I want to teach you a bit more about determinants because this is
a very useful function for engineers and for scientists.
So let's go back to the beginning.
Let's start talking about two-by- two and three-by-three determinants.
We had already introduced the two-by-two matrix a,
b, c, d, where we calculated the inverse of this matrix.
So the inverse I wrote as 1 divided by the determinant of a,
and the inverse matrix was obtained by switching d and a.
So d and a on the diagonal and negating the off-diagonal elements of the matrix.
So, A times A inverse is equal to I.
The important point here is that we divide through by the determinant of A.
So, the determinant of A turned out to be a times d,
multiplying the diagonal elements,
minus b times c the off diagonal elements.
In order for the inverse to exist,
the determinant of A needed to be non-zero.
What happens if there is inverse to the matrix A?
Then if you have a system of linear equations A_x equals b,
if it turned out that the determinant of A was not 0,
then you can find the inverse of A and you can multiply on the left by A inverse.
So, you can solve this equation, at least formally,
by x equals A inverse times b,
and that would be a unique solution.
So, the existence of the inverse gives you
a unique solution to a system of n equations and n unknowns.
For the inverse to exist,
this function, determinant of a,
which takes an n-by-n matrix and returns a number,
that number that it returns has to be non-zero.
If that number is not 0,
then the matrix A is invertible,
and A_x equals b has a unique solution.
In particular, the equation A_x equals 0,
if A is invertible,
has only one solution which is x equals 0.
That implies x equals 0,
the 0 column vector,
if determinant of A was not equal to 0.
That actually will be a very important result when we talk about
eigenvalues and eigenvectors later in this course.
This was the two-by-two matrix.
You can play this game with a three-by-three matrix.
So, here, let's write the three-by-three matrix,
A equals a, b,
c, d, e, f,
g, h, i, and one can write the equation A_x equals 0,
and come up with a condition so that x equals 0 is the only solution.
That's an exercise in algebra,
would fill up half of the light board but it's absolutely doable.
So if you did that,
you would get determinant of A.
So there's a nice trick for
remembering what is the determinant of a three-by-three matrix.
So let me show you the trick and wrote down the determinant.
So this is for a three-by-three matrix.
The trick is that you draw diagonals in this matrix.
So the first diagonal is aei,
and then the second diagonal is bf,
but then right you run out of the third row,
so you just pretend that this matrix is written again on the right hand side.
So you go bf, and then you would repeat the matrix.
So then this diagonal goes to g. So it
would be aei plus bfg,
and then one more term which would be cdh,
So plus cdh, and that gives
all of the diagonals sloping from upper left to lower right.
That's three terms and then there are
three more terms with diagonals going in the other direction.
So then we have this diagonal here, ceg.
This one, the diagonals going in the other direction gets a minus ceg
and then we have two more of those diagonals.
So we have bdi.
So minus bdi, and one
more which would be afh.
So minus afh.
So all these six terms form the determinant of a three-by-three matrix.
This is a very nice,
it would say a mnemonic.
It's a way of remembering what is the determinant of a three-by-three matrix.
Unfortunately this trick of drawing diagonals doesn't generalize to n by n,
so this only works for three- by-three matrices.
It's a nice thing to remember.
If the determinant of A is not equal to 0,
then A is an invertible matrix,
and A_x equals b has a unique solution.
So that's true for the two-by-two matrix,
that's true for the three-by- three matrix.
So what we're going to want to do then is to generalize this to
n-by-n matrices and we will do that in the next video.
So mainly, the thing to remember here is that the two-by-two determinant is very simple,
ad minus bc, and the three-by-three determinant isn't that much harder.
There are six terms instead of two,
but it's very easy to remember what those six terms
are if you remember that you just draw these diagonals.
I'm Jeff Chasnov. Thanks for watching,
and I'll see you in the next video.