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##### Linear Algebra: Formula for 2x2 inverse : Figuring out the formula for a 2x2 matrix. Defining the determinant.

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- 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.