x  
y  
z  
u  
v  
w  
r  
s  
t 

Solutions of the linear system of equations  

x =  y =  z =  
u =  v =  w =  
r =  s =  t = 


A system of m linear equations in n unknowns has a solution if and only if the
rank r of the augmented matrix equals that of the coefficient matrix. If the two matrices have the same rank r and r = n, the solution is unique. If the two matrices have the same rank r and r < n, then at least one set of r of the unknowns can be solved in terms of the remaining (n − r) unknowns.



Another way to solve linear system of equation is by Cramer's rule which
involves only determinants. Consider the system of equations:



Example: Solve the system of linear equations of three variables x, y and z:

Consider the system of linear equations containing three variables x, y and z.
this equations cab be written in matrix notation as: One way to solve the equations is by row transformation on the augmented matrix.
Solving the above equations by Cramer' s rule.


Solve the above equations by Cramer' s rule.  
Example: Solve the system of the linear equations:

After performing rows operations, we get the upper triangular matrix:
It is seen that r = n (rank = number of variables) therefore the equations have a unique solution: z = 1 y = 3 x = 1 or in vector representation (1, 3, 1) 

Example: Solve the system of the linear equations:

The result is that r < n (r = 2 n = 4)
for example: w = a and y = b (a, b are any number) then:
for example if we choose: a = 1 b = − 1 then the solution is: (7, − 1, 3, 1). 

Example: Solve the system of the linear equations:

From the last row it is observed that: 0 = 8 certainly, this is not possible, hence this set of equations don't have a solution. 


Solve the following system of the linear equations by the methods a) Cramer's rule, b) by row transformation, c) by variable elimination.


a) Solution by Cremer's rule And the solution is: b) Solution by row transformation
Now it is clear that 64z = 192 and z = 192/64 = 3 For the second row we have: −5y + 4z = 17 and y = (17 − 4z)/−5 = (17 − 4 · 3)/−5 = −1 From the first row we have 2x − y + 2z = 11 and x = (11 + y − 2z)/2 = (11 − 1 − 2 · 3)/2 = 2 And we get the solution as before: x = 2 y = −1 z = 3 c) Solution by elimination From the first equation we find the value of y:
And again, the solution is as before: x = 2 y = −1 z = 3 


Solve the following system of the linear equations, investigate the answers. 

To make solution easier we will subtract 3rd row from the 2nd row to get the following system of equations By Cremer's rule we have the following determinants:
We can see that for valid solution we have the condition: b ≠ 0 and a ≠ 1
Now we can clearly see that when a = 1 then the 1st and 3rd equations are the same. And we get only two equations for three variables as follows We can now subtract 2nd row from the 1st row to get the set of equations: From the second row we immediately see that the value of y is: y = 2 and we get from the first row the solution x = 2 − z and we see that there is infinite number of solutions. 


Solve the following system of linear equations.


The augmented matrix (including free term) of this set of equations is: by row transformation method we get:
From the result we see that the rank of the coefficient matrix (3 columns from the left  variable matrix) is R_{c} = 2 while the rank of the augmented matrix is R_{g} = 3 therefore there is no solution to this problem. We can summarize the rules of the existence of solutions by the following facts:



Solve the following homogeneous system of linear equations.


It is obvious that the trivial solution is: x = y = z = 0, we can say that every system of linear equations have that solution. Notice that the rank of the coefficients and the augmented matrix are the same.
The rank of the coefficients and the augmented matrices are the same and equal to, R_{c} = R_{g} = 2 Because the number of variables is 3 then R is less then n (2 < 3) and we have a non zero solution which depends on the value of the third variable that can get infinite number of values: y = −z and x = −z for example if z = −1 then the result is (1, 1, −1) 




We will find the determinant value D by the diagonal method: Copy the first two columns to the end of the matrix and the result is the multiply of the 3 values alone the line so that the multiplication D = 1· 1 · (− 2) + 0 · 0 · 0 + 2 · 1 · 1 − 2 · 1 · 0 − 1 · 0 · 1 − 0 · 1 · (− 3) = 0 The determinant is equal to 0 so we have at least two identical rows.
Because R_{c} = R_{g} = 2 and number of variables is n = 3 < R_{g} then 2 variables can be expressed by the third variable, and we get infinite number of solutions:
Notice that the solution of x and y are a function of any value of z. 