### Video Transcript

In this video, we will learn how to
interpret augmented matrices and represent systems of linear equations as an
augmented matrix. One of the oldest general problems
in mathematics is to be able to solve a system of linear equations in multiple
variables. The simplest, nontrivial example of
this would be a system of two linear equations in two variables. This can be shown as follows.

Letβs consider the two equations π₯
plus three π¦ equals one and two π₯ minus π¦ equals three. In this case, π₯ and π¦ are the
variables to be found, with the numbers multiplying them being referred to as the
coefficients. In this case, the coefficient of
the π₯-term in the first equation is one, and the coefficient of the π¦-term is
three. For the second equation, the
coefficient of the π₯-term is two, and the coefficient of the π¦-term is negative
one. This system, with two equations and
two variables, is often referred to simply as simultaneous equations.

The methods we have seen for
solving these simultaneous equations become more complicated when we extend our
system to more equations and more variables. For example, consider the system of
three linear equations with three variables as shown. This system of linear equations
would require many more steps to solve than the first example. In order to try and prevent any
mistakes, a much neater method is used. This is called row reduction or
GAUSSβJordan elimination. This method aims to remove all
extraneous detail by first placing the coefficients from a system of linear
equations within a matrix, where each entry of this matrix corresponds to exactly
one coefficient.

Letβs consider the system of linear
equations five π₯ plus two π¦ equals two and three π₯ minus three π¦ equals six. We note that the alignment of the
π₯-terms and their coefficients appear directly underneath each other. And the same is true of the
π¦-terms. Given that this is the case, we can
write the coefficients in a two-by-two matrix as shown, where the left column
relates to the coefficient of the π₯-terms and the right column relates to the
coefficients of the π¦-terms. This works because when we multiply
this two-by-two matrix by the column matrix with elements π₯ and π¦, respectively,
the result is a two-by-two matrix with the elements five π₯ plus two π¦ and three π₯
minus three π¦ as in the original equations.

Suppose now that we wish to include
all information about the system of equations. The terms on the right-hand side of
our equations are also aligned. And we can represent this in the
matrix as shown, where the vertical bar represents the equal sign. This approach to solving equations
becomes more valuable when there is an increase either in the number of equations or
number of variables. Letβs now consider a formal
definition.

The two matrices we previously saw
are known as the coefficient and augmented matrices. We begin by considering a general
system of linear equations in the variables π₯ sub one, π₯ sub two, and so on and
the coefficients π ππ. This can be written as shown. This leads us to the following
coefficient matrix and also the augmented matrix as shown. Whilst it is outside the scope of
this video to demonstrate the full process, we can actually use these to solve
systems of equations. Letβs now look at some specific
examples.

Find the augmented matrix for the
following system of equations: π₯ plus five π¦ equals three, and three π₯ plus five
π¦ equals one.

We begin by rewriting the system of
equations by highlighting the π₯- and π¦-terms. We have two equations in two
variables, and these equations have already been ordered so that the π₯-terms appear
first followed by the π¦-terms and then the equal sign of each equation. This means that the augmented
matrix has two rows and three columns as shown. We begin by looking at the
coefficients of the π₯-terms. For the first equation, the
coefficient of the π₯-term is one. And in the second equation, the
coefficient is three. This means that the left column of
the matrix is populated with the numbers one and three. The next column is populated by the
π¦-coefficients, in this case five and five. Finally, the values on the
right-hand side of our equations, three and one, complete the matrix. Reading row by row, the completed
augmented matrix is one five, three, three, five, one. This corresponds to the system of
equations π₯ plus five π¦ equals three and three π₯ plus five π¦ equals one.

In this example, the system of
equations was written in a convenient form, with the first term of each equation
being the π₯-term followed by the π¦-term and then with an equal sign immediately
following this. Given that this was already the
case, writing out the corresponding augmented matrix was a fairly straightforward
exercise. The same is true if we are
presented with the augmented matrix and asked to write out a corresponding system of
linear equations.

Letβs now consider an example of
this type.

Find the system of equations from
the following augmented matrix: seven, two, negative seven, negative five, four,
six.

We begin by assuming that the
variables of this system are π₯ and π¦, with π₯ corresponding to the first column
and π¦ corresponding to the second column. If this is the case, this system of
linear equations will take the form shown, where the missing values will be
populated using the augmented matrix. The first column of the augmented
matrix features the coefficients seven and negative five. And these are the coefficients of
the π₯-terms in the system of equations. The middle column of the augmented
matrix contains the values two and four, which are the coefficients of the
π¦-terms. Finally, we use the right most
column of the augmented matrix to populate the remaining blank entries of the
system. These are negative seven and six,
respectively.

The augmented matrix seven, two,
negative seven, negative five, four, six corresponds to the system of equations
seven π₯ plus two π¦ equals negative seven and negative five π₯ plus four π¦ equals
six. It is important to note here that
since the example did not specifically state we should have used the variables π₯
and π¦, we could easily have used other variables, for example, π and π. Our equations would then have been
seven π plus two π equals negative seven and negative five π plus four π equals
six.

In our first two examples, we have
only worked with augmented matrices where all entries are nonzero. We have also seen the equations
written in a particularly helpful way. We will now consider an example
where neither of these are the case.

Find the augmented matrix for the
following system of equations: eight π₯ minus three π§ minus seven equals zero, six
π¦ plus three π₯ equals zero, and seven π§ minus six π¦ plus eight equals zero.

By examining the system of
equations given, we can see that this is not written in the most helpful format. In order to minimize our chance of
making a mistake when creating the augmented matrix, we begin by rewriting the
system of linear equations. We will do this by lining up the
π₯, π¦, π§, and constant terms underneath each other. Adding seven to both sides of our
first equation, we have eight π₯ minus three π§ is equal to seven. And as there is no π¦-term, we will
leave a gap here. The second equation has no
π§-term. And this can be rewritten as three
π₯ plus six π¦ equals zero. Subtracting eight from both sides
of the third equation and then rearranging the left-hand side gives us negative six
π¦ plus seven π§ is equal to negative eight. We notice that in this equation
there is no π₯-term.

Having rewritten our equations, we
see that the system has three equations in three variables. And hence, the augmented matrix is
of dimension three by four. The first column of the augmented
matrix contains the coefficients of π₯. These are eight, three, and
zero. The second column contains the
coefficients of π¦: zero, six, and negative six. Next, we have the π§-components:
negative three, zero, and seven. And finally, we enter the constants
on the right-hand side of our equations: seven, zero, and negative eight. The augmented matrix for the system
of equations eight π₯ minus three π§ minus seven equals zero, six π¦ plus three π₯
equals zero, and seven π§ minus six π¦ plus eight equals zero is eight, zero,
negative three, seven, three, six, zero, zero, zero, negative six, seven, negative
eight.

In our final example, we will look
at the reverse of this process, where once again we are given an augmented matrix
and need to find the system of equations.

From the augmented matrix two,
zero, negative nine, five, zero, four, negative nine, five, negative four, negative
nine, zero, zero, find the system of equations.

We begin by assuming that the
variables corresponding to the first, second, and third columns should be labeled as
π₯, π¦, and π§, respectively. This means that we need to populate
the missing entries in the following system of three equations in three
unknowns. The first column of our augmented
matrix corresponds to the π₯-coefficients. These are two, zero, and negative
four. The second column corresponds to
the π¦-coefficients: zero, four, and negative nine. The third column, negative nine,
negative nine, and zero are the π§-coefficients. Finally, the elements in the
right-hand column of our matrix correspond to the entries on the right-hand side of
our equations. These are five, five, and zero.

We can simplify these equations by
ignoring any term that has a coefficient of zero. As adding negative nine is the same
as subtracting nine. The first equation can be rewritten
as two π₯ minus nine π§ is equal to five. In the same way, the second
equation becomes four π¦ minus nine π§ equals five and the third equation becomes
negative four π₯ minus nine π¦ equals zero. The augmented matrix two, zero,
negative nine, five, zero, four, negative nine, five, negative four, negative nine,
zero, zero corresponds to the system of equations two π₯ minus nine π§ equals five,
four π¦ minus nine π§ equals five, and negative four π₯ minus nine π¦ equals
zero.

We have seen in this video that it
is usually a simple process to switch between a system of linear equations and the
corresponding augmented matrix. Once a system of linear equations
is written into the correct augmented matrix, then we can undertake solving the
system of equations by manipulating the augmented matrix using row operations as
part of the GaussβJordan elimination. However, as previously mentioned,
this is outside of the scope of this video. Letβs now summarize the key
points.

We saw in this video that we can
write a general system of linear equations in terms of its augmented matrix or vice
versa. We saw that for a system of linear
equations with π equations in π variables, the coefficient matrix has π rows and
π columns and, therefore, has dimension π by π. The augmented matrix of such a
system has π rows and π plus one columns, meaning it has a dimension of π by π
plus one.

When writing a system of linear
equations as an augmented matrix, it is essential that the variables appear in the
same order for each question before the augmented matrix is populated with the
coefficients of this system. One way of helping with this
process is to color the variables. If a quantity appears that is
neither a variable nor a coefficient of a variable, then it should appear on the
right-hand side of the equation before the augmented matrix is created. Finally, if a variable does not
appear in one of the equations, then the coefficient of this variable is zero and
the augmented matrix should have a value of zero in the corresponding entry.