-
-
So far we've been able to define
the determinant for a
-
2-by-2 matrix.
-
This was our definition right
here: ad minus bc.
-
And then we were able to broaden
that a bit by creating
-
a definition for the determinant
of a 3-by-3
-
matrix, and we did that right
here, where we essentially
-
said the determinant is equal
to each of these terms-- you
-
could call these maybe the
coefficient terms-- times the
-
determinant of the matrix-- you
can kind of view it as the
-
submatrix produced-- when you
get rid of each of these guys'
-
column and row.
-
So when you got rid of this
guy's column and row, you're
-
left with this matrix.
-
So we said this guy times
the determinant of this.
-
And we kept switching signs,
minus this guy times the
-
determinant, if you move
his column and his row.
-
So it was left with these terms
right there to get that
-
determinant.
-
Then finally, you switched
signs again.
-
So plus this guy times the
determinant of the 2-by-2
-
matrix if you get rid of this
row and this column.
-
So this thing right here,
which was this matrix.
-
Now let's see if we can
extend this to a
-
general n-by-n matrix.
-
So let's write out our n-by-n
matrix right over here.
-
I'll do it in blue.
-
So let's say I have some matrix
A that is an n-by-n
-
matrix, so it's going
to look like this.
-
This would be a11, that would
be a12, and we would go all
-
the way to-- you're going
to have n columns, a1n.
-
And when you go down, this is
going to be your second row:
-
a21, and it's going to go all
the way down to an1, because
-
you have n rows as well.
-
And then if you go down the
diagonal all the way, this
-
right here would be ann.
-
So there is my n-by-n matrix.
-
Now, before I define how to find
the determinant of this,
-
let me make another
definition.
-
Let me define-- so this
is my matrix A.
-
Let me define a submatrix Aij to
be equal to-- see this is n
-
by n, right?
-
So this is going to be an n
minus 1 by n minus 1 matrix.
-
So if this is 7 by 7, the
submatrix is going to be 6 by
-
6, one less in each direction.
-
So this is going to be the n
minus 1 by n minus 1 matrix
-
you get if you essentially
ignore or if you take away--
-
maybe I should say take away.
-
Let's say ignore, like
the word ignore.
-
If you ignore the i-th row, this
right here is the row,
-
the i-th row and the
j-th column of A.
-
So, for example, let's go back
to our 3 by 3 right here.
-
This thing could be denoted
based on that definition I--
-
we could have called
this, this was a11,
-
this term right here.
-
We could denote the matrix when
you get rid of the first
-
column and the first row or the
first row and the first
-
column, we could call this thing
right here, we could
-
call that big matrix A11.
-
So this was big matrix A11.
-
This is big matrix A21, or
actually, this matrix was
-
called C, so this would
be C11 right there.
-
We could call this one, this
would be matrix C12.
-
Why is that?
-
Because if you get rid of the
first row, let me get rid of
-
the first row, right?
-
The first term is your row.
-
If you get rid of the first
row and the second column,
-
this is the matrix that's
left over: 2, 3, 4, 1.
-
So this is this guy
and this guy.
-
2, 3, 4, 1.
-
So this is the submatrix C
because this is the big matrix
-
C, But this one is C12.
-
I know it's a little
bit messy there.
-
So that's all we mean
by the submatrix.
-
Very similar to what we did
in the 3 by 3 case.
-
You essentially get rid of--
so if you want to find out
-
this guy's submatrix, you would
call that A11, and you
-
would literally cross out the
first row and the first
-
column, and everything
left over here
-
would be that submatrix.
-
Now, with that out of the way,
we can create a definition,
-
and it might seem a little
circular to you at first, and
-
on some level it is.
-
We're going to define the
determinant of A to be equal
-
to-- this is interesting.
-
It's actually a recursive
definition.
-
I'll talk about that
in a second.
-
It's equal to-- we start
with a plus.
-
It's equal to a11 times the
submatrix if you remove this
-
guy's row and column.
-
So that, by definition, is
just A, big capital A11's
-
determinant.
-
So that's exactly what we did.
-
Let me write that a
little bit neater.
-
So times the determinant of
its submatrix, so the
-
determinant of A11.
-
So you take A11, you get rid of
its column and its row or
-
its row and its column, and
everything else, you find the
-
determinant of that.
-
Actually, let me write it
in terms of-- let me
-
write it this way.
-
a11 times the determinant
of the submatrix A11.
-
And then we switch signs.
-
We're just going to go along
this row, and then you do
-
minus a12 times the determinant
of its submatrix,
-
which we'll just call A12.
-
We would get rid of this row
and this column, and
-
everything left would
be this matrix A12.
-
We want to find its
determinant.
-
And then we'll take the next
guy over here would be a13.
-
So we switch signs with minus.
-
Now, you go plus,
so a13 times the
-
determinant of its submatrix.
-
So if this is n by n, these each
are going to be n minus 1
-
by n minus 1.
-
So the determinant of A13.
-
And you're just going to keep
doing that, keep switching
-
signs, so it's going to be a
minus and then a plus and you
-
keep going all the way--
and then I don't know.
-
It depends on whether an,
whether we're dealing with an
-
odd number or an even number.
-
If we're dealing with an even
number, this is going to be a
-
minus sign.
-
If it's an odd number, it's
going to be a plus sign, but
-
you get the idea.
-
It's either going to be a plus
or a minus, not just-- if it's
-
odd, this is going
to be a plus.
-
If it's an even n, it's going to
be a minus, All the way to
-
a1n, the n-th column times
its submatrix, A1n.
-
With that submatrix, you get rid
of the first row and the
-
n-th column, and it's going
to be everything
-
that's left in between.
-
And you immediately might
say, Sal, what kind of a
-
definition is this?
-
You defined a determinant for an
arbitrary n-by-n matrix in
-
terms of another definition
of a determinant.
-
How does this work?
-
And the reason why this works
is because the determinant
-
that you use in the definition
are determinants
-
of a smaller matrix.
-
So this is a determinant of an n
minus 1 by n minus 1 matrix.
-
And you're saying hey, Sal, that
still doesn't make any
-
sense because we don't know how
to find the determinant of
-
an n minus 1 by n
minus 1 matrix.
-
Well, you apply this definition
again, and then
-
it's going to be in terms of n
minus 2 times n-- or n minus 2
-
by n minus 2 matrices.
-
And you're like how
do you do that?
-
Well, you keep doing it, and
you're going to get all the
-
way down to a 2-by-2 matrix.
-
-
And that one we defined well.
-
We defined the determinant of
a 2-by-2 matrix not in terms
-
of a determinant.
-
We just defined it in terms of
a times-- we defined it as--
-
let me write it up here.
-
It was a times d minus
b times c.
-
And you can see.
-
I mean, we could just go down to
the 3 by 3, but the 2 by 2
-
is really the most fundamental
definition.
-
And you could see that the
definition of a 3-by-3
-
determinant is a special
case of the general
-
case for an n by n.
-
We take this guy and we
multiply him times the
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determinant of his submatrix
right there.
-
Then we take this guy where
we switch signs.
-
We have a minus.
-
And we multiply him times the
determinant of his submatrix,
-
which is that right there.
-
Then you do a plus.
-
You switch signs and then you
multiply this guy times the
-
determinant of his submatrix,
which is that right there.
-
So this is a general case
of what I just defined.
-
But we know it's never that
satisfying to deal in the
-
abstract or the generalities.
-
We want to do a specific case.
-
And actually, before I do that,
let me just introduce a
-
term to you.
-
This is called a recursive
formula.
-
And if you become a computer
science major,
-
you'll see this a lot.
-
But a recursive function or a
recursive formula is defined
-
in terms of itself.
-
But the things that you use in
the definition use a slightly
-
simpler version of it, and as
you keep going through, or you
-
keep recursing through it, you
get simpler and simpler
-
versions of it until you get
to some type of base case.
-
In this case, our base case is
the case of a 2-by-2 matrix.
-
You keep doing this, and
eventually you'll get to a
-
determinant of a 2-by-2 matrix,
and we know how to
-
find those.
-
So this is a recursive
definition.
-
-
But let's actually apply it
because I think that's what
-
actually makes things
concrete.
-
So let's take-- this is going
to be computationally
-
intensive, but I think if we
focus, we can get there.
-
So I'm going to have a 4-by-4
matrix: 1, 2, 3, 4.
-
1-- throw some zeroes in there
to make the computation a
-
little bit simpler, 0, 1, 2,
3, and then 2, 3, 0, 0.
-
So let's figure out this
determinant right there.
-
This is the determinant
of the matrix.
-
If I put some brackets
there that would
-
have been the matrix.
-
But let's find the determinant
of this matrix.
-
So this is going to be equal
to-- by our definition, it's
-
going to be equal to 1 times the
determinant of this matrix
-
right here if you get rid of
this row and this column.
-
So it's going to be 1 times the
determinant of 0, 2, 0; 1,
-
2, 3; 3, 0, 0.
-
That's just this guy right here,
this matrix right there.
-
Then I'm going to have a 2, but
I'm going to switch signs.
-
So it's minus 2 times the
determinant if I get rid of
-
that row and this column,
so it's 1, 2, 0.
-
I'm ignoring the zero because
it's in the same column as the
-
2: 1, 2, 0; 0, 2, 3,
and then 2, 0, 0.
-
-
And then I switch signs again.
-
It was a minus, so now
I go back to plus.
-
So I do that guy, so
plus 3 times the
-
determinant of his submatrix.
-
Get rid of that row and get
rid of that columm,
-
I get a 1, 0, 0.
-
-
I get a 0, 1, 3.
-
-
I skip this column every time.
-
Then I get a 2, 3, 0,
just like that.
-
We're almost done.
-
One more in this column.
-
Let me switch to
another color.
-
I haven't used the
blue in this yet.
-
So then I'm going
to do a minus 4.
-
Remember, it's plus, minus,
plus, minus 4 times the
-
determinant of its submatrix.
-
That's going to be
that right there.
-
So it's 1, 0, 2; 0, 1, 2;
2, 3, 0, just like that.
-
And now we're down to
the 3-by-3 case.
-
We could use the definition of
the 3 by 3, but we could just
-
keep applying this recursive
definition.
-
-
So this is going to be equal
to-- let me write it here.
-
It's 1 times-- what's
this determinant?
-
This determinant's going to be
0 times the determinant of
-
that submatrix, 2, 3, 0, 0.
-
-
That was this one right here.
-
And then we have minus 2, minus
this 2-- remember, we
-
switched signs-- plus, minus,
plus, so minus 2 times its
-
submatrix, so it's 1, 3, 3, 0.
-
-
And then finally plus 0 times
its submatrix, which is this
-
thing right here: 1, 2,
3, 0, just like that.
-
And then we have this
next guy right here.
-
As you can see, this can get a
little bit tedious, but we'll
-
keep our spirits up.
-
So minus 2 times 1 times its
submatrix, so that's this guy
-
right here-- times the
determinant of its submatrix
-
2, 3, 0, 0.
-
Then minus 2 times-- get
rid of that row, that
-
column-- 0, 3, 2, 0.
-
-
And then plus 0 times
0, 2, 2, 0.
-
-
That's this one right there.
-
Halfway there, at
least for now.
-
And then we get this next one,
so we have a plus 3.
-
Bring out our parentheses.
-
And then we're going to have 1
times its sub-- I guess call
-
it sub-determinant.
-
So 1 times the determinant
1, 3, 3, 0, right?
-
You get rid of this guy's row
and column, you get this guy
-
right there.
-
And then minus 0-- get rid
of this row and column--
-
times 0, 3, 2, 0.
-
-
Then you have plus 0 times its
sub-determinant 0, 1, 2, 3.
-
-
Three-fourths of
the way there.
-
One last term.
-
Let's hope we haven't made
any careless mistakes.
-
Minus 4 times 1 times 1,
2, 3, 0 right there.
-
-
Minus 0 times-- get rid of those
two guys-- 0, 2, 2, 0.
-
-
And then plus 2 times
0, 1, 2, 3, right?
-
Plus 2-- get rid of these
guys-- 0, 1, 2, 3.
-
-
Now, we've defined or we've
calculated or we've defined
-
our determinant of this matrix
in terms of just a bunch of
-
2-by-2 matrices.
-
So hopefully, you saw in this
example that the recursion
-
worked out.
-
So let's actually find what
this number is equal to.
-
A determinant is always just
going to be a number.
-
So let me get a nice
vibrant color.
-
This is 0 times--
I don't care.
-
0 times anything's
going to be 0.
-
0 times anything is
going to be 0.
-
0 times anything's
going to be 0.
-
0 times anything's
going to be 0.
-
Just simplifying it.
-
These guys are 0 because
it's 0 times that.
-
0 times this is going
to be equal to 0.
-
So what are we left with?
-
-
This is going to be equal to 1
times-- this is all we have
-
left here is a minus 2 times--
and what is this determinant?
-
It's 1 times 0, which is 0.
-
It's 0-- let me write this.
-
This is going to be 1 times 0
is 0, minus 3 times 3 is 0
-
minus 9, so minus 9.
-
This right here is
just minus 9.
-
So minus 2 times minus 9.
-
That's our first thing, I'll
simplify it in a second.
-
Now let's do this
term right here.
-
So it's minus 2 times-- now
what's this determinant?
-
2 times 0 minus 0 times 3.
-
That's 0 minus 0.
-
So this is 0.
-
That guy became 0, so we
can ignore that term.
-
This term right here is
0 times 0, which is 0,
-
minus 2 times 3.
-
So it's minus 6.
-
So it's minus 2 times-- so this
is a minus 6 right here.
-
You have a minus 2 times a minus
6, so that's a plus 12.
-
So I'll just write
a plus 12 here.
-
This minus 2 is that minus
2 right there.
-
And then we have a plus 3.
-
And then this first term is 1
times 0, which is 0, minus--
-
let me make the parentheses
here-- 1 times 0, which is 0,
-
minus 3 times 3, which
is minus 9 times 1.
-
So it's minus 9.
-
Everything else was a 0.
-
We're in the home stretch.
-
We have a minus 4.
-
Let's see, this is 1 times 0,
which is 0, minus 3 times 2,
-
so minus 6.
-
So this is minus 6 right here.
-
Minus 6, this is 0, and then we
have this guy right here.
-
So we have 0 times 3, which
is 0, minus 2 times 1.
-
So that's minus 2, and then
you have a minus 2 times a
-
plus 2 is minus 4.
-
So now we just have
to make sure we do
-
our arithmetic properly.
-
This is 1 times plus 18,
so this is 18, right?
-
Minus 2 times minus 9.
-
This right here is minus 24.
-
This right here is minus 27.
-
And then this right here, let's
see, this is minus 10
-
right here.
-
That is minus 10.
-
Minus 4 times minus
10 is plus 40.
-
Let's see if we can simplify
this a little bit.
-
If we simplify this a little
bit-- I don't want to make a
-
careless mistake right
at the end.
-
So 18 minus 24, 24 minus 18 is
6, so this is going to be
-
equal to minus 6, right?
-
18 minus 24 is minus 6.
-
And then-- let me do it in
green-- now what's the
-
difference?
-
If we have minus 27 plus
40, that's 13, right?
-
It's positive 13.
-
So minus 6 plus positive
13 is equal to 7.
-
And so we are done!
-
After all of that computation,
hopefully we haven't made a
-
careless mistake.
-
The determinant of this
character right
-
here is equal to 7.
-
The determinant is equal to 7.
-
And so the one useful takeaway,
we know that this is
-
invertible because it has
a non-zero determinant.
-
Hopefully, you found
that useful.
-