Answer:The system could have no solution or n number of solution where n is the number of unknown in the n linear equations.
Step-by-step explanation:
To determine if solution exist or not, you test the equation for consistency.
A system is said to be consistent if the rank of a matrix (say B ) is equal to the rank of the matrix formed by adding the constant terms(in this case the zeros) as a third column to the matrix B.
Consider the following scenarios:
(1) For example:Given the matrix A=![\left[\begin{array}{ccc}1&2\\3&4\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%262%5C%5C3%264%5Cend%7Barray%7D%5Cright%5D) , to transpose A, exchange rows with columns i.e take first column as first row and second column as second row as follows:
, to transpose A, exchange rows with columns i.e take first column as first row and second column as second row as follows:
Let A transpose be B.
∵B=![\left[\begin{array}{ccc}1&3\\2&4\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%263%5C%5C2%264%5Cend%7Barray%7D%5Cright%5D)
the system Bx=0 can be represented in matrix form as:
![\left[\begin{array}{ccc}1&3\\2&4\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%263%5C%5C2%264%5Cend%7Barray%7D%5Cright%5D)
![\left[\begin{array}{ccc}x_{1} \\x_{2} \end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Dx_%7B1%7D%20%5C%5Cx_%7B2%7D%20%5Cend%7Barray%7D%5Cright%5D) =
=![\left[\begin{array}{ccc}0\\0\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D0%5C%5C0%5Cend%7Barray%7D%5Cright%5D) ................................eq(1)
 ................................eq(1)
Now, to determine the rank of B, we work the determinant of the maximum sub-square matrix of B. In this case, B is a 2 x 2 matrix, therefore, the maximum sub-square matrix of B is itself B. Hence,
|B|=(1*4)-(3*2)= 4-6 = -2 i.e, B is a non-singular matrix with rank of order (-2).
Again, adding the constant terms of equation 1(in this case zeros) as a third column to B, we have  :
:       
 =
=![\left[\begin{array}{ccc}1&3&0\\4&2&0\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%263%260%5C%5C4%262%260%5Cend%7Barray%7D%5Cright%5D) . The rank of
. The rank of  can be found by using the second column and third column pair as follows:
 can be found by using the second column and third column pair as follows:
| |=(3*0)-(0*2)=0 i.e,
|=(3*0)-(0*2)=0 i.e,  is a singular matrix with rank of order 1.
 is a singular matrix with rank of order 1.
Note: a matrix is singular if its determinant is = 0 and non-singular if it is  0.
0.
Comparing the rank of both B and  , it is obvious that
, it is obvious that
Rank of B Rank of
Rank of  since (-2)<1.
 since (-2)<1. 
Therefore, we can conclude that equation(1) is <em>inconsistent and thus has no solution.     </em>
(2) If B=![\left[\begin{array}{ccc}-4&5\\-8&10&\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-4%265%5C%5C-8%2610%26%5Cend%7Barray%7D%5Cright%5D) is the transpose of matrix A=
 is the transpose of matrix A=![\left[\begin{array}{ccc}-4&-8\\5&10\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-4%26-8%5C%5C5%2610%5Cend%7Barray%7D%5Cright%5D) , then
, then
Then the equation Bx=0 is represented as:
![\left[\begin{array}{ccc}-4&5\\-8&10&\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-4%265%5C%5C-8%2610%26%5Cend%7Barray%7D%5Cright%5D)
![\left[\begin{array}{ccc}x_{1} \\x_{2} \end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Dx_%7B1%7D%20%5C%5Cx_%7B2%7D%20%5Cend%7Barray%7D%5Cright%5D) =
=![\left[\begin{array}{ccc}0\\0\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D0%5C%5C0%5Cend%7Barray%7D%5Cright%5D) ..................................eq(2)
..................................eq(2)
|B|= (-4*10)-(5*(-8))= -40+40 = 0  i.e B has a rank of order 1.
 =
=![\left[\begin{array}{ccc}-4&5&0\\-8&10&0\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-4%265%260%5C%5C-8%2610%260%5Cend%7Barray%7D%5Cright%5D) ,
, 
| |=(5*0)-(0*10)=0-0=0   i.e
|=(5*0)-(0*10)=0-0=0   i.e  has a rank of order 1.
 has a rank of order 1.
we can therefor conclude that since 
rank B=rank  =1,  equation(2) is <em>consistent</em> and has 2 solutions for the 2 unknown (
=1,  equation(2) is <em>consistent</em> and has 2 solutions for the 2 unknown ( and
 and  ).
).
<u>Summary:</u>
- Given an equation Bx=0, transform the set of linear equations into matrix form as shown in equations(1 and 2).
- Determine the rank of both the coefficients matrix B and  which is formed by adding a column with the constant elements of the equation to the coefficient matrix. which is formed by adding a column with the constant elements of the equation to the coefficient matrix.
- If the rank of both matrix is same, then the equation is consistent and there exists n number of solutions(n is based on the number of unknown) but if they are not equal, then the equation is not consistent and there is no number of solution.