system of linear equations by diler

System of linear equations

Fulfillment

Of

Active learning Participation in the Course

(2110015)

Prepared by:

Diler Mohmadsharif Sidi 140 150 106 099

Rajan Nitinbhai Chhatrada 140 150 106 084

Guided by:

Prof. K. K. Pokar

•Common terms related with the chapter

•Echelon form

•Reduced row echelon form

•Rank of a matrix

•Solution

•Consistent & Inconsistent system

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•Echelon form

Any matrix is said to be in echelon form if it satisfies following three properties…

1) All non-zero rows must be above the “zero” row

2) Each leading entry is in a column to the right of the leading entry in the previous row.

3) The first non-zero element in each row, called the leading entry, is 1

Reducing any matrix to echelon form by doing row transformations, we achieve a staircase shape…

It should look like following for 1)Rectangular matrix

2)Square matrix

1 1 0 4 1 3 9 7

0 1 0 1 3 0 4 1

0 0 0 1 1 1 2 2

0 0 0 0 0 0 0 1

•Common terms related with the chapter

•Echelon form

•Reduced row echelon form

•Rank of a matrix

•Solution

•Consistent & Inconsistent system

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•Reduced row echelon form

A matrix is in reduced row echelon form when it satisfies the following conditions.

1) The matrix satisfies conditions for a row echelon form.

2) The leading entry in each row is the only non-zero entry in its column.(Means rest of the elements are “0”.)

•Common terms related with the chapter

•Echelon form

•Reduced row echelon form

•Rank of a matrix

•Solution

•Consistent & Inconsistent system

Note: Please use mouse clicks to navigate through slides.

•Rank of a matrixThe maximum number of linearly independent rows in a matrix A is called the row rank of A

For any square matrix, the rank can be found very easily1)Reduce the matrix into reduced row echelon form2)Count the non-zero rows of matrix

That’s it, the number of non-zero rows in a reduced row echelon form matrix is the Rank of that matrix.

E.g.

Here, number of nonzero row is 1. So the rank of a matrix is 1.

•Common terms related with the chapter

•Echelon form

•Reduced row echelon form

•Rank of a matrix

•Solution

•Consistent & Inconsistent system

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•Solution

Solution to any linear equation system is the value of unknowns that satisfies all the equation, graphically as shown in figure its an intersection of two or more lines. Solution to a linear system can be unique, infinite or an empty set(no solution).

•Common terms related with the chapter

•Echelon form

•Reduced row echelon form

•Rank of a matrix

•Solution

•Consistent & Inconsistent system

Note: Please use mouse clicks to navigate through slides.

•Consistent & Inconsistent system

A system of equations is said to be consistent if that has at least one solution

otherwise (if it has no solution) the system is said to be inconsistent.

Consistent ConsistentInconsistent

•Click here to jump to the Theory.

•Common terms and their meaning

•Cramer’s rule for solving linear equations

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•If the system has nonzero coefficient determinate D = det (A), then the system has unique solution and this solution is of the form

X1= X2= ,…,Xn=

Cramer’s rule can be used only when the Matrix is a Square Matrix, suppose there are n equations in the number of variables X1,X2,X3,…,Xn then the solution of the system has the following cases

Where Di is the determinant obtained from D by replacing in D the ith column by the column with the entries b1,b2,…,bn

e.g.

Now, If the system has zero coefficient determinant D = det(A), then we have two possibilities as discussed below:

1. If at least one of Di is nonzero then the system has no solution.

2. If all Di’s are zero, then the system has infinite number of solutions.

If the system is homogeneous, then we have the following two possibilities of its solution.

1. If D ≠ 0, then the system has only trivial solution.

X1=0 ,X2=0,…, Xn=02. If D = 0,then the system has also non trivial solutions.

•Use Cramer’s rule to solve:

Here, A= ,X= ,b=

Here, matrix A is a square matrix of order 3,so Cramer’s rule can be appliedNow,

D=det(A)=|A|=

=1(-4-1)-2(12-1)+1(3+1)=1(-5)-2(11)+1(4)=-23

Therefore, the given system has unique solution.For finding unique solution, let us first find D1,D2 and D3.It can be easily verified that

D1= =5(-4-1)-2(24-7)+1(6+7)

=5(-5)-2(17)+1(13)=-46

D2= =1(24-7) - 5(12-1) + 1(21-6)=1(17) -5 (11) + 15=-23

D3= =1(-7-6)-2(21-6)+5(3+1)

=-13-2(15)+5(4)=-23

Therefore the unique solution of the given system is

X= = =2 , y= = =1

Outline :Square matrixNon-homogeneous matrixdet (A) ≠ 0None of Di = 0

Unique solution

D3= =1(-7-6)-2(21-6)+5(3+1)

=-13-2(15)+5(4)=-23

Therefore the unique solution of the given system is

X= = =2 , y= = =1 , z= = = 1

•Use Cramer’s rule to find the solution of the system

In matrix form, the given matrix can be written as Ax=b,Where,

A= , X= , b=

Here, matrix A is a square matrix of order 3, so Cramer’s rule can be applied.Now,

D=|A|= =1(2-12) – 2(4-6) +1(8 - 2)=1(-10) – 2(-2) + (6)=0

Therefore, either system has no solution or infinite number of solutions. Let us check for it.

D1= = 3(2 - 12) – 2(10 - 21) + 1(20 - 7)

=3(-10) –2(-11) + 1(13)=5

0

Therefore, the system has no solution as at least one Di , i=1, 2,3 (Here D1) is nonzero.

Out line:Square matrix , Non-homogeneous matrix, det (A) = 0, At least one of Di = 0 No solution

•Use Cramer's rule to solve:

Here, matrix A is a square matrix of order 3, so Cramer’s rule can be appliedNow,

D=|A|= =1(45 - 48) -2 (36 - 42) + 3(32 - 35)

=-3 -2(-6) +3(-3)=-3+12-9=0

Also, D1=

=6(45 - 48) – 2(135 - 144) + 3(120 - 120)=-18 + 18=0

Therefore, the system has infinite number of solutions.Now,

=5-8=3

Therefore,

Omitting m-r=3-2=1 equation (here, we have omitted third equation but it is not necessary), we get system as

Considering n-r =3-2=1 variable as arbitrary (here, we considered x as arbitrary but it is not necessary), the remaining system becomes

Where x is arbitrary.Now,

Therefore,

Let x=k, where k is arbitrary, then the infinite number of solutions of the given system is

Where k is an arbitrary constant.

•Solve: -2X1+ X2 - X3=0 X1+ 2X2+ 3X3=0 3X1 +X3 =0

In matrix form, the given system of equations can be written as

Where,

,

Here, matrix A is a square matrix of order 3, so Cramer’s rule can be applied.Now,

Therefore, the given system has only trivial solution that is,

•Use Cramer’s rule to solve

Solution: In matrix form, the given system of equations can be written as

Where,

Here, matrix A is of order 3, so Cramer’s rule can be applied.Now,

Therefore, the system has nontrivial solution in addition to trivial solution.Now,

Therefore,

Omitting equation, are get

Considering variable arbitrary, the remaining system becomes

Where z is arbitrary,

Now,

Therefore,

Let z=k, where k is arbitrary, then the nontrivial solution (in addition to trivial solution x=0, y=0, z=0) of the system becomes,

Where k is an arbitrary constant.

System of Equations

Linear System Non-Linear System

For a system involving two variables (x and y), each linear equationdetermines a line on the XY-plane. Because a solution to a linear system must satisfy all of the equations, the solution set is the intersection of these lines, and is hence either a line, a single point, or the empty set. And thus they can be classified differently.

System Of Linear Equations

Homogeneous System Non-Homogeneous System

•Non Linear System

• Non linear system is the one that contains ,

maximum power in its equations more than 1.

E.g. X2+Y2=R2 A Circle, has power 2. Simply Non-linear

X1Y1=1 A Hyperbola, Its power is 2. (1+1),Non-linear

X2+X-1=0 The Maximum power is 2. So, this is a non-linear equation

E.G

This is a non-linear matrix system.

•Non-Homogeneous System

A Matrix equation

AX=B

is said to be Non-homogeneous. If B is a Non-zero matrix.

E.g.

Non-homogeneous System

Non-Trivial Solution

Unique Non-Trivial Solution

Infinite Solutions

No Solution

System of equations

Unique solutions exist

Infinite solutions exist

Inconsistent system(No solution)

• Solve the system of equations:

By Gaussian elimination method.

Solution: The augmented matrix of the given solution is

Operating R12, we get

Operating R13 (-2), we get

Operating R23 (-3), we get

Which is the required row echelon form.

By going back to equations, we get

The third equation is incorrect and so the system has no solution.

•Solve the following system of equations

Solution: The augmented matrix is

By following the row operations R12(1), R13(-2), R2(-1), we get

Which is the required row echelon form for Gaussian elimination.

Let us apply Gauss Jordon method for simplification of equations.Operating R12 (2) on (i), we get

Which is reduced row echelon form.By going back to equations, we have

This is underdetermined system.Let us write it as

For various choice of free variable z, we get different solutions to the system.Let z=t, where t is any number, then

Thus infinite number of solutions exists for infinitely many choices of t.

•Use Gaussian elimination and Gauss-Jordon elimination to solve the following system of equations:

Solution: The augmented matrix of the given system is

Note To make the leftmost nonzero entry in the top row as one, the following are the various procedures for (i)•Divide the top row of (i) by -2•Doing operation R31(1) on (i)•Doing operation R21(3) on (i)•Interchanging two rowsHere, we have adopted (4)Again, operating R12(2), R13(-3), we get

Operating R2 (1/5), we get

Operating R23(6), we get

Operating R3 (-1/2), we get

Which is the row echelon form of the augmented matrix.For Gaussian Elimination method we will stop here and go back to equations.

This gives

Which implies

So, we have a unique non-trivial solution for the equation system.

•Homogeneous System• A System of linear equations is said to be Homogeneous ,

if it is in the form of, AX=0

Homogeneous system always has at least one solution

a11 a12 … a1n

a21 a22 … a2n

… … … …

am1 am2 … amn

x1

x2

…

xn

=

0

0

…

0

a11x1 + a12x2 + … + a1nxn=0a21x1 +a22x2 + … + a2nxn=0

… … … …am1x1 + am2x2 +… + amnxn=0

Homogeneous System

Only Trivial Solution

Non-Trivial Solution

Infinite Solutions

•Trivial Solution

• A solution of a set of homogeneous linear equations in which all the variables have the value zero.

and|A| ≠ 0

For a trivial solution, to any matrix,

X1=X2=X3=…=Xn=0

•Problem related with this topic.

•Non-Trivial SolutionA solution or example that is not Trivial. Often, solutions or examples involving the number zero(0) are considered trivial. Nonzero solutions or examples are considered nontrivial.

In a given homogeneous system..

If Number of Unknowns = Number of Equations

There may be many nonzero solutions in addition to the trivial solution

If the Rank of a matrix = The number unknowns,

Then the solution is a Unique Non-trivial solution.

E.g.

•Infinite Solutions

If ,Rank of a matrix< Number of Unknowns

There will be infinite solutions.

•Problem related with this topic.

•No Solution (Inconsistence)

If any equation system has no solution,It is said to be inconsistence.

It can be also explained as below.

Here 0=6. Which is not possible, so the system has no solution.

•Solve the following homogeneous system of linear equations by using Gauss Jordan method.

-2X1+ X2 - X3=0 X1+ 2X2+ 3X3=0 3X1 +X3 =0

So, the augmented matrix is

Let us reduce this into reduced row echelon form,

Here,

Therefore we have unique trivial solution.

•Solve the following homogeneous system of linear equations by using Gauss-Jordan method.

Solution The augmented matrix of the given system is

Operating R14 we get (Interchanging the row 1 and row 4)

Operating R12 (1), R14 (1), we get

Operating R2 (-1), we get

Operating R21 (2), R23 (1), R24 (2), we get

Which is reduced row echelon form.

By going back to equations, we get

Let , then

Thus, infinite number of solutions exists for infinitely many choices for k1 and k2.