In previous videos, we've seen
how to find the determinant of a

two by two matrix. We've also
seen the role that the

determinant plays in solving
simultaneous linear equations,

and in finding the inverse of a

matrix. And what we're going to
do in this video is we're going

to look at three by three
determinant statements of square

matrices with three rows and
three columns. Now, if you can

remember the rule for two by
two, determinant's was very

straightforward, but perhaps not
very intuitively obvious.

With three by three matrices, it
gets a little bit more

complicated, and, again not very
intuitively obvious and to

workout the determinant of three
by three matrices, we need to

first of all introduce some
other ideas, called minors and

cofactors, and will start by
doing that. And throughout this

video will always use the same
matrix A. So this is the matrix.

Say that we're going to use and
you can see that it's got three

rows and three columns. We can
only find determinants when we

have a square matrix.

The matrix that has the same
number of rows as it has

columns. And to find the
determinant of this three by

three matrix, we first of all
need to understand what's meant

by a minor. Every element within
our matrix has its own minor, so

we spell minor like this MINOR,
and the minor of an element is

the value of the determinant we
get when we cross out the row

and the column containing.

The element we're looking at, so
let's look at this element 7.

Element 7 is in the first row
and the first column. So if we

cross out that throw in that

column. I'm going to hide them
rather than across the Mount.

You see that we are left with a
2 by two matrix and we can find

the determinant of that two by

two matrix. So the minor of
the element 7 in Matrix A is

this two by two determinant
with entries 3 -- 1 four and

minus 2.

Remember the rule for
evaluating a two by two

determinant is to multiply
the two elements on the

leading diagonal and then
subtract the product of the

two elements that aren't on
the leading diagonal. So we

do 3 * -- 2.

And then we take away 4 * -- 1.

So that gives us minus 6
-- -- 4.

She's minus 6 + 4, which gives
us minus 2.

So the minor of our elements 7
from our original three by three

Matrix is the value minus 2.

Now we can do the same thing for
any element in the matrix and

find its minor. So let's pick
this four. The four that's in

the 3rd row and the second
column, and we're going to find

the minor of four.

So we go
through the same

process. Before is in the bottom
row, so we cross that out and

it's in the second column. So we
cross that out.

And so we see we're left
with a 2 by two matrix.

And so we workout the
determinant of that two by

two matrix.

These are two by two Matrix 7,
one, 0 -- 1.

So we do 7 * -- 1, which is
minus 7 and then we take away

nought times one we're taking
away nought. And so we're

getting an answer of minus 7.

So the minor of the element for
in this matrix is minus 7.

Now we could, but I'm not going
to now. We could workout the

minor of every single element in
our matrix A and we get a value

of the minor. So we get 9
values, one for each element of

the matrix A.

Now you can see now what the
process is to work those apps.

I want to go now from minus to

cofactors. And to get
from minors cofactors,

I need the idea of what
we call a place sign.

Every place within the Matrix
has a sign associated with it.

We always start with a plus in
the top left hand corner and

then we alternate the signs. So
as we go across the road, we're

going to go plus minus plus.

As we go down the column, we're
going to go plus, minus plus.

You can see that if we fill this
in, it will look like this.

This is the matrix of place
signs. So for instance, the

element 4 we've looked at
before has a place sign of

minus, whereas the seven that
we've looked at as a place

sign of plus and we used to
play signs to convert minors

into what we call cofactors.

So the cofactor of the element
7 is equal to the place

sign. Times

the minor.
That's true for all the

elements. Every element has a
cofactor which we find by doing

the product of its place sign
with the value of its minor.

So if the element 7, the place
sign was plus.

And then at 7 the minor was
minus two. So we get plus minus

two, which is just equal to

minus 2. To the cofactor of the
element 7 in this matrix A is

minus 2. If you take the
element 4, the cofactor.

A full Is equal to its place
sign times it's minor, so the

minor of four we saw was minus
seven. The place sign is a

minus, so we don't play sign
minus times the value of the

minor minus seven, so minus
minus seven is plus seven. So

the cofactor of Element 4 in
this matrix is 7 when they are

going to calculate the
determinant of our matrix A by

using the cofactors.

So if we look at what we've
written down here, what I've

done is I've taken our matrix A
and I've worked out the 9 minus.

So the minor of seven. We've
already seen his minus two and

the minor of four we've seen as
minus seven. I've worked out all

the other minors and I've put
them into this matrix here that

I've called M. Remember that we
have a place sign matrix that

starts with a plus in the top
left hand corner and alternate

sign as you go down or across.

And then we calculate the
cofactors by taking the minor

and multiplying by its place
site. So I've assembled into

the Matrix, see the cofactors
of all the elements of our

matrix A and you can see that
you get the cofactors from the

minors simply by multiplying
by the place site.

Now to workout the determinant

of a. What we have to do is we
can pick any row or any column.

So let's say we were to pick the
first row and you see that the

row has three elements in it
would pick the column that would

have had three elements in it as

well. What we do is we pick
our row. We've got three

elements and they've each got
their own cofactor.

So we're going to do is going to
form the product of element

times its cofactor, so we'll be
doing 7 * -- 2 two times, three

under 1 * 9.

I've just explained what the
steps are for calculating

determinant, so let's now write

it all down. We've seen that the
only things that really matter

are the matrix we started with
and its cofactors, so that's all

we've got on on the sheet.

So here's our matrix A.

And here's the matrix of

cofactors. Remember what we need
to do is we were picking the

first row. We could pick any row
or column, but we've chosen to

pick the first row.

And workout the determinant of

a. Or write Det the determinant,
the T brackets a meaning of the

determinant of a.

And sometimes you'll see this
written with vertical lines.

To mean the determinant of a.

And what we do, because we've
chosen the first row, we write

down the elements of the row
7211, and we multiply each

element by its cofactor. So
introduce 7 * -- 2.

We're going to do 2 * 3.

And we're going to do 1 * 9.

So we can get these three
products and then we add them

up. So 7 * -- 2 is minus
14. Two times three is 6 of 1

* 9 is 9, so we workout minus
14 + 6 + 9. We get the

answer one. So the determinant
of a is equal to 1.

Now you're probably thinking,
well, what if I chosen a

different row or a different
column? Surely I would have got

a different answer, so let's
choose a different row, a

different column, see what
happens. Let's choose the second

column. So if we're going to use
the second column to workout the

determinant of a, we've got
elements 2, three and four, 2,

three, and four.

And we multiply these by their
cofactors. 2 gets multiplied by

three, 3 gets multiplied by
minus 11, and four gets

multiplied by 7.

We got the three products.

And we add them up.

So 2 * 3 is six
3 * -- 11 is minus

33 and 4 * 7 is

28. Work this out. We get one.

So we get the same value.

And we will always get the same
value whichever row or column

that we choose to use.

Doesn't matter, it's completely
our choice. So what this means

is that if we're working out the
determinant of a matrix, we

actually don't need to workout
all nine cofactors.

You see, when we use the first
row, we only needed these three

cofactors. When we use the
second column, we only needed

those three cofactors.

So before we go ahead and work
out all nine, if all we want to

do is workout the determinant of
a, we should decide which row or

column we're going to use first,
and then only workout the

cofactors that correspond to the
row or the column that we've

chosen. And in general it's
handy idea. Good idea to keep

your workload down is if you've
got zeros in a row or column.

That's a good row or column to
choose, because you don't need

to actually workout the cofactor
that goes with the zero element,

because it's going to get end up
being multiplied by that zero

element. So if we use the first

column. The first value is 7,
the next value is zero and the

last value is minus 3.

We then have to multiply these
elements by their corresponding

cofactors 7. * -- 2.

Nought Times 8 and 3 * -- 5.

Have them up. We get 7 * -- 2
is minus 14 nought. Times 8 is

nought minus 3 * -- 5 is plus

15. An minus 14 plus not plus 15
is equal to 1. The same answers

before we knew it was going to
come out to be the same answers

before. The important thing is
that we really didn't need to

know what the code factor of 0
was. So once we've chosen to use

this column, we actually need to
find the cofactors of seven and

minus three because the Co
factor of nought is going to get

multiplied by normal. So what
we've seen then is how to

workout the determinant of a
three by three matrix. What we

do is we pick Aurora column and
keep our work as low as

possible. We choose the row or
column with the most zeros, and

then what we do is we workout
the cofactors of every element

in the row or column that we've
chosen, but we don't have to

workout the cofactors of zero

elements. Because what we do
then is we multiply each

element by its cofactor, and
then we add up the products of

element times, code factor, and
that's how we workout the

determinant. Now this process
works for any size square

matrix, although once you get
above 3, the amount of

arithmetic involve gets to be
very large.