
   Three intersecting lines have  equations :
    u x + v y + w  = 0
    u'x + v'y + w' = 0
    u"x + v"y + w" = 0

   The lines are  concurrent if and only if

       | u   v  w |
       | u'  v' w'| = 0
       | u"  v" w"|

Investigation of systems with a parameter

Example 1

Take the system with parameter m

/mx + y + z = m+1
| x + my + z = 0
\ x + y + mz = 1

De determinant of the matrix formed by the coefficients is (m - 1)2 .(m+2)

First case: m is different from 1 and -2

With Cramer's Rule, we find for x, y and z the solutions :

 
     m(m-1)(m+2)
x = ------------------  = m/(m-1)
     (m - 1)2 (m+2)

     -(m-1)(m+2)
y = -----------------  = 1/(1-m)
      (m - 1)2 (m+2)

z = 0

Second case: m = 1

Then, the system is:

/ x + y + z = 2
| x + y + z = 0
\ x + y + z = 1

We see immediately that the system has no solutions

Third case m = -2

We replace m by -2

/-2x + y + z = -1
| x - 2y + z = 0
\ x + y + -2z = 1

The rank of the matrix formed by the coefficients is 2.
We choose as the first and the third equation as main equations. x and y are the main unknowns.

We calculate the characteristic determinant of the side equation.

 
| -2  1  -1 |
|  1  1   1 | = 0
|  1 -2   0 |
so, the side equation may be deleted and we bring the terms in z to the right side.
 
/ -2x + y = -1 -z
\ x + y   = 1 + 2z
With Cramer's Rule, we find

x = z + 2/3
y = z + 1/3

z is arbitrary. There are infinitely many solutions.

Example 2

Take the system with parameter m
 
/(m-1)x + (2m+1)y + z = m
\ x + y + 2z = m+1
The matrix of the coefficients is
 
[ m-1    2m+1   1]
[  1      1     2]
Choose the determinant
 
| m-1 1 |
|  1  2 |
The determinant = 2m-3

First case: m is different from 1.5

We choose the matrix

 
[ m-1 1 ]
[  1  2 ]
as main matrix. The main unknowns are x and z. The side-unknown is y.
We bring the terms with y to the right side. With Cramer's Rule, we find ...
 
     m-1 -4my -y
x = ---------------
     2m-3

     m2 -m - 1 -3my + 2y
z = ----------------------
       2m-3
There are infinitely many solutions. y is arbitrary.

Second case: m = 1.5

 
/ 0.5 x + 4 y + z = 1.5
\  x + y + 2z    = 2.5
Choose y and z as main unknowns. We bring the terms with x to the right side. With Cramer's Rule, we find for each x just one solution.

Example 3

A system with parameter m
 
/ 2x -my + z = 0
| x + y + 2z = 0
\ 3x - y + z = 0
The system is homogeneous. It always has the solution (0,0,0).
Al the characteristic determinants are 0.
The determinant of the matrix of the coefficients is -5m+2.

First case: m is different from 2/5.

There is just one solution : (0,0,0).

Second case: m = 2/5

The rank of the matrix of the coefficients is now 2. We choose the main matrix as simple as possible.

 
[1  2]
[-1 1]
y and z are main unknowns. The first equation is de side equation. We delete this equation. We bring the terms in x to the right side. With Cramer's Rule, we find y = 5x/3 en z = -4x/3. We find for each x just one solution.

Example 4

A system with parameter a
 
/ ax + y = 0
| 4x - y = 1
\ (a + 2) x + 3 y = 0
This is a system with more equations than unknowns. The rank of the matrix of the coefficients. Its maximum is 2.

First case: a is different from -4.

Now the rank of the matrix of the coefficients is 2. We choose the first two equations as main equations. The characteristic determinant of the side equation is :

 
| a    1  0|
| 4   -1  1| = -2 (a - 1)
|a+2   3  0|
If a = 1 then the characteristic determinant is 0. The side equations may be omitted. We have a Cramer system. We find x=1/5 and y = -1/5

If a is not 1, then the characteristic determinant is not 0. There are no solutions.

Second case: a = -4 We substitute -4 in the system. We immediately see that there are no solutions.

Example 5

Find a and b such that the system has infinitely many solutions.
 
/ ax - y + az = 4
| -x + 3y + z = b
\ 3x + y + 5z = 1
The determinant of the matrix of the coefficients is 4(a-2).
So, a has to be 2.
We take the first equation as side equation z as side unknown. Now the characteristic determinant must be 0. Then b = -7.

Conclusion : There are infinitely many solutions for (a=2 and b=-7)

Example 6

Find the real values of a and b such that the system has at least one solution.
 
 / ax + by  = 1
 | ax +  y  = b
 \  x + by  = a
First we see that: If a = b= 1 the system is equivalent with the equation x + y = 1. There are infinitely many solutions. With each x corresponds exactly one y value. Suppose further that a and b are not both equal to 1.

The three 2 x 2 determinants that we can construct with the aid of the matrix of the coefficients are equal to a-ab ; ab-1 and ab-b.

Case 1: a.b is not equal to 1.
The rank of the matrix of the coefficients is two. We consider the second and third equation as main equations. The characteristic determinant of the side equation is :

 
  | a  1  b|
  | 1  b  a| = (b + a + 1) (1 - a) (b - 1)
  | a  b  1|
This determinant is 0 if and only if ( a = 1 of b = 1 of b = -a -1).
In these cases there is a solution to the system.

Case 2: a.b = 1 ( a not 1 )
The three 2 x 2 determinants that we can construct with the aid of the matrix of the coefficients are a-1 ; 0 ; 1-b.
As a is not equal to 1, b is different from 1. Then the rank of the matrix of the coefficients is two. We consider the first and second equation as main equations. The characteristic determinant of the side equation is :

 
  | a  b  1|
  | a  1  b| = (b + a + 1) (1 - a) (b - 1)
  | 1  b  a|
The system has a solution if and only if b = - a -1. But :
 
  b =  - a -1 en a.b = 1

 =>   a(- a -1) = 1

 =>   a2 + a + 1 = 0
There is no real value of a such that a2 + a + 1 = 0. So this second case can not occur.

Summary:
If (a = b= 1) there are an infinite number of solutions.
If (a.b not 1) the system has a solution on condition that ( a = 1 or b = 1 or b = -a -1).

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