Alcances del derecho a la Libertad Cultural; de acuerdo a la Corte Inter

There is an ideal form for the rearranged section of the augmented matrix to the left of the line. We again begin by looking at the simple example represented by the system we just examined.

 3 2 4

We stopped at this point because the second row represented an almost direct readout for the variable x. We could have gone further

R2=R2÷7

−→

 3 2 4

1 0 2



as the second row now really does give a direct readout for x = 2. However, we could have continued again

R1=R1−3R2

−→

 0 2 −2

1 0 2



which now means the first row gives 2y = −2. We could take two more steps.

R1=R1÷2

6.9 Solving systems of equations 112

The second step is just to make it very clear that the left closely resembles the identity matrix. If we now examine this set of equations we see it is simply

0x + y = −1; 1x + 0y = 2 or

x = −1, y = 2.

Therefore the most ideal solution might be to rearrange the left side of the augmented matrix to the identity matrix, whereupon the right hand side becomes the solutions for the variables. To accomplish this we attempt to produce zeros in the left hand column everywhere except the top entry, and then we try to produce zeros everywhere in the next column except for the second entry and so on.

There are two reasons why we do not always do this:

• the left hand side may not be square, such as when we have fewer equations than unknowns;

• the extra operations aren’t really necessary to solve the equations.

In practice therefore we try to reduce the cells to the bottom and left of the leading diagonal to zero. It is nice, but not essential if the first number on each row is a 1. This will allow us to begin with the bottom row and solve for one unknown and then we move upwards through the row substituting the known variables to find each next unknown.

This process is known as Gaussian Elimination.

6.9.2 Example

Use Gaussian elimination to solve the following system of equations:

x + 2y + z = 1

4x + 2y − 3z = 2

2x + 3y + z = 3

Solution

The augmented matrix for this system is as follows

AB =





1 2 1 1

4 2 −3 2

2 3 1 3





6.9 Solving systems of equations 113

and we now begin working to produce zeros in the first column, except for the top number. We use the top number to let us produce the zeros. In this case the top number is 1 which is very convenient. If that had occurred in another row an initial swap might have been a good idea.

R2=R2−4R1 so that’s a great start. For our purposes row 1 is fine as it is now, and we won’t touch it in subsequent work. Now we work only with rows 2 and 3.

We now want to introduce a zero at the bottom of column 2, to continue our job of zeroing the elements to the bottom and left of the leading diagonal.

Now we could add a sixth of row 2 to row 3 to achieve this, but it’ll be easier to swap rows 2 and 3 first. I have also taken the liberty to multiply through row 2 by −1 as it makes our life a little easier. It’s much easier to make the bottom of column 2 a zero now, and doesn’t introduce fractions so early. If we simply add row 2 to row 3 six times. and we need not go further. Row 3 now gives

−z = −8 ⇒ z = 8 and Row 2 now gives

y + z = −1 ⇒ y + 8 = −1 ⇒ y = −9 and Row 1 now gives

x + 2y + z = 1 ⇒ x − 18 + 8 = 1 ⇒ x = 11 and so we have a unique solution

x = 11, y = −9, z = 8.

6.9 Solving systems of equations 114

6.9.3 Example

Use Gaussian elimination to solve the following system of equations:

3x − 2y + z = 1

2x + 3y + z = 2

4x − 7y + z = 3

Solution

The augmented matrix for this system is

A_B =

We begin by producing a 1 somewhere in the first column, in order to make subsequent work more simple. This isn’t necessary, but just easier.

R2=R2÷2

This zero can now be used to produce an elimination of the other numbers in the column. Now we would next work to produce a zero at the bottom of the right hand column. To reduce the fractions, let’s multiply row 2 by 2.

R2=R2×2

Now we can create a zero in the bottom of column 2 by subtracting row 2 from row 3:

Now actually this represents a problem. The bottom row is now equivalent to the equation

0x + 0y + 0z = 3 ⇒ 0 = 3

6.9 Solving systems of equations 115

and this is clearly impossible. This indicates that our system of equations is inconsistent and there is in fact no solution. A quick calculation of the determinant of the coefficient matrix will reveal it to be zero.

6.9.4 Example

Use Gaussian elimination to solve the following system of equations:

3x − 2y + z = 1

2x + 3y + z = 2

4x − 7y + z = 0

Solution

You may note this is extremely similar to the previous example. Again the inverse of the coefficient matrix is zero. This means there is no inverse, but critically it does not mean there are no solutions in itself, it just means there is not a unique solution.

The augmented matrix for this system is

A_B =

We begin by producing a 1 somewhere in the first column, in order to make subsequent work more simple. This isn’t necessary, but just easier.

R2=R2÷2

This zero can now be used to produce an elimination of the other numbers in the column. Now we would next work to produce a zero at the bottom of the right hand column. To reduce the fractions, let’s multiply row 2 by 2.

R2=R2×2

6.9 Solving systems of equations 116

Now we can create a zero in the bottom of column 2 by subtracting row 2 from row 3:

Compare this with example 6.9.2 at this point. It might be tempting to think this is the same situation, but it is not. This time the bottom row is

0x + 0y + 0z = 0 ⇒ 0 = 0

which is true. Therefore there is no inconsistency, it’s just that one equation out of the original three is essentially worthless, it’s a copy of the original two. (If you look you will see that R3 = 2R1 − R2 in the original system).

This means there will be no unique solution. In this case we proceed as follows. There is no constraint on z, so z can take any value, say t for example, where t is any real number.

Then row 2 gives (multiplying by −1 first for clarity) 13y + z = 4 ⇒ 13y = 4 − t ⇒ y = 1

represents all the (infinite!) solutions of this problem. This is called a para-metric solution since the solution depends on one or more parameters (in this case t), and as t takes all the values a real number can, we obtain all the solutions of this system.