# Systems of Linear Equations

Now let us eliminate the appearance of x from the third
equation.

To do this add the first equation to the third equation.

x - 2y - z = 4

y + 3z = 2

3y + 10z = 7.

The next step is to eliminate y from the third equation by
using the

second equation. To do this, take the second equation, multiply it by

-3 and add it to the third equation,

x - 2y - z = 4

y + 3z = 2

z = 1.

This completes the elimination step. Notice the characteristic upside

down staircase shape of the equations. Now we do backwards substi-

tution. Since the last equation does not involve either x or y, we read

o the value for z from the last equation, to get z = 1. Now take this

value for z and use the second equation, which does not involve x to

solve for y ,

y + 3 = 2.

This gives y = -1. Finally substitute the values for x and y to deter-

mine x,

x + 2 - 1 = 4,

to get x = 3. The solution is (x, y, z) = (3,-1, 1) ∈ R^{3}. It is easy
to

check that this is indeed a solution. Strictly speaking we should also

check that there are no other solutions. We will come to this point

later.

Let us introduce some notation, which for the time being we will

think of as just being a convenience. Instead of carrying around the

variables x , y, z etc, let us just put the information into a table. We

represent the system

x - 2y - z = 4

2x - 3y + z = 10

-x + 5y + 11z = 3.

by the** augmented matrix**

The ** coefficient matrix ** is

A is a matrix with three rows and three columns, denoted 3×
3. Note

that the rows represent equations and the columns variables. Let

b is also known as a column vector. Clearly b, as a
matrix, is 3×1.

Then the augmented matrix is formed from the two matrices A and b

If we let

then we may also write down the system of equations very
compactly

as

Av = b.

It is possible to undefirstand the previous method of solving linear equa-

tions in terms of the augmented matrix only. We start with,

The first step is to take the first row, multiply it by -2
and add it

to the second row,

This puts a zero in the second row, first column. The next
step is to

take the first row, multiply it by 1 and add it to the third row,

This puts a zero in the third row, first column. The next
step is to

take the second row, multiply it by -3 and add it to the third row,

This puts a zero in the third row, second column. At this
stage, we

think of this augmented matrix as representing a system of equations

and use the old method of back substitution to solve this system.

Let us look at a slightly more complicated example.
Suppose that

we start with a system of four equations in four unknowns ,

x + 3y - 2z - w = -1

-6x - 15y + 9z + 9w = 9

-x - z + 4w = 5

4x + 10y - 5z - 2w = -3.

We first replace this by the augmented matrix,

The first step is to use the one in the first row, first
column to eliminate

the -6, -1 and 4 from the first column. To do this we add 6, 1, -4

times the first row to the second, third and fourth row, to get

The next step is to multiply the second row by 1/3 to get
a one in the

second row, second column,

Now we eliminate the 3 and the -2 in the second column. To
do this

we add -3 and 2 times the second row to the third and fourth row,

The next step is to get a one in the third row, third
column. We cannot

do this by rescaling the third row. We can do it by swapping the third

and fourth rows,

Consider what happens when we try to solve the resulting
linear equa -

tions by back substitution. The last equation reads

0x + 0y + 0z + 0w = 2.

There are no values for x, y, z and w which work. The original system

has no solutions. Now suppose we start with the system

x + 3y - 2z - w = -1

-6x - 15y + 9z + 9w = 9

-x - z + 4w = 4

4x + 10y - 5z - 2w = -5.

The only thing we have changed is 5 to 4 = 5 - 1. If we follow the

same steps as before, we get down to the same matrix, except that the

last entry is 0 = 1-1 (remember at some point we swapped two rows)

The last equation

0x + 0y + 0z + 0w = 0,

places no restriction on x, y, z and w. The previous equation reads,

z + 4w = 1.

Using this equation, we can solve for z in terms of w, to get z = 1-4w.

We can use this value for z in the second equation to determine y in

terms of w,

y - (1 - 4w) + w = 1,

to get y = 2 - 5w. Finally, we can use this value for y and z in the

first equation to solve for x,

x + 3(2 - 5w) - 2(1 - 4w) - w = -1,

to get

x = -5 - 8w.

We get the family of solutions,

(x, y, z, w) = (-5 + 8w, 2 - 5w, 1 - 4w,w).

Note that no matter the value of w, we get a solution for the original

system of linear equations . For example if we pick w = 0, we get the

solution

(x, y, z, w) = (-5, 2, 1, 0),

but if we choose the value w = 1, we get the solution

(x, y, z, w) = (3,-3,-3, 1).

In particular this system has infinitely many solutions. As before the

reason for this is because the original system of equations was not

independent. In fact even the first three equations are not independent.

We can think of this as being the same thing as the first three rows are

not independent. The third row minus four times the first row is the

same as the second row plus the first row (we can see this by following

the steps of the Gaussian elimination). This is the same as saying the

third row is equal to the second row plus three times the first row. It is

then automatic that any solution to the first equation and the second

equations is a solution to the third equation. Note also that we can

represent the solutions in a slightly different way ,

(x, y, z, w) = (-5, 2, 1, 0) + w(8,-5,-4, 1).

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