##### System of homogeneous linear equation

**Homogeneous linear equations:**

a_{11}x_{1} + a_{12}x_{2} + .................................a_{1n}x_{n}=0

a_{21}x_{1 }+ a_{22}x_{2} +................................a_{2n}x_{n}=0

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a_{m1}x_{1 }+ a_{m2}x_{2} + ..........................a_{mn}x_{n}=0

is a system of m homogeneous equation in 'n' unknown x_{1},x_{2},....................x_{n}

Let

A= X= 0=

then ** AX=0**

The number of linearly independent solutions of 'm' homogeneous linear equation in 'n' variable ,AX=0 is (n-r) where'r' is the rank of the matrix 'A'.

**Conclusion:- **

Case1:if r=n,the equation AX=0 will have n-n i.e no linearly independent solutions.

Case2:Ifr<n, we have 'n-r' linearly indpendent solutions. Any linear combination of these 'n-r' solutions willalso be a solution of AX=0

∴infinite no. of solutions possible .

__Working rule to find solutions of equations AX=0__

- Reduce the cofficent matrix 'A' to Echelon form by applying row transformation only.
- If rank =r ,then it means that in this process of transformatons (m-r) equations are eliminated.
- The given system of 'm' equation will thus be replaced by an equivalent system of 'r' equations.
- Solving these 'r' equation, we can express the values of some 'r' unknown in terms of the remaining 'n-r' unknowns.
- if r=n , the zero solution will be the only solutions
- if r<n, there will be infinite no of solutions.

Example:

x+2y+3z=0

3x+4y=4z=0

7x+10y+12z=0

Solution:

make a cofficient matrix A from given equations

we know that,for homogeneous equation to have solution

AX=0

=

R2-->R2-3R1

R3-->R3-7R1

=

now we get rank 'r' =3 and we know that no of variable is also 3.

therefor n-r=0

thereis no linearly independent solution. The only solution possible is

x=0 y=0 z=0