Mathematics of Linear Algebra

Definition of Linear Equation

noreply@blogger.com (Rashad Hassan) — Mon, 03 Nov 2008 20:49:00 +0000

Any straight line in the XY plane can be represented algebraically by an equation of the form

a1x1 + a2x2 + - - - - - - + anxn = b ----------------------- (1)

Where a1, a2 - - - - - - an and b are real constants. An equation of this form is called Linear Equation in the variables x1, x2 - - - - - - xn . The variables in a linear equation are sometimes called Unknown.

Types of Linear Equation

noreply@blogger.com (Rashad Hassan) — Mon, 03 Nov 2008 20:26:00 +0000

There are two types of linear equation such as Homogeneous Linear Equation and Non-homogeneous Linear Equation.

If b = 0 then (1) is called Homogeneous Linear Equation and if b is not equal 0 the (1) is called Non-homogeneous Linear Equation.

System of Linear Equation

noreply@blogger.com (Rashad Hassan) — Fri, 31 Oct 2008 17:13:00 +0000

A finite set of equation with the variables x1, x2- - - - -xn is called a System of Linear Equation.

a1x1 + a2x2 + - - - - - - + anxn = b ----------------------- (1)

And there are two types of linear equation such as Homogeneous Linear Equation and Non-homogeneous Linear Equation.

If b = 0 then (1) is called Homogeneous Linear Equation and if b is not equal 0 the (1) is called Non-homogeneous Linear Equation.

Solution of a System of Linear Equation by Gaussian Elimination

noreply@blogger.com (Rashad Hassan) — Fri, 31 Oct 2008 16:59:00 +0000

First Step: -

Consider the following system of m linear equations (or set of m simultaneous linear equations) in n unknown x1, x2 - - - - - - xn.

a11x1 + a12x2 + - - - - - - + a1nxn = b1

a21x1 + a22x2 + - - - - - - + a2nxn = b2

… …. …. ….. … … .. .. . .. . .. .. . .. .. .. .. . .. System (1)

. … . .. .. . .. .. . .. .. . .. . .. . .. .. . .. . .. . .. ..

am1x1 + am2x2 + - - - - - - + amnxn = bm

We reduce the System (1) to a simpler system as follows:

Step 1: -

Elimination of x1 from the second, third……..mth equations. We may assume that the order (rule) of the equations and the order (rule) of the unknowns in each equation such that a11 is not equal 0. The variables x1 can then be eliminated from the second, third ………………..mth equations by subtracting.

a21/a11 times the first equation from the second equation

a31/a11 times the first equation from the third equation

.. ………….. …….. ………… … .. .. . .. .. . .. .. .. . .. .

.. . .. . .. . .. … .. … . . .. . . . . . .. … . . . . .. . . . . . ...

am1/a11 times the first equation from the mth equation.

This gives a new system of equations of the form

a11x1 + a12x2 + - - - - - - + a1nxn = b1

c22x2 + - - - - - - + c2nxn = b’2

… …. …. ….. … … .. .. . .. . .. .. . .. .. .. .. . .. System (2)

. … . .. .. . .. .. . .. .. . .. . .. . .. .. . .. . .. . .. ..

cm2x2 + - - - - - - + cmnxn = b”m

Any solution of the system (1) is a solution of the system (2) and conversely.

Second Step: -

Elimination of x2 form the third, fourth, ………. …. mth equations in the system (2). If the co-efficient c22, c23, …… cmn, in the system (2) are not all zero. We may assume that the order (rule) of the equations and the unknowns such that c22 is not equal zero. Then we may eliminate x2 from the third, fourth, ……… mth equations of the system (2) by subtracting.

c32/c22 times the second equation from the third equation

c42/c22 times the second equation from the fourth equation

………….. ………………. ……………………

…………… ……………… ………………….

cm2/c22 times the second equation from the mth equation

The further steps are now obvious. In the fourth step we eliminate x3, in the fourth step we eliminate x4 etc.

This process will terminate only when no equations are left or when the co-efficient of all the unknowns in the remaining equations are zero. We have a system of the from.

a11x1 + a12x2 + - - - - - - + a1nxn = b1

c22x2 + - - - - - - + c2nxn = b’2

… …. …. ….. … … .. .. . .. . .. .. . .. .. .. ..

. … . .. .. . .. .. . .. .. . .. . .. . .. .. . .. . .. . ..System (3)

crsxr + - - - - - - + kmxn = br

0 = b r+1

…………

0 = bm

Where r< m. We see that there are three possible cases

(1) No solution if r <>b r+1 ………. bm is not zero.

(2) Precisely one solution if r = n and b r+1 ………. bm if present are zero.

This solution is obtained by solving the nth equation of the system (3) for xn, then the (n-1)th equation for xn-1 and so on up to the line.

(3) Infinitely many solutions if rb r+1 ………. bm if present is zero. Then any of this solutions is obtained by choosing values at pleasure for unknowns xr+1, …….. xn solving the rth equation for xr then (r-1)th equation for xr-1, and so on up to the line.

Problem (1)

noreply@blogger.com (Rashad Hassan) — Fri, 31 Oct 2008 16:55:00 +0000

Solve the Flowing homogeneous system of linear equation by Gauss-Jordan elimination

2x – y – 3z = 0

-x +2y – 3z = 0

x + y + 4z = 0

Solution: -

The given system is

2x – y – 3z = 0

-x +2y – 3z = 0 System (1)

x + y + 4z = 0

Let us represent the three linear equation of the system (1) by L1, L2, and L3 respectively. Reduce the system to echelon form by elementary operation.

Apply L1ßà L2

x + y + 4z = 0

-x +2y – 3z = 0 System (2)

2x – y – 3z = 0

Apply L2 à L2 +L1 and L3à L3 – 2L1

Thus we obtain the equivalent system is

x + y + 4z = 0

3y + z = 0 System (3)

–3 y – 11z = 0

Now apply L3à L3 + L2

Thus we obtain the equivalent system is

x + y + 4z = 0

3y + z = 0 System (4)

10z = 0

System (4) which is the echelon form of the given system (1). It has three equations with three variables. So the system has only tribal solution. The solution of the given system is

(x,y,z) = (0,0,0)

Problem (2)

noreply@blogger.com (Rashad Hassan) — Fri, 31 Oct 2008 16:52:00 +0000

Solve the Flowing homogeneous system of linear equation by Gauss-Jordan elimination

Solution: -
The given system is

2I1 – I2 + 3I3 + 4 I4 = 9

I1 - 2I3 + 7I4 = 11 System (1)

3I1 – 3I2 + I3 + 5I4 = 8

2I1 + I2 + 4I3 + 4I4 = 10

Let us represent the four linear equation of the system (1) by L1, L2, L3 and L4 respectively. Reduce the system to echelon form by elementary operation.

Apply L2 → L1 – 2L2 , L3 → 3L1 – 2L3 and L4 → L4 – L1

Thus we obtain the equivalent system is

2I1 – I2 + 3I3 + 4I4 = 9

– I2 - 7I3 - 10I4 = 13 System (2)

3I2 + 7 I3 + 2I4 = 11

2I2 + I3 + = 1

Apply L3 → 3L2 + L3 and L4 → 2L2 + L4

Thus we obtain the equivalent system is

2I1 – I2 + 3I3 + 4I4 = 9

– I2 - 7I3 - 10I4 = 13 System (3)

- 14 I3 - 28I4 = 50

- 13I3 - 20I4 = 27

Apply L4 → 13L3 – 14L4

Thus we obtain the equivalent system is

2I1 – I2 + 3I3 + 4I4 = 9

– I2 - 7I3 - 10I4 = 13 System (3)

- 14 I3 - 28I4 = 50

- 84I4 = 272

The system (3) is in echelon form. Now is has four equations with four variables. It has a unique solution.

Now in L4 of system (3)

-84I4 = 272

.: I4 = 68/21

And in L3 of system (3)

-14I3 – 272/3 =50

.: I3 = -211/21

In L2 of system (3)

– I2 - 211/3 - 680/21 = 13

.: I2 = 810/7

And in L1 of system (3)

2I1 – 810/7 -211/7 + 272/21 = 9

.: I1 = 2980/21

So the solution of the given system is I1 = 2980/21, I2 = 810/7, I3 = -211/21, I4 = 68/21.

Problem (3)

noreply@blogger.com (Rashad Hassan) — Fri, 31 Oct 2008 16:48:00 +0000

Solve the Flowing homogeneous system of linear equation by Gauss-Jordan elimination

2x1 + 3x2 + 5x3 – x4 = 8

3x1 + 4x2 + 2x3 – 3x4 = 8

x1 + 2x2 + 8x3 – x4 = 8

7x1 + 9x2 + x3 – 8x4 = 8

Solution: -

The given system is

2x1 + 3x2 + 5x3 – x4 = 8

3x1 + 4x2 + 2x3 – 3x4 = 8 System (1)

x1 + 2x2 + 8x3 – x4 = 8

7x1 + 9x2 + x3 – 8x4 = 8

Let us represent the three linear equation of the system (1) by L1, L2, L3 and L4respectively. Reduce the system to echelon form by elementary operation.

Apply L1↔L3

Then we have the equivalent system

x1 + 2x2 + 8x3 – x4 = 8

3x1 + 4x2 + 2x3 – 3x4 = 8 System (2)

2x1 + 3x2 + 5x3 – x4 = 8

7x1 + 9x2 + x3 – 8x4 = 8

Now Apply L2 → L2 - 3L1, L3→ L3 – 2L1 and L4→ L4 – 7L1.

Thus we obtain the equivalent system

x1 + 2x2 + 8x3 - x4 = 8

- 2x2 -22x3 + 6x4 = -26 System (3)

- x2 - 11x3 + 3x4 = -13

- 5x2 - 55x3 +15x4 = -56

Apply L2 → L2/2

x1 + 2x2 + 8x3 - x4 = 8

- x2 -11x3 + 3x4 = -13 System (4)

- x2 - 11x3 + 3x4 = -13

- 5x2 - 55x3 +15x4 = -56

Apply the operations L3 → L2 + L3 and L4 → 7L2 + L4

Thus we obtain the equivalent system

x1 + 2x2 + 8x3 - x4 = 8

x2 +11x3 -3x4 = 13 System (5)

0 = 0

0 = 9

The given system has been reduced to echelon form and contains an equation of the form 0 = 9 which

is not true: hence the given system is inconsistent. i.e. the system has no solution.

Problem (4)

noreply@blogger.com (Rashad Hassan) — Fri, 31 Oct 2008 16:45:00 +0000

For which value of a will be following system have no solution? Exactly one solution? Infinitely many solutions.

x + 2y- 3z = 4

3x - y + 5z = 2

4x + y +(a2 - 14)z = (a+2)

Solution: -

The given system is

x + 2y- 3z = 4

3x - y + 5z = 2 System (1)

4x + y +(a2 - 14)z = (a+2)

Let us represent the three linear equation of the system (1) by L1, L2, and L3 respectively. Reduce the system to echelon form by elementary operation.

Apply L2 → L2 - 3L1 and L3→ L3 – 4L1

Thus we obtain the equivalent system is

x + 2y- 3z = 4

- 7y + 5z = -10 System (2)

- 7y + (a2 - 2) z = (a -14)

Now apply L3→ L3 - L2

Thus we obtain the equivalent system is

x + 2y- 3z = 4

- 7y + 5z = -10 System (3)

(a2 - 42) z = (a - 4)

The system is now

x + 2y- 3z = 4

- 7y + 5z = -10 System (4)

(a + 4) (a - 4) z = (a - 4)

The system (4) is in echelon form. Now it has three equations with three variables. If a ≠ 4 and a ≠ -4 then we get the unique solution.

If a = 4 we get two equations with three free variables. So then we get more then one solution.

And if a = -4 we get degenerate linear equation and in that case we get no solution.

Problem (5)

noreply@blogger.com (Rashad Hassan) — Fri, 31 Oct 2008 16:40:00 +0000

For what value of λ, the following linear equations have a solution and solve them completely in each case:

x + y + z = 1

x + 2y + 4z = λ

x + 4y + 10z = λ2

Solution: -

The given system is

x + y + z = 1

x + 2y + 4z = λ System (1)

x + 4y + 10z = λ2

Let us represent the three linear equation of the system (1) by L1, L2, and L3 respectively. Reduce the system to echelon form by elementary operation.

Apply L2 → L2 - L1 and L3→ L3 – L1

Thus we obtain the equivalent system is

x + y + z = 1

y + 3z = λ - 1 System (2)

3y + 9z = λ2 – 1

Now apply L3→ 3L2 - L3

x + y + z = 1

y + 3z = λ - 1 System (3)

0 = λ2 – 3λ + 2

If λ2 – 3λ + 2 ≠ 0, the given system will be in echelon form having two equations with three variables. So the system has 3-2 = 1 free variable which is z. So the system has non-zero solution for λ2 – 3λ + 2 = 0

that is (λ-1)(λ-2) = 0

So, λ = 1 or λ = 2

Thus the given system is consistent for λ = 1 and λ = 2.

Case İ when λ = 1

The system (3) will be

x + y + z = 1

y + 3z = 0

Which is in echelon form where z is a free variable.

Let z = a where a is arbitrary real number.

So y = -3a and x = 1 + 2a

Hence the given system of linear equations has infinite number of solutions for λ = 1. In particular, Let, a = 1, then x = 3, y = -3 and z = 1, is a particular solution of the given system.

Case İİ when λ = 2

The system (3) will be

x + y + z = 1

y + 3z = 1

Which is in echelon form where z is a free variable.

Let z = b where b is arbitrary real number. So y = 1- 3b, x = 2b. Hence the system has infinite number of solutions for solutions for λ = 2. In particular, Let, b = -1, then x = 4, y = -2 and z = -1, is a particular solution of the given system.