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Linear Homogeneous and Non-Homogeneous Equation

Lecture on Linear Equation
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The system of linear equation, Homogeneous Equation, Non-homogeneous Equation and it's solution.

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Lecture on Linear Independence
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The concept of linear independence.

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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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System of linear non homogeneous equations

Non Homogeneous Equations

a1x+b1y=c1

a2x+b2y=c2

If there is n equation & n unknown you will not find the unique solution always.

eg. x+y=1

       x+y=10 there is no solution for these equation.

Working rule for finding the solution of linear non homogeneous equation (AX=B)

  1. Suppose the coefficient matrix 'A' is of type M*N
  2. Write the augmented matrix [AB] and reduce to echlon form by applying only row transformations.
  3. This echelon form will enable us to know the ranks of the augmented matrix [AB] and the coefficient matrix A. Then the following different cases arise

Case1:  Rank A < Rank [AB] ⇒AX=B is inconsistant   ⇒ No solution

 

Case2: Rank A = Rank [AB]=r(say)

a)  r=n  ⇒ unique solution

b) r<n   ⇒ n-r linearly independnt solution

                ⇒n-r variable will be assigned random values 

                ⇒infinite no of  solutions

Example:

x+y+z=9

2x+5y+7z=52

2x+y-z=0

 

Solution:

[AB] = \begin{bmatrix} 1 & 1 &1 &: & 9\\ 2& 5 &7 & : & 52\\ 2& 1 &-1 & : & 0 \end{bmatrix}

 

R2-->R2-2R1

R3-->R3-2R1

\begin{bmatrix} 1 & 1 &1 &: & 9\\ 0& 3 &5 & : & 34\\ 0& -1 &-3 & : & 18 \end{bmatrix}

R2⇔R3

\begin{bmatrix} 1 & 1 &1 &: & 9\\ 0& -1 &-3 & : & 18\\ 0& 3 &5 & : & 34\\ \end{bmatrix}

 

 

R3-->R3+3R2

\begin{bmatrix} 1 & 1 &1 &: & 9\\ 0& -1 &-3 & : & 18\\ 0& 0 &-4 & : & -20\\ \end{bmatrix}

 

r(A)=r(AB)=3

n=3

n=r

therefor ,unique solution

x+y+z=9

-y-3z=-18

-4z=-20

z=5 , y=3, x=1

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