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## Linear algebra

### Course: Linear algebra > Unit 2

Lesson 5: Finding inverses and determinants# Formula for 2x2 inverse

Figuring out the formula for a 2x2 matrix. Defining the determinant. Created by Sal Khan.

## Want to join the conversation?

- Why do determinants work? Also, where would you use a determinant in real life?(14 votes)
- i'd buy every mathematics textbook you'll ever write(8 votes)

- Sal sure tripped me up with this one! The transformations at the beginning are quite nuanced and actually use elementary matrix multiplication - am I right? .

It took me some time to work through the workings of this for the general case. It was skipped over quickly in the video.

If I am missing the ' elephant in the room ' on this could somebody point it out to me!(18 votes)- This can be more easily seen by an example: try to zero out:

| 4 8 |

| 2 2 |

To zero out the second row on the first column, you need to do:

4(2) - 2(4) = 0

And then the same thing to the second column, second row:

4(2) - 2(8) = -8

Therefore, the more generalized transformation is:

4(c2) - 2(c1)

That makes sense, right? The c2 and the c1 changes for every column.

However, the 4 and 2, stay the same in each calculation. And what is 4 and 2? It's just 4=a and 2=b, right? That makes sense.

So if you can generalize even more to say that 4=a and 2=b, then the even more generalized equation is:

a(c2) - b(c1)

If that makes sense, then maybe you can think about is as just multiplying the top and bottom and then subtracting it from the product of the bottom and the top. That's the basic idea, if it makes sense to you: (top)*(bottom) - (bottom)*(top) = 0(9 votes)

- Is the determinant useful in any other way besides understanding if we can find the inverse or not?(7 votes)
- Yes! In the 2x2 case the determinant of A gives the signed area of a parallelogram formed by the two column vectors of A. I made a 2x2 determinant exercise that demonstrates this with an illustration. Sorry for late answer, but did it also for current users with the same question. Here's a link:

https://www.khanacademy.org/computer-programming/2x2-determinant-exercise/5285171024232448(6 votes)

- Why does Sal cancels out the a's at9:27without admitting that a is different than zero? This would result in division by zero, I guess.(6 votes)
- Sal should have stated explicitly that a is not equal to zero. That said, a is not equal to zero because a is the first vector in the first row, meaning a is a pivot variable in reduced row echelon form, which can't be equal to zero. If a is zero you can easily interchange the first row and the second row. If a is zero, then c certainly is not equal to zero because that would mean the two row vectors (or column vectors if you'd like) would not be linearly independent. If the two rows (or columns) are not linearly independent, the matrix is not invertible. To recap, for matrix A to be invertible, at least one of a and c is non-zero and you can just assume that a is non-zero. Hope that helps.(7 votes)

- I don't know why Sal keeps referring to column vectors when he does row operations...I recognize that a matrix can always be visualized as a bunch of column vectors, but that is not at all helpful when he's about to do a bunch of ROW operations.(6 votes)
- I also find it a huge distraction. He's obviously performing operations on row vectors. It's work enough understanding what he's trying to say without having to decode his language.(2 votes)

- Where does c1 and c2 come from, what must I read to understand the T symbol? Are c1 and c2 scalars?(3 votes)
- I think Sal means row1 and row2 or r1 and r2, not c1 and c2. He leaving row1 unchanged and changing row2, not column c1.(3 votes)

- If the determinant of a square matrix is zero, does that imply that there is NO inverse on that matrix?(3 votes)
- Yes, it does. Let
**A**be any n x n matrix for which det**A**= 0. Then**A**is singular (not invertible).**Proof**

Suppose**A**is*not*singular, and let**B**denote the inverse of**A**. That is, if**I**is the n x n identity matrix, then**BA**=**I**. By the product formula for determinants, we have det**A**= 1 / det**B**≠ 0. But this contradicts the fact that det**A**= 0, and the proof is complete.(3 votes)

- Does this method work to find the inverse of bigger square matrices?(2 votes)
- Yes, it does work. If you augment the matrix with the identity and when you put the new matrix into Reduce Row Echelon from you get the identity on the left side, the right side will be the inverse of the matrix, no matter the dimensions of it.(4 votes)

- is there any formula for 3x3 inverses ?(2 votes)
- The next video on 3x3 determinants may have something to say about it, but I don't know, as I haven't watched yet. But to answer your question, yes there is a formula. How useful such a formula would be to us is questionable, because finding the inverse is in practice done on a computer or by hand using essentially the same algorithm Sal applied in the 2x2 case here.(3 votes)

- So what would be a basic definition of the determinant? I don't really get it.(3 votes)
- Up to this point, it's the denominator of a common factor of all the terms in the inverse matrix for an arbitrary 2x2 matrix.(0 votes)

## Video transcript

I've got a 2 by 2 matrix here. Let's say A is equal
to a, b, c, d. So I'm going to keep
it really general. So this is really any
2 by 2 matrix. What I want to do is use our
technique for finding an inverse of this matrix to
essentially find a formula for the inverse of a
2 by 2 matrix. So I want to essentially find a
inverse, and I want to do it just using a formula that it
just applies to this matrix right here. So how can I do that? Well, we know a technique. We just create an augmented
matrix. So let's just create an
augmented matrix right here. So we have a, b, c, d, and then
we augment it with the identity in R2, so 1, 0, 0, 1. And we know if we perform a
series of row operations on this augmented matrix to get the
left-hand side in reduced row echelon form. The right-hand side, if the
reduced row echelon form here gets to the identity, then the
right-hand side is going to be the inverse. So let's do it in this general
case, not dealing with particular numbers here. So the first thing I want to do,
or I would like to do, is I would like to zero
this guy out. What we want to do is we want
to zero that out, zero that out, and then these two terms
have to become equal to 1. So the best way to zero this
out, let's perform a little transformation here. So if I perform the
transformation on the columns, C1, so those are the entries
of a column-- this would be one column right here, that
would be another column right there, that's the third column, that's the fourth column. But the transformation I'm going
to perform on each of these columns, and we know this
is equivalent to a row operation, is going to be equal
to-- since I want to zero this one out, I'm going to
keep my first row the same, so it's going to be C1, and I'm
going to replace my second row with a times my second row
minus c times my first row. Now why am I doing that? Because a times c minus c times
a is going to be 0. So this guy's going to be 0. That's the row operation
I'm going to perform. And I'm doing this so we can
kind of keep track, account for what we're doing because
the algebra's going to get hairy in a little bit. So let me perform
this operation. So if I perform that operation
on our matrix, what do we have. So our first row's going
to be the same. Let me start with our second
row because that's a little bit more complicated. So I'm going to replace c with
a times c minus c times a. That's ac-- so let me put it
this way-- so that's going to be 0 right there. I'm going to replace d with d
times a or a times d minus c times C1 in this
column vector. So minus c times b. Let me write this as bc. And then let me augment it. And then this guy's going to be
a times 0, because he's C2 minus c times C1. So it's going to be minus c. And then finally, this guy right
here is going to be a times 1-- a times this 1 right
here-- minus c times 0. So that's just going
to be an a. And then the first row is
pretty straightforward. We know that the first row or
the first entries in our column vectors just stay
the same through this transformation. So it's a, b, 1, 0. And just to make sure you're
clear what we're doing, when you perform this transformation
on this column vector right here, you
got this column vector right there. When you perform the
transformation on this column vector right there, you
get this column vector right there. Now I just want to make that
clear because I did all of the second entries of all of the
column vectors at once because we all were essentially
performing the same row operation, so that just helped
me simplify at least my thinking a little bit. Let me stay in this mode. So let's continue to get this
in reduced row echelon form. The next thing we want to do
is, let's make another transformation, we'll
call this T1. That was our first
transformation. Let's do another
transformation. T2, or another set of
row operations. So if I start with the column
vector C1, C2, what I want to do now is I want to keep my
second row the same and I want to zero out this character
right here. I want to zero him out. I know I'm going to keep my
second row the same, so C2 is just going to still be C2. But in order to zero this out,
what I can do is I can replace the first row with this scaling
factor times the first row minus this scaling factor
times the second row. So it'll be ad minus bc times
your first entry in your column vector minus b times
your second entry. And the whole reason why I'm
doing that is so that this guy zeroes out. So if we apply that to this
matrix up here-- let's do the first row first. So this first entry right here
is going to be ad minus bc times a, because that's C1--
let me write that down. So it's ad minus bc times a
minus b times C2 minus 0. So it's just going to be--
that second term just becomes 0. Fair enough. Now what is this guy
going to be? He's going to be-- I'll
write it out. He's going to be ad minus bc
times b minus b times your C2 in this column vector, minus
b times ad minus bc. And you can see immediately that
these two guys are going to cancel out and you're
going to get a 0 there. And then we've got
to augment it. I want to make sure I don't run
out of space, I should've started to the left
a little bit more. So what's this guy
going to be? Well I'm going to have this guy
times ad minus bc-- I'll do it in pink-- so you're going
to have ad minus bc times 1, which is just ad minus
bc minus b times C2, so minus b times minus c. So that's plus bc. So 1 times ad minus bc
minus b times minus c is equal to that. And you can immediately see
that these two guys will cancel out. You're just [INAUDIBLE] ad. And then this guy over here,
you're going to have 0 times ad minus bc, which is just
a 0, minus b times a. So you have minus ab-- just
squeeze it in there. And we know that our second
row just stays the same. Our second row just stays the
same in this transformation. So we had a 0 here, we're
still going to have a 0. We had an ad minus bc. We'll still have
an ad minus bc. We had a minus c. Then we had an a. Just like that. Now let me re-write this matrix
just so it gets cleaned up a little bit. So let me re-write
it right here. I'll do it in my orange--
well, let me do it in this yellow. So I have ad minus bc times a. And then this term right
here just became a 0. This term right here is a 0. This term right here
is an ad minus bc. And then our augmented part,
this part was just an ad. This was a minus ab. This is a minus c. And then this is an a. Now we're almost at reduced row
echelon form right here. These two things just have to be
equal to 1 in order to get reduced row echelon form. So let's define a transformation
that'll make both of these equal to 1. So if this was T2 let me define
my transformation T3. You give it a column
vector, C1, C2. And it's just going to scale
each of the column vectors. So what I want to do is I want
to divide my first entries by this scaling factor right here
so that this becomes a 1. So I'm essentially going to
multiply 1 over ad minus bc times a, so 1 over ad, bc, a
times my first entry in each of my column vectors. And then my second one I
want to divide by this. So that this guy becomes a 1. So I'm doing two scalar
divisions in one transformation. So this one's going to be 1
over ad minus bc times C2. So I'm just scaling everything
by these two scaling factors. So if you apply this
transformation to that right there, what do we get? We get a matrix. And this guy, I'm going to
divide him by ad minus bc times a, so I'm dividing
it by itself, so that guy's going to be 1. I'm going to divide 0 by
this, but 0 divided by anything is just 0. Then we're in an
augmented part. ad divided by-- so let me
write it like this. So ad-- I'm going to divide by
this-- so it's going to be ad minus bc times a-- you
immediately see that the a's cancel out. This is going to be minus
ab divided by ad minus bc times a. Once again, the a's
cancel out. And then in my second row are my
second entries in my column vectors, 0 divided
by anything is 0. So 0 divided by this thing is
going to be 0, assuming we can divide by that, and we're
going to talk about that in a second. This guy divided by this guy,
we're just dividing by himself, so it's going
to be equal to 1. Now we have minus c divided
by this, or ad minus bc. And then we have an a. a divided by ad, ad minus bc. And we're done. We put the left-hand side of
our augmented matrix into reduced row echelon form. And now this is going
to be our inverse. So let me clean it
up a little bit. So, so far we started off with
a matrix-- I'll do it in purple-- we started off
with a matrix a is equal to a, b, c, d. And now just using our technique
we figured out that a inverse is equal to this
thing right here. And just to simplify-- well let
me just write it the way I have it there, because I don't
want to skip any steps-- this is equal to d over
ad minus bc. Right, this guy and that
guy canceled out. And then we have a minus b over
ad minus bc because that guy and that guy canceled out. Then you have a minus
c over ad minus bc. And then finally you have
an a over ad minus bc, which is our inverse. But one thing might just pop out
at you immediately is that everything in our inverse is
being divided by this. So maybe an easier way
to write our inverse. We could also write our
inverse like this. We could just write it as 1
over ad minus bc times the matrix d minus b
minus c and a. And just like that we have come
up with a formula for the inverse of a 2 by 2 matrix. You give me any real numbers
here and I'm going to give you its inverse. That straightforward. Now one thing you might be
saying, hey, but not all 2 by 2 matrices are invertible. How can this be the case
for all of them. And I'll give you a question,
when will this thing right here not be defined? When is this thing
not defined? Every operation I did, I can do
with any real numbers, and this applies to any
real numbers. But when is this thing
not defined? Well it's not defined
when I divide by 0. And when would I divide by 0? Everything else you can multiply
and subtract and add a zero to anything, but you
just can't divide by 0. We've never defined what it
means when you divide something by 0. So it's not defined if ad
minus bc is equal to 0. So this is an interesting
thing. I can always find the inverse of
a 2 by 2 matrix as long as ad minus bc is not equal to 0. We came up with all of these
fancy things for invertability, you've got to
put it into reduced row echelon form, and before
that we talked being on 2 and 1 to 1. For at least a 2 by
2 matrix we've really simplified things. As long as ad minus bc does not
equal 0, we can use this formula and then we know that
a-- and it goes both ways-- a is invertible. And not only is it invertible,
but we can just apply this formula to it. So immediately something
interesting might-- you might say hey, this is an interesting
number. We should come up with
some name for it. And lucky for us, we have come
up with a name for it. This is called the
determinant. Let me write it in pink. Determinant. So the determinant of a, and
it's also written like this with these little straight lines
around a, and you could also write it like
this, a, b, c, d. But most people kind of think
this is redundant to have brackets and these lines. So then they just write it like
this, this is equal to just, they just write the
lines, a, b, c, d. I want to make this
very clear. If you have the brackets you're
dealing with a matrix. If you have just these straight
lines you're talking about the determinant
of the matrix. But this is defined for
the 2 by 2 case to be equal to ad minus bc. This is a definition
of the determinant. So we can re-write, if we have
some matrix here, we have some matrix a which is equal
to a, b, c, d. We can now write its inverse,
a inverse is equal to 1 over this thing, which we've defined
as the determinant of a times-- and let's just
see a good way of kind of memorizing this. We're swapping these two
guys, right, the a and the d get swapped. So you get a d and an a. And then these two guys
stay the same, they just become negative. So minus b and minus c. So that's the general
formula for the determinant of a 2 by 2 matrix. Let's try to do a couple. Let's try to find the
determinant of the matrix 1, 2, 3, 4. Easy enough. So the determinant of-- let's
say this is the matrix B. So the determinant of B, or we
could write it like that, that's equal to the
determinant of B. That is just equal to-- that's
this thing right here-- 1 times 4 minus 3 times 2, which
is equal to 4 minus 6, which is equal to minus 2. So the determinant is minus
2, so this is invertible. Not only is it invertible, but
it's very easy to find its inverse now. We can apply this formula. The inverse of B in this case--
let me do it in this color-- B inverse is equal to
1 over the determinant, so it's 1 over minus 2 times the
matrix where we swap-- well, this is the determinant of B. I want to be careful. B is the same thing,
but with brackets. 1, 2, 3, 4. So B inverse is going to be 1
over the determinant of B, which is equal to minus 2. So 1 over minus 2. We swap these two guys, so they
get a 4 and a 1, and then these two guys become
negative-- minus 2 and then minus 3. And then if we were to multiply
this out it would be equal to minus 1/2 times
4 is minus 2. Minus 1/2 times minus 2 is 1. Minus 1/2 times minus
3 is 3/2, minus 1/2 times 1 is minus 1/2. So that there is the
inverse of B. Now let's say we have
another matrix. Let's say we have the
matrix C, and C is equal to 1, 2, 3, 6. What is the determinant of C? It is equal to-- we could write
this way-- 1, 2, 3, 6. And it is equal to 1 times 6
minus 3 times 2, which is equal to 6 minus 6, which
is equal to 0. And there you see it's equal to
0, so you cannot find-- so this is not invertible. So we can't find its inverse
because if we would try to apply this formula right here
you'd have a 1 over 0. But we know this formula just
comes out-- that attempt to put it into reduced row echelon
form, and in that last step we just had to essentially
divide everything by these terms. So these terms
would be 0 in this matrix C that I just constructed
for you. And the reason why I knew-- I
just pulled this out of my brain-- I knew this wasn't going
to be invertible because I constructed a situation where
I have columns that are linear combinations
of each other. I have 1, 3-- you multiply that
by 2 you get 2 and 6. So I knew that these aren't
linearly independent columns. So you know that its rank wasn't
going to be equal to, so I knew it wasn't going to be
invertible, but we see that here by just computing
its determinant.