System of homogeneous linear equation

Homogeneous linear equations:

a11x1 + a12x2 + .................................a1nxn=0

a21x1 + a22x2 +................................a2nxn=0

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am1x1 + am2x2 + ..........................amnxn=0

is a system of m homogeneous equation in 'n' unknown x1,x2,....................xn

Let 

A=  \begin{bmatrix} a11 &a12 &....... &a1n \\ a21&a22 &....... &a2n \\ . &. &. &. \\ am1&am2 &am3 &amn \end{bmatrix}         X=\begin{bmatrix} x1\\ x2\\ .\\ xn \end{bmatrix}           0=\begin{bmatrix} 0\\ 0\\ .\\ 0 \end{bmatrix}

 

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

  1. Reduce the cofficent matrix 'A' to Echelon form by applying row transformation only.
  2. If rank =r ,then it means that in this process of transformatons (m-r) equations are eliminated.
  3. The given system of 'm' equation will thus be replaced by an equivalent system of 'r' equations.
  4. Solving these 'r' equation, we can express the values of some 'r' unknown in terms of the remaining 'n-r' unknowns.
  5. if r=n , the zero solution will be the only solutions
  6. 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

\begin{bmatrix} 1 & 2 &3 \\ 3& 4 &4 \\ 7&10 &12 \end{bmatrix}

 

we know that,for homogeneous equation to have solution

AX=0

\begin{bmatrix} 1 & 2 &3 \\ 3& 4 &4 \\ 7&10 &12 \end{bmatrix}   \begin{bmatrix} x\\ y\\ z \end{bmatrix}    =\begin{bmatrix} 0\\ 0\\ 0 \end{bmatrix}

 

R2-->R2-3R1

R3-->R3-7R1
\begin{bmatrix} 1 & 2 &3 \\ 0& -2 &-5\\ 0&-4 &-9 \end{bmatrix}   \begin{bmatrix} x\\ y\\ z \end{bmatrix}    =\begin{bmatrix} 0\\ 0\\ 0 \end{bmatrix}

 

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

 

 

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