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Let's now compute the determinant using
the previous formula of a 3 by 3 matrix.

So there's my matrix A, let's say.

And, once again, I'm allowed to expand
along any row or column and I'll get

the equivalent result when I compute
this determinant.

And then in conclusion,
depending on whether that number is

zero or not, that tells me whether or not
the original matrix was invertible.

I'm just going to use just that first
general formula we wrote here

for a determinant by expanding along
the first row.

So I take the number 5 and then I
multiply by the

determinant of the corresponding submatrix
which is now 0, 2, 1, 3. And then I go

minus, I alternate signs, so check this
minus 3, goes +3 so we flip that and

then I remove again row 1 column 2,
and I just plug in the determinant here,

1,2,2,3, the corresponding submatrix.

And then lastly I pluck out the 2, the
last entry in that row, and then I

multiply that by the corresponding
submatrix here, that

determinant here 1, 0, 2, 1.
So there's my determinant formula.

And again, just as a friendly reminder,
the determinant of a 2x2 matrix is this

sort of cross multiplication process,
adbc.

So let's go ahead and clear this up here.

So we have 5, adbc, so this is 0(2) so
this is going to be a +2, so 5(2).

So plus 3 times ad, which is
3(4) is 1.

And lastly we have +2, adbc, is 1.

So we have that here. And let's add
these numbers up.

We get 1032 and that results in 5.

That means that my original matrix A
here, is invertible.

So there's some matrix out there I
could find and multiply by that inverse

matrix and then produce the identity.

Since we haven't done a column yet,
let's expand along column 2 here.

And we will compute the determinant
that way and we see that we also get 5,

but we'll have to do little less

work because of the 0 here.

I just want to note right we have this
checkerboard pattern with the signs.

I'm just going to remind you for
the determinant we have a negative with

that term, a positive and then a negative,
we alternate signs.

Let's now compute the determinant of A
one more time, just expanding along this

time column 2. Ok, so I have, negative 3
which is +3 times the determinant of

the submatrix, the same as it was before.

So 1, 2, 2, 3. Ok, now alternate sign +0,
ok I'll just leave this as a place holder

so we can see it. Plus zero. Of course,
that'll go away.

But the submatrix associated with zero
when I remove now column 2 and row 2 is

going to be 5, 2, 2, 3. Ok so that just
goes away. And then I go minus 1

which is plus 1 when I alternate sign,
and then I multiply by the determinant

of the submatrix corresponding with the
1 here, so I'm going to remove row 3

and column 2 and that leaves me
with 5, 2 and 1, 2.

Let's compute that determinant.
So again, I use the formula

for a 2x2 determinant adbc here. Right,
so 3 times 1(3),ad, 34 is 1. Ok. plus 0,

and then plus 1 times 102 is going to be
8. So what I have when the dust settles

is 3+8 sure enough results in 5.
Point being the determinant can be

ascertained by expanding along any
row or column we get a consistent result.