Video Transcript
Using Determinants to Calculate
Areas
In this video, we’re going to learn
how we can use determinants to calculate the areas of various polygons. We’ll discuss how we can use
determinants to calculate the areas of triangles given the coordinates of their
vertices and how we can calculate the areas of parallelograms given their vertices
or vectors defining them. We’ll also discuss how we can
extend this to more complicated polygons.
To start discussing this topic, we
first need to answer a very important question. What does the determinant of a
matrix actually mean? We know the definition of a
determinant, and we know how to calculate this. However, we don’t know what this
actually means. There’s actually a lot of different
interpretations about what the determinant of a matrix can mean. We’ll only go through one of
these.
We’ll start with what we mean by
the determinant of a two-by-two matrix, the matrix 𝑎, 𝑏, 𝑐, 𝑑. The determinant of this two-by-two
matrix is the signed area of the parallelogram with sides defined by either the rows
or columns of this matrix. Now, this is a very
complicated-sounding statement. However, if we break it down piece
by piece, it can be a lot easier to understand. We’ll start by choosing to define
our parallelogram by the rows of our matrix. And we’re going to discuss how we
create this parallelogram.
So our matrix has two rows. The first row is 𝑎, 𝑏. And the first thing we’ll do is
define a vector 𝐕 one, which will have horizontal component 𝑎 and vertical
component 𝑏. We’ll then do the same for our
second row. This gives us our second vector 𝐕
two. And it’s worth pointing out we can
also choose the columns of our matrix to define this parallelogram. The only difference is we would
choose the columns of our matrix and we’d end up with column vectors. However, we’ll just go through the
example of rows.
And now, with these two vectors, we
can define a parallelogram. If we start at a point and arrange
our vectors in the following form, we will get a parallelogram. Now, our definition of the
determinant tells us that the determinant of this matrix is the signed area of this
parallelogram. In other words, if we call this
parallelogram 𝑃, then the area of 𝑃 is equal to either the determinant of this
matrix or negative one multiplied by the determinant of this matrix. And that’s because we know the
determinant can give us a negative value. But we don’t want a negative value
for our area.
This gives us a useful formula for
calculating the area of any parallelogram. And you’ll often see this written
in two different ways. First, we’ll take the absolute
value of the right-hand side of this equation. This then removes the need for the
positive or negative value. However, some people prefer to use
this notation for the determinant of a matrix. So you can also see this written as
positive or negative times the determinant of our matrix. And these all mean exactly the same
thing. We just calculate the determinant
of this matrix and take the positive value.
And there’s two things worth
pointing out here. First, remember, we could’ve taken
column vectors instead of row vectors. And this would give us exactly the
same result where we look at the columns of our matrix instead. The second thing worth noting is
what would happen if 𝐕 one and 𝐕 two did not form a parallelogram. For example, if 𝐕 one and 𝐕 two
point in exactly the same direction or in opposite directions or one of 𝐕 one or 𝐕
two are the zero vector, then we wouldn’t end up with a shape with area. It would have zero area. And this is exactly what we mean by
the determinant being equal to zero.
And it also turns out we can extend
this definition of the determinant to larger matrices. The determinant of a three-by-three
matrix is the signed volume of the parallelepiped, that’s the three-dimensional
version of a parallelogram, with sides defined by the rows or columns of that
matrix. Now, it’s beyond the scope of this
video to know exactly why this is true. However, by assuming this is true
and using some algebraic manipulation and geometry, we can arrive at a very useful
result.
If we know the coordinates of the
vertices of a triangle 𝑇, we can calculate its area. Its area is equal to one-half times
the absolute value of the determinant of a three-by-three matrix where each row is
the coordinate pair and an extra row for the value of one. Once again, the proof of this
statement is beyond the scope of this video. However, we’ll see how to use this
in a few examples.
Find the area of the triangle below
using determinants.
The question wants us to find the
area of the triangle given to us in the diagram. And it wants us to do this by using
determinants. So the first thing we’re going to
need to do is recall how we find the area of a triangle by using determinants. We recall we just need to find the
coordinates of its three vertices and then use the following formula. The area of 𝑇 is equal to one-half
multiplied by the absolute value of the determinant of a three-by-three matrix where
each row in our matrix is formed by the coordinates of the vertex and an extra
one.
So the first thing we need to do is
find the three coordinates of the vertices of our triangle. We get negative one, negative four;
zero, four; and four, negative two. And there’s something worth
pointing out here. It doesn’t matter which of these
points we label one, two, or three. All this would do is change the
sign of our determinant. But we’re taking the absolute value
of this, so it won’t change our answer.
So we’ll just choose to number our
points from left to right. This gives us the area of the
triangle 𝑇 is equal to one-half multiplied by the absolute value of the determinant
of this three-by-three matrix. So what we need to do now is
calculate this determinant. And there’s a few different ways of
doing this. We’ll use matrix minors. And we’re going to expand out the
first column because it contains zero.
First, we’ll start by recalling the
signs we need to multiply by. We get positive, negative,
positive. Next, we need to find our matrix
minors. We’ll start by finding the one by
removing the first row and the first column. So we get negative one multiplied
by the determinant of the two-by-two matrix four, one, negative two, one. Next, the second term in our
expansion has a coefficient of zero. So we don’t need to do this. So we’ll just move on to the third
term.
First, we know we need to multiply
this by four. And we need to find the matrix
minor by removing the first column and third row. So we get four multiplied by the
determinant of the two-by-two matrix negative four, one, four, one. Now, all we need to do is evaluate
these two-by-two determinants. First, we get four times one minus
negative two times one. Next, we get negative four
multiplied by one minus four times one.
Now, all that’s left to do is
evaluate this expression. First, negative one multiplied by
four times one minus negative two times one is equal to negative six. Next, four times negative four
times one minus four times one is equal to negative 32. So this just simplified to give us
one-half times the absolute value of negative six minus 32. Negative six minus 32 is negative
38. The absolute value of negative 38
is 38. Then we multiply this by one-half
to get 19. But remember, this gives us the
area of a triangle, so we can give this units. These must be given in square
units.
Therefore, we were able to show the
area of the triangle given to us in the diagram is 19 square units. It’s also worth pointing out we
could’ve found the area of this triangle by using geometry. And this would give us a good way
of checking our final answer.
Let’s now go through an example of
how we could use determinants to help us find the coordinates of the vertex of a
triangle.
If the area of a triangle whose
vertices are ℎ, zero; six, zero; and zero, three is nine square units, then ℎ is
equal to (A) zero or 12, (B) zero or negative 12, (C) negative six or six, (D)
negative 12 or 12.
We’re given that a triangle has an
area of nine square units. And we’re given the three
coordinates of its vertices. We need to use this information to
find the possible values of ℎ. It’s worth pointing out here one
method to do this is to plot all three of these points on a diagram and then just
calculate the possible areas. And this definitely works and can
be used to get the correct answer. However, we’re going to do this by
using determinants.
Recall, we can find the area 𝐴 of
a triangle given its vertices by using the following formula. 𝐴 is equal to one-half multiplied
by the absolute value of the determinant of a three-by-three matrix, where each row
in our matrix is the coordinate pair with an extra value of one. And it doesn’t matter which order
we choose our points. So we’ll just order them from left
to right. So by using the vertices given to
us in the question, we get the area of our triangle is equal to one-half times the
absolute value of the determinant of the following three-by-three matrix.
The next thing we’re going to need
to do is find the value of this determinant. There’s a few different ways of
doing this. We can see that one of the columns
of our matrix has two values of zero. So we’ll find this determinant by
expanding over the second column. So we need to do this by finding
the matrix minors. Remember, our first two values are
zero. So we don’t need to calculate
these. We only need to do the final one
with the coefficient of three. And remember, we multiply this by
negative one to the power of three plus two because it’s in row three, column
two.
Now, remember, we need to find the
determinant of the matrix minor we get by removing the third row and second
column. We can see this is the two-by-two
matrix ℎ, one, six, one. So we’ve shown the area of our
triangle is equal to one-half times the absolute value of negative one to the power
of three plus two times three multiplied by the determinant of the two-by-two matrix
ℎ, one, six, one.
Now, all we need to do is evaluate
this expression. First, negative one to the power of
three plus two is just negative one. Next, if we evaluate the
determinant of our two-by-two matrix, we see it’s equal to ℎ minus six. So the area of our triangle is
one-half times the absolute value of negative three times ℎ minus six. And we can simplify this
further. We can take the negative three
outside of our absolute value. And remember, this means it becomes
positive three. So we get three over two times the
absolute value of ℎ minus six.
Remember, in the question, we’re
told the area of our triangle is nine. So, in fact, this expression must
be equal to nine. So we just need to solve an
equation involving the absolute value symbol. We’ll start by dividing both sides
of our equation through by three over two. Nine divided by three over two is
six. We get six is equal to the absolute
value of ℎ minus six.
Remember, to solve equations
involving the absolute value symbol, we need to consider the positive and negative
solution for our absolute value. And by taking the positive and
negative, we get two equations we need to solve. Six is equal to ℎ minus six, and
six is equal to negative one times ℎ minus six. And these are both linear
equations, and we can solve them. We get ℎ is equal to 12 or ℎ is
equal to zero. And this is option (A). Therefore, we were able to show
that ℎ is equal to zero or 12.
And it’s also worth pointing out we
could check our answer by substituting our values of ℎ into this coordinate pair,
plotting these points, and then calculating the area of the triangle and seeing that
we get nine. And this can be a good way to check
that we got the correct answer.
Let’s now see an alternate method
of calculating the area of a triangle by using determinants.
Use determinants to work out the
area of the triangle with vertices two, negative two; four, negative two; and zero,
two by viewing the triangle as half a parallelogram.
We want to find out the area of a
triangle by using determinants. And we’re given the coordinates of
its vertices. So we could just do this directly
from our formula. However, the question doesn’t want
us to do this. It wants us to do this by viewing
the triangle as half of a parallelogram. So to do this, we’ll start by
sketching our three points to get an idea of what our parallelogram could look
like. If we plot the three points and
connect them, we get a triangle which looks like this. The question then becomes, how are
we going to turn this into a parallelogram?
And although it may not seem like
it, there’s actually three different ways we could do this. There’s a few different ways of
seeing this. One way to do this is
geometrically. Let’s draw an exact copy of our
triangle. And we can consider the shape if we
were to glue two of the sides together. If we were to do this, we would get
a shape which looks like this. And we could find the coordinate of
this vertex by using what we know about vectors. This is a parallelogram with twice
the area of our original triangle because it’s made up of two triangles of equal
area. But this wasn’t the only choice for
the side. For example, we could’ve chosen
this side. If we were to glue these two
triangles together along this edge, we would get something which looks like the
following diagram. And once again, we could find the
coordinates of this vertex by using what we know about vectors.
Once again, the area of this
triangle plus the green triangle is still twice the area of our original
triangle. So it’s a parallelogram with twice
the area. Finally, we could do exactly the
same thing by combining the last two edges. And if we did this, we would get a
similar story. We get a third parallelogram. We can find the coordinates of its
vertex. And this parallelogram is twice the
area of our original triangle. It doesn’t matter which of these we
choose. For simplicity, we’ll choose the
following example. And it’s worth pointing out we
didn’t need to do this geometrically. You can also do this by choosing
two of the sides of the triangle as a vector. This would give us the same
result.
The question now wants us to find
the area of this parallelogram. And it wants us to do this by using
determinants. Recall, we know how to find the
area of a parallelogram by using determinants. If we define our parallelogram by
the vectors of its sides, 𝐕 one and 𝐕 two, then its area 𝐴 is equal to the
absolute value of the determinant of the two-by-two matrix 𝑎, 𝑏, 𝑐, 𝑑, where our
row vectors are the rows of our matrix. And we could’ve also used column
vectors and used the columns of our matrix instead. It wouldn’t change our answer.
So to answer this question, we need
to find the vectors 𝐕 one and 𝐕 two. There’s a few different ways of
doing this. Let’s start by finding 𝐕 one. In 𝐕 one, our 𝑥-component starts
at two and ends at zero. So the horizontal component or
change in 𝑥 of 𝐕 one is zero minus two, which is of course just negative two. We can do the same for the vertical
component. We end at a 𝑦-coordinate of two
and begin at negative two. So we get two minus negative two,
which is four. So 𝐕 one is the vector negative
two, four. We can then do exactly the same to
find vector 𝐕 two. 𝐕 two is two, zero.
We’re now ready to use our
formula. The area of our parallelogram is
the absolute value of the determinant of the two-by-two matrix negative two, four,
two, zero, where negative two, four is 𝐕 one and two, zero is 𝐕 two. And we can evaluate this
expression. We get the absolute value of
negative two times zero minus four times two, which is just equal to eight. But remember, this is the area of
our parallelogram. This is twice the area of our
triangle. So we need to divide this value by
two. So we divide through by two. And we get the area of our triangle
is equal to four.
And one thing worth pointing out
here is we never needed to calculate the fourth coordinate of our parallelogram. This is because the area of a
parallelogram is entirely defined by the vectors which make up the
parallelogram.
Let’s now see an example of how we
would calculate the area of a more complicated polygon.
Consider the quadrilateral with
vertices 𝐴 one, three; 𝐵 four, two; 𝐶 4.5, five; and 𝐷 two, six. By breaking it into two triangles
as shown, calculate the area of the quadrilateral using determinants.
We’re given a quadrilateral
determined by four vertices which we’re given. We want to calculate its area by
using determinants. We can see from the diagram this is
not a parallelogram. So, instead, we’re going to break
this up into triangles and calculate the area of each triangle by using
determinants. So let’s start by recalling how we
calculate the area of a triangle by using determinants.
We recall, if we know the
coordinates of the three vertices of our triangle, then its area is equal to
one-half times the absolute value of the determinant of the three-by-three matrix,
where each row in this matrix is the coordinate pair of a vertex and then an extra
coordinate of one. And we’re told to split our
quadrilateral into two triangles: triangle 𝐴𝐷𝐶 and triangle 𝐴𝐵𝐶. So we’ll need to apply this formula
to both of these.
So let’s start by finding the area
of triangle 𝐴𝐵𝐶, which we’ll just write as 𝐴𝐵𝐶. From our formula, we need to know
the coordinates of these three vertices, which we’re given in the question. So we write these into our matrix
and then add an extra column of one. This gives us the following
expression for the area of triangle 𝐴𝐵𝐶.
Now, to evaluate this expression,
we’re going to need to calculate its determinant. We’ll do this by expanding the
third column. First, we’re going to want to find
the sign we need to multiply by each of these columns. We get positive, negative,
positive. So our three coefficients are
positive one, negative one, and positive one. Next, we need to find the
determinants of the three matrix minors. Removing the first row and third
column, we get the two-by-two matrix four, two, 4.5, five. We then need to subtract the
determinant of our second matrix minor. Then, finally, we need to add the
determinant of our third matrix minor. This gives us the following
expression, which is equal to the area of triangle 𝐴𝐵𝐶.
All we need to do now is evaluate
each of these three determinants. Let’s start with our first
determinant. We get four times five minus 4.5
times two. That’s 20 minus nine, which is
equal to 11. Calculating the second determinant
and simplifying, we get five minus three times 4.5, which is negative 17 over
two. But remember, we’re subtracting
this value. Finally, we need to calculate the
third determinant. It’s equal to two times one minus
three times four. That’s two minus 12, which is equal
to negative 10. So this gives us the area of 𝐴𝐵𝐶
is one-half times the absolute value of 11 minus negative 17 over two plus negative
10. And if we evaluate this expression,
it’s equal to 19 over four.
So let’s keep track of this and now
do exactly the same thing to find the area of 𝐴𝐷𝐶. This time, our three vertices are
𝐴, 𝐶, and 𝐷. Remember, to use our formula, we
add these coordinate pairs into our matrix and then add an extra coordinate of
one. This gives us the area of 𝐴𝐷𝐶 is
the following expression. And we’ll evaluate this in the same
way. We’ll expand over our third
column. Of course, the third column is
still going to go positive, negative, positive. So we need to start by adding the
determinant of our first matrix minor by removing the first row and third
column. That’s the two-by-two matrix 4.5,
five, two, six. Then we need to subtract the
determinant of our second matrix minor by removing the second row and third
column. That’s the two-by-two matrix one,
three, two, six. Then we add on the determinant of
our third matrix minor, giving us the following expression for the area of
𝐴𝐷𝐶.
Now, all we need to do is evaluate
these determinants. Our first determinant is six times
4.5 minus two times five, which is 17. Our second determinant is six times
one minus two times three, which we can calculate is zero. And calculating our third
determinant, we get negative 17 over two. Finally, we can just evaluate this
expression. It’s one-half times the absolute
value of 17 minus 17 over two, which is equal to 17 over four.
Remember, the area of our
quadrilateral will be the sum of these two values. And if we add these two values
together and simplify, we get the area of our quadrilateral is nine. And there is one interesting thing
worth noting about this formula. What would happen if the area of
our triangle was zero?
Well, for the area to be zero, it
must not be a triangle at all. And since a triangle is three
points which don’t lie on the same line, that means our three points must lie on the
same line. In other words, they’re
collinear. But now look at our formula. The only part of our formula which
can be equal to zero is the determinant of this matrix. This gives us a test for
collinearity of three points.
Let’s now go over the key points of
this video. First, we showed we can find the
area of a parallelogram if its sides are the vectors 𝑎, 𝑏, 𝑐, 𝑑 as the absolute
value of the determinant of the two-by-two matrix 𝑎, 𝑏, 𝑐, 𝑑. We also saw we can instead use
column vectors. We also saw how to find the area of
a triangle if we’re given the coordinates of its vertices by using determinants. We also know if a triangle’s area
is zero, its vertices must be collinear. So this gives us a test for
checking if three points are collinear. Finally, we saw we can find the
area of more complicated polygons if we can split these into triangles.
