### Video Transcript

In this video, we will learn how to
represent a system of two equations as a matrix equation. We will only be dealing with
matrices up to two by two. However, the method can be extended
for matrices of higher order.

A system of linear equations can be
represented in matrix form using a coefficient matrix, a variable matrix, and a
constant matrix. If we consider the two equations
ππ₯ plus ππ¦ is equal to π and ππ₯ plus ππ¦ is equal to π, where π, π, π,
π, π, and π are all constants, then the coefficient matrix is π, π, π, π. This is formed by aligning the
coefficients of the variables of each equation in a row. It is important that each equation
is written in standard form with the constant term on the right.

We have two variables π₯ and
π¦. Therefore, the variable matrix can
be written as shown. Whilst we will only deal with two
variables in this video, this can be extended to multivariable equations. On the right-hand side of our
equations, we have the constant terms π and π. These correspond to the first and
second equations, respectively. Therefore, the constant matrix is
π, π. The order of our three matrices are
two by two, two by one, and two by one, respectively.

The system of linear equations ππ₯
plus ππ¦ is equal to π and ππ₯ plus ππ¦ is equal to π can be rewritten in
matrix form as π, π, π, π multiplied by π₯, π¦ is equal to π, π. We can use matrix multiplication to
see that the matrix representation is equivalent to the system of equations. As previously mentioned, we could
extend this to a system of three equations in three variables and actually
generalize the result to π variables. Letβs now consider some specific
examples.

Express the simultaneous equations
three π plus two π equals 13 and two π plus three π equal seven as a matrix
equation.

We recall that we can represent a
system of linear equations in matrix form using a coefficient matrix, a variable
matrix, and a constant matrix. Before doing this, we need to
ensure that our equations are written in standard form, ππ₯ plus ππ¦ is equal to
π, where π, π, and π are constants. In this question, both of our
equations are written in standard form. However, it is important to note
that π and π are variables and not constants.

We know that the coefficient matrix
can be formed by aligning the coefficients of the variables of each equation in a
row. The coefficients of our first
equation are three and two. In the second equation, we have two
and three, giving us the two-by-two coefficient matrix three, two, two, three. The variables here are π and
π. So we can write the variable matrix
as the two-by-one matrix π, π. On the right-hand side, we have the
constant terms 13 and seven. As these correspond to the first
and second equation, respectively, the constant matrix is 13, seven.

We now have a matrix equation made
up of a coefficient matrix, a variable matrix, and a constant matrix. The simultaneous equations three π
plus two π is equal to 13 and two π plus three π is equal to seven can be
rewritten as a matrix equation three, two, two, three multiplied by π, π is equal
to 13, seven.

In our next example, we will begin
by rearranging one of the equations.

Express the simultaneous equations
one-third π₯ minus two-thirds π¦ is equal to five-thirds and three-quarters π¦ plus
one-quarter π₯ is equal to seven-quarters as a matrix equation. In order to rewrite a system of
linear equations as a matrix equation, we need to find a coefficient matrix, a
variable matrix, and a constant matrix. Before starting, however, we need
to ensure that all of our equations are written in standard form.

As the π₯-term comes first in our
first equation, we can rewrite the second equation as a quarter π₯ plus
three-quarters π¦ is equal to seven-quarters. This is because addition is
commutative. The coefficients of our first
equation are one-third and negative two-thirds, whilst the coefficients of our
second equation are one-quarter and three-quarters. This means that the two-by-two
coefficient matrix is one-third, negative two-thirds, one-quarter, three
quarters. The two variables here are π₯ and
π¦. Therefore, the variable matrix is
simply π₯, π¦.

On the right-hand side of our
equations, we have the constant terms five-thirds and seven-quarters. These make up the constant
matrix. We now have a matrix equation made
up of a coefficient matrix, a variable matrix, and a constant matrix as
required. The two simultaneous equations
expressed as a matrix equation is one-third, negative two-thirds, one-quarter,
three-quarters multiplied by π₯, π¦ is equal to five-thirds, seven-quarters.

In our next example, we will need
to write a matrix equation as a set of simultaneous linear equations.

Write down the set of simultaneous
equations that could be solved using the given matrix equation. Three, three, two, four multiplied
by π, π is equal to 10, 12.

In order to answer this question,
we need to perform matrix multiplication. We begin by multiplying the
elements in the first row of the coefficient matrix by the elements in the column
variable matrix. Three multiplied by π is equal to
three π, and three multiplied by π is equal to three π. The sum of these terms will be
equal to the element in the first row of the constant matrix. This gives us the equation three π
plus three π is equal to 10.

We then repeat this process for the
second row of the coefficient matrix. Two multiplied by π is two π, and
four multiplied by π is four π. This gives us the equation two π
plus four π is equal to 12. We now have a set of simultaneous
equations that could be solved. Three π plus three π is equal to
10, and two π plus four π equals 12. Whilst we do not need to solve the
equations in this video, we could do so using the elimination or substitution
methods.

We will now try and solve a similar
problem where one of our coefficients is negative.

Write down the set of simultaneous
equations that could be solved using the given matrix equation. 11, negative three, nine, four
multiplied by π₯, π¦ is equal to eight, 13.

As with any problem of this type,
we can solve it using matrix multiplication. When multiplying matrices, we need
to multiply each row of the first matrix by each column of the second matrix. Multiplying 11 by π₯ gives us
11π₯. Negative three multiplied by π¦ is
equal to negative three π¦. This will be equal to the element
in the top row of our constant matrix, in this case, eight. Our first equation is 11π₯ minus
three π¦ is equal to eight.

We then repeat this process with
the second row of our two-by-two coefficient matrix. Multiplying nine by π₯ gives us
nine π₯. Four multiplied by π¦ is four
π¦. Our second equation is therefore
nine π₯ plus four π¦ is equal to 13. We now have a pair of linear
simultaneous equations that could be solved using the elimination or substitution
methods. These would give us the values of
π₯ and π¦ that solve the matrix equation.

In our final question, we will
begin by rearranging our equations so theyβre in standard form.

Express the simultaneous equations
three π₯ minus 24 is equal to negative eight π¦ and π₯ is equal to three minus π¦ as
a matrix equation.

In order to answer this question,
we need to ensure that both of our equations are written in standard form. We need to rewrite the first
equation so it is in the form ππ₯ plus ππ¦ is equal to π and the second equation
so it is in the form ππ₯ plus ππ¦ is equal to π, where π, π, π, π as well as
π and π are constants.

Letβs begin with the equation three
π₯ minus 24 is equal to negative eight π¦. We can add 24 and eight π¦ to both
sides of our equation. This gives us three π₯ plus eight
π¦ is equal to 24. This equation has now been written
in standard form. Our second equation stated that π₯
is equal to three minus π¦. Adding π¦ to both sides of this
equation and ensuring that our π₯- and π¦-terms are in the same order gives us π₯
plus π¦ is equal to three.

We now have a pair of linear
simultaneous equations written in standard form. Our matrix equation will consist of
a two-by-two coefficient matrix. The coefficients of π₯ and π¦ in
our first equation are three and eight. These will make up the top row. The coefficients of π₯ and π¦ in
our second equation are one and one. We therefore have the two-by-two
coefficient matrix three, eight, one, one.

Our variables are π₯ and π¦. This means that the coefficient
matrix can be multiplied by this column variable matrix. On the right-hand side of our two
equations, we have the constants 24 and three. Our simultaneous equations can be
expressed as the matrix equation three, eight, one, one multiplied by π₯, π¦ is
equal to 24, three.

We will know summarize the key
points from this video. We saw in this video that a matrix
equation consists of a coefficient matrix, a variable matrix, and a constant
matrix. We saw that we can express a system
of linear simultaneous equations as a matrix equation and vice versa. The linear simultaneous equations
ππ₯ plus ππ¦ equals π and ππ₯ plus ππ¦ equals π can be rewritten as the matrix
equation π, π, π, π multiplied by π₯, π¦ is equal to π, π. As previously mentioned, this can
be extended to equations with three or more variables, matrices larger than two by
two, and also help us solve the simultaneous equations.