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## PowerPoint Slideshow about ' Inconsistent Systems' - colby-parsons

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An inconsistent system is one which has either no solutions, or has infinitely many solutions.

Example of No solutions:

-x + y = -22

3x + 4y = 4

4x - 8y = 32

In this example we have only two unknowns but three equations. To have a solution the third equation must intersect the first two in the exact same location.

The graph of each equation would be a straight line. As you can see, the three do not share a common intersection point.

3x + 4y = 4

4x - 8y = 32

Let’s multiply equation 1 by -1 to get a +1 in the first position and then write the augmented matrix.

x - y = 22

3x + 4y = 4

4x - 8y = 32

-3R1 +R2 = -3 +3 -66

3 +4 +4

0 7 -62

-4R1 + R3 = -4 +4 -88

4 -8 32

0 -4 -56

This final matrix says y = 14 and it says y = -8.85

which is not possible. Thus no solutions.

By multiplying the last row by -1 and adding to row 2, our last row says 0 = -22.857 which is a false statement and hence no solutions.

Having too many equations can be a problem but having the wrong equations can also be a problem.

Example 2 of No solutions:

x + y + z = 3

2x - y - z = 5

2x + 2y + 2z = 7

Set up the augmented matrix and multiply row 1 by -2

-2 -2 -2 -6 then add row 2

2 -1 -1 5

0 -3 -3 -1

Next multiply row 1 by -2 again and add row 3

-2 -2 -2 -6 add row 3

2 2 2 7

0 0 0 1

Since the coefficients of x, y and z are all 0 in the last row, this makes the equation 0 = 1 which is a false statement. Hence No Solutions.

Graphically in 3D,

this would imply

parallel planes cut by the transversal plane.

i.e. no common intersection and hence no solutions.

Too many equations can cause no solutions, but too few equations can cause infinitely many solutions as shown in the next example.

x + y - 5z =3

x - 2z = 1

Form the augmented matrix and multiply R1 by -1 and add row 2.

-1 -1 +5 -3

1 0 -2 1

0 -1 3 -2

Since we have done all we can, we must conclude that the remaining equations are -y +3z = -2 and x + y - 5z = 3

Graphically we know that the intersection of two planes is a straight line.

-y +3z = -2 and x + y - 5z = 3

How can we turn this into an answer?

We decide that one variable (in this case, either y or z) will be a free variable in that it can take on all values in the Reals.

If we decide it should be z then we write the other variables in terms of z as follows

-y + 3z = -2

3z = y - 2

3z + 2 = y or y = 3z + 2

Now we write x in terms of z:

x + (3z + 2) - 5z = 3

x - 2z +2 = 3

x = 2z + 1

The solution is z = any real number

y = 3z + 2 and x = 2z +1 and hence infinitely many solutions.

Remember this example from no solutions?

x + y + z = 3

2x - y - z = 5

2x + 2y + 2z = 7

By changing one number we can go to infinitely many solutions.

x + y + z = 3

2x - y - z = 5

2x + 2y + 2z = 6

By changing the 7 to a 6 I have created a duplicate equation to equation 1. (Multiply equation 1 by 2)

As we work on the matrix

we would get the final row to

be all zeroes which means

0x + 0y + 0z = 0 which is true.

Thus infinitely many solutions.

-3y - 3z = -1 and x + y + z = 3

should be treated as in the last example.

Let z be any real number

(1 - 3z)/3 = y or y = (1 - 3z)/3

x + (1-3z)/3 + z = 3

3x + 1 - 3z + 3z = 9

3x = 8

x =8/3

(8/3, (1-3z)/3,z) form the points of this solution.

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