Video Transcript
In this video, we will learn how
to use Cramer’s rule to solve a system of linear equations. So what we’re gonna look to do is
use determinants to solve systems of two linear equations, use determinants to solve
systems of three linear equations, and understand and use Cramer’s rule.
So Cramer’s rule is a useful way of
helping us to solve simultaneous equations. And one handy thing about it is
that it allows us to solve for one variable, not a whole system of equations. So how did it come about? Well, Cramer’s rule is named after
Gabriel Cramer. He was a Genevan mathematician. And what he devised was a way of
solving simultaneous equation using matrices or matrix equations and actually
determinants of these matrices. Now, before we have a look at some
examples of how to use Cramer’s rule, we’re just gonna have a little look at the
rule itself and how it works and what it means.
So it’s also worth noting at this
point that often different notation can be seen and used. So on this page, you can even see
that we’ve got in the actual Cramer’s rule in the bubble, we have the notation which
uses Δ for our matrix so it says the determinant of the matrix Δ sub 𝑥, whereas
also you can write this as D sub 𝑥. This means the determinant of the
matrix.
So if we take a look at the rule,
then what Cramer’s rule tells us is that if we have a system of equations — so in
this case we’ve got three variables, 𝑥, 𝑦, and 𝑧, but we’re gonna look at two and
three variables in this lesson — then we can find the solutions to the equations
using 𝑥 is equal to the determinant of the matrix Δ sub 𝑥 over the determinant of
the matrix Δ. 𝑦 is equal to the determinant of
the matrix Δ sub 𝑦 over the determinant of the matrix Δ. And 𝑧 is equal to the determinant
of the matrix Δ sub 𝑧 over the determinant of the matrix Δ. So this is all great. But what does this actually
mean? Well let’s have a look.
Well, let’s imagine we had a system
of two equations and the variables were 𝑥 and 𝑦. So we’ve got three 𝑥 plus two 𝑦
equals 23 and two 𝑥 minus four 𝑦 equals negative 22. Well, then, what we could do is, in
fact, set this up as a matrix equation. So we’d have the matrix three, two,
two, and negative four. And this is the matrix of our
coefficients, with our 𝑥-coefficients being the first column and our
𝑦-coefficients being the second column. They’re multiplied by the matrix
𝑥, 𝑦, which is our variables. And then this would be equal to an
answer matrix 23, negative 22. And we get these because these are
our constant values or our answers to our equations. Okay, great. So we’re now at this stage, still
not quite at our rule. So what do we need to you now?
Well, what we could imagine is that
the matrix is matrix Δ. So we’ve got a coefficient matrix
here. So therefore, we’d know as the
denominator, if we wanted to find our 𝑥-variable, we’d have the determinant of the
matrix three, two, two, negative four. Okay, so this makes sense. But what’s our numerator going to
be? What is this matrix Δ sub 𝑥? Well, the matrix Δ sub 𝑥 is what
we would get if we substitute our answers, so the values from our answer matrix,
instead of the column, which contains our 𝑥-coefficients, which would give us the
matrix 23, two, negative 22, negative four cause we can see that the 𝑦-coefficients
would stay the same. So therefore, what we could say is
that our 𝑥-solution would be equal to the determinant of the matrix 23, two,
negative 22, negative four over the determinant of the matrix three, two, two,
negative four.
And what we could do then is apply
Cramer’s rule to find the 𝑦-solution as well. And we’ll do that in some of the
examples we’re gonna come onto. This is just to show how it all
works. Now, before we get straight into
some examples, obviously here we’re talking a lot about determinants. What I want to do is quickly run
through how to find the determinant of two-by-two and three-by-three matrices. This is something you should
already know, so this is just gonna be a very quick recap.
Well, first of all, if we think
about a two-by-two matrix, if we have the two-by-two matrix 𝑎, 𝑏, 𝑐, 𝑑, then the
determinant of this is gonna be equal to 𝑎𝑑 minus 𝑏𝑐. So what we do is cross multiply and
subtract. And then for a three-by-three
matrix, if we wanna find the determinant, for instance, of the matrix 𝑎, 𝑏, 𝑐,
𝑑, 𝑒, 𝑓, 𝑔, ℎ, 𝑖, this is equal to 𝑎 multiplied by the determinant of the
submatrix 𝑒, 𝑓, ℎ, 𝑖 minus 𝑏 multiplied by the determinant of the submatrix 𝑑,
𝑓, 𝑔, 𝑖 plus 𝑐 multiplied by the determinant of the submatrix 𝑑, 𝑒, 𝑔, ℎ.
So an important key point,
remember, is that we’ve got positive, negative, positive when we’re looking at the
coefficients before we multiply the determinants of our submatrices. And also we’re gonna quickly remind
ourselves how we find the two-by-two submatrix. So if we take the element 𝑎, then
we delete the column and row that it’s in. Then what we’re left with is the
two-by-two submatrix 𝑒, 𝑓, ℎ, 𝑖. So now, we’ve recapped those and
we’ve also looked at how to use Cramer’s rule. We’re now gonna have a look at some
examples. And in our first example, what
we’re gonna be taking a look at is one of the conditions for Cramer’s rule.
Is Cramer’s rule useful for finding
solutions to systems of linear equations in which there is an infinite set of
solutions?
Well, we could answer this question
very quickly cause we could just say no, because Cramer’s rule is not useful for
when there is a system of linear equations where there is an infinite set of
solutions. And that’s because if we are making
a matrix equation, then it is not viable when the matrix is singular. And a matrix is singular when there
is an infinite set of solutions. But that’s a bit of a short
answer. Let’s have a look to why this is
the case. So, as we’ve already said, if a
system of equations has a singular matrix, then there is a solution set with an
infinite set of solutions. But how do we know or what are the
properties of a singular matrix?
Well, what we know about a singular
matrix is that it has a determinant equal to zero. So then, if we take a look at
Cramer’s rule and we want to define one of the three variables 𝑥, 𝑦, or 𝑧, then
what we can see is that in the rule we’ve got the determinant of the matrix as the
denominator. Well, if it was a singular matrix,
this would be zero. And if we have anything divided by
zero, then that means that our answers cannot be defined. So this is why we can say that no,
Cramer’s rule would not be useful for finding solutions to systems of linear
equations in which there is an infinite set of solutions.
Okay, so, great. We’ve looked at the conditions for
which Cramer’s rule can be applied. So now let’s take a look at some
examples of how we use Cramer’s rule.
Use determinants to solve the
system negative eight 𝑥 minus four 𝑦 equals negative eight and nine 𝑥 minus six
𝑦 equals negative nine.
So the first thing we want to do
with this problem is actually set up a matrix equation of our system of
equations. And when we do that, what we’re
gonna have is the matrix negative eight, negative four, nine, negative six
multiplied by the matrix 𝑥, 𝑦 is equal to the answer matrix and this is negative
eight, negative nine. So now what we’re gonna do, because
we want to use determinants to solve the system, then what we’re gonna do is use
Cramer’s rule. And Cramer’s rule tells us that 𝑥
is equal to the determinant of the matrix Δ sub 𝑥 over the determinant of the
matrix Δ. And then to find 𝑦, it’s equal to
the determinant of the matrix Δ sub 𝑦 over the determinant of the matrix Δ.
But we might look at this and
think, “Well, what is the matrix Δ sub 𝑥?” Well, in fact, what this is is the
matrix that’s formed when we substitute the answer matrix for the column of
𝑥-coefficients in our original matrix. So, for example, in our problem, we
swap the first column in our matrix for the answer matrix. So instead of reading negative
eight and then nine, it read negative eight and then negative nine. So now let’s move straight on and
find our determinants. So first of all, we want the
determinant of the matrix negative eight, negative four, nine, negative six. So when we work this out, we’re
gonna get negative eight multiplied by negative six minus negative four multiplied
by nine, which is gonna give us 48 plus 36. And this is equal to 84.
And a good thing about this is that
it also helps us to check that we can solve our system of equations because if our
matrix was singular, then the determinant will be equal to zero. So we can see that it isn’t the
case in this problem here. So next, what we’re gonna have a
look at is the determinant of the matrix Δ sub 𝑥. Well, what we already said here is
what Δ sub 𝑥 is, is the matrix we get when we substitute in negative eight and
negative nine, our answer matrix, instead of our 𝑥-coefficients. So we’re gonna get the matrix
negative eight, negative four, negative nine, negative six. So for this determinant, what we’re
gonna get is negative eight multiplied by negative six minus negative four
multiplied by negative nine, which is gonna be equal to 12. Okay, great. So one more determinant to work
out.
So now what we’re looking for is
the determinant of the matrix Δ sub 𝑦. What this is gonna be equal to is
the determinant of the matrix negative eight, negative eight, nine, negative
nine. And as is before, what we’ve got
this by is substituting in our answer matrix for the coefficients of 𝑦. So this is gonna be equal to
negative eight multiplied by negative nine minus negative eight multiplied by nine,
which is gonna be equal to 144. Okay, great. So now we’ve got everything we need
to use Cramer’s rule to solve our system of equations. So, using Cramer’s rule, what we’re
gonna get is 𝑥 is equal to 12 over 84. But then what we can do is divide
the numerator and denominator by 12. And when we do that, we get 𝑥 is
equal to one over seven.
Okay, great. We found the solution for 𝑥. Now let’s move on to 𝑦. So once again, using Cramer’s rule,
we’re gonna get 𝑦 is equal to and we’ve got the determinant of the matrix Δ sub 𝑦
over the determinant of the matrix Δ, so this is gonna give us 144 over 84. So once again, what we can do is
simplify our fraction. We could do that by dividing the
numerator and denominator both by 12. And when we do that, we get 𝑦 is
equal to 12 over seven or twelve-sevenths. So therefore, we can say the
solutions to our equation are 𝑥 equals a seventh and 𝑦 equals twelve-sevenths.
Okay, so, great. We’ve looked at an example of how
to solve a system of two equations. So now, what we’re gonna do is take
a look at a system of three equations with three variables 𝑥, 𝑦, and 𝑧.
Use determinants to solve the
system five 𝑥 equals negative two 𝑦 minus five plus three 𝑧, negative three 𝑥
minus 𝑦 plus one equals two 𝑧, and two 𝑦 minus 𝑧 equals negative five 𝑥 plus
three.
So, in a problem like this, the
first thing we want to do is actually rearrange our equations so that we have our
variables on the left-hand side. And then we have our answers on the
right-hand side, which are numerical values or constants. So our first equation rearranged is
gonna be five 𝑥 plus two 𝑦 minus three 𝑧 equals negative five. Then, for the second equation,
we’re gonna have negative three 𝑥 minus 𝑦 minus two 𝑧 equals negative one. And then finally, five 𝑥 plus two
𝑦 minus 𝑧 equals three.
Okay, great. We’ve got it like this, but why do
we want it in this form? We want it in this form so we can
set up a matrix equation. And when we do, what we get is the
matrix five, two, negative three, negative three, negative one, negative two, five,
two, negative one multiplied by the matrix for our variables, which is 𝑥, 𝑦,
𝑧. Then this is equal to our answer
matrix negative five, negative one, three. Okay, great. But how does this help us meet our
objective, which is to solve the system of equations using determinants? Well, what we’re gonna do is use
Cramer’s rule. And what Cramer’s rule tells us is
that we can find the variables or solutions to our system of equations using, for
example, 𝑥 is equal to, then we’ve got the determinant of the matrix Δ sub 𝑥 over
the determinant of the matrix Δ. And then this pattern continues for
𝑦 and 𝑧.
Okay, to use this then, what we
need to do is work out our determinants. The first determinant we’re gonna
work out is the determinant of Δ, which is gonna be our coefficient matrix. So what we’re gonna do is find out
the determinant of the matrix five, two, negative three, negative three, negative
one, negative two, five, two, negative one. So this is gonna be equal to five
multiplied by the determinant of the submatrix negative one, negative two, two,
negative one minus two multiplied by the submatrix negative three, negative two,
five, negative one minus three multiplied by the submatrix negative three, negative
one, five, two, remembering that when we find the determinants, the coefficients go
positive, negative, positive. And to find our submatrices, we
delete the column and row that our coefficient is in.
Okay, great. So now, we calculate this. And then remembering that when we
work out determinants of two-by-two matrices, what we do is cross multiply and then
subtract, we’re gonna get five multiplied by one plus four minus two multiplied by
three plus 10 minus three multiplied by negative six plus five which is equal to
two. So this is great cause it also
tells us that the matrix is nonsingular. So therefore, we know that there is
not gonna be an infinite number of solutions. And that’s because if there was,
then the determinant would be equal to zero. So now, what we’re gonna do is
clear a space and work out the other determinants we need to find.
So next, what we want to do is find
the determinant of Δ sub 𝑥. And the way we do that is by
substituting in the answer matrix values for the coefficient of 𝑥-value, so the
first column in our matrix. So now what we’re gonna want to do
is find the determinant of this matrix. And to do that, what we’re gonna do
is to use the same methods as we used for the previous determinant, which is gonna
give us a determinant value of negative 42. And you can see the working
there. Okay, great. So once again, we’re gonna clear a
bit of space and look at our next determinant.
So now what we’re gonna find is the
determinant of the matrix Δ sub 𝑦. And this is gonna be where we
substitute in our answer matrix for the 𝑦-coefficients in the matrix. So then once again, using the same
method to find the determinant, we’re gonna have a determinant of 112. And again, the working is shown
here. So once again, what we’re gonna do
is clear a space for the final determinant. So for the final one, it’s gonna be
the determinant of the matrix Δ sub 𝑧. So then once again, we go through
the same method to find the determinant of our three-by-three matrix. And what it gives us is a value of
eight.
So now what we have are all the
determinants we need to use Cramer’s rule to find out our variables 𝑥, 𝑦, and
𝑧. So, first of all, we’re gonna start
with 𝑥, which is gonna be equal to negative 42 over two. And we get that because it’s the
determinant of Δ sub 𝑥 over the determinant of Δ. So this is gonna give us a value of
𝑥 equals negative 21. And then for 𝑦, we’re gonna have
112 over two which is gonna give us a 𝑦-value of 56. And then finally, we’re gonna get
𝑧 is equal to eight over two, and this is gonna give us 𝑧 is equal four. So therefore, we can say the
solutions to our systems of equation are 𝑥 equals negative 21, 𝑦 equals 56, and 𝑧
equals four.
Okay, so we’ve now looked at three
different examples, one that helped us identify one of the properties of Cramer’s
rule. Then we looked at solving a system
of equations with two equations. And now we’ve just looked at
solving a system of three equations. So now let’s have a recap over the
key points of the lesson.
And the first key point is, if we
have a system of equations, then what we want to do is have it so that we’ve got the
answers on their own on the right-hand side. And this is so that we can write it
as a matrix equation with the answer matrix on the right-hand side of the equal
sign. Then we also saw that to be able to
use Cramer’s rule, the matrix must not be singular. So that’s the coefficient
matrix. So that means that the determinant
is not equal to zero.
And then we have Cramer’s rule and
this tells us how we would find our unknowns using determinants. So, for instance, if we want to
find 𝑥, this would be equal to the determinant of Δ sub 𝑥 over the determinant of
Δ, also remembering we might see different notation here because instead of the
determinant of Δ sub 𝑥, we might just see D sub 𝑥. And similarly, we’d have D sub 𝑦
and D. So then finally, what we’ve got is
the determinant of Δ sub 𝑥. And we’d find this by substituting
in the answer matrix in instead of the 𝑥-coefficients, so we’d have the determinant
of negative eight, negative eight, seven, and six for our example.
