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# Inconsistent Systems - PowerPoint PPT Presentation

Inconsistent Systems. By Dr. Julia Arnold. 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.

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Presentation Transcript

### Inconsistent Systems

By

Dr. Julia Arnold

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.

-x + y = -22 or has infinitely many solutions.

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 or has infinitely many solutions.

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.

Infinitely many solutions: row, this makes the equation 0 = 1 which is a false statement. 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 remaining equations are -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? remaining equations are -y +3z = -2 and x + y - 5z = 3

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.

The two equations remaining: remaining equations are -y +3z = -2 and x + y - 5z = 3

-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.