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 sides. 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 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.

i appreciate your vids : )

thank you Sal you are amazing

It's Sal!

God bless you dude I really mean it ! =)

Lol I tried solving this via transforming it into an triangular matrix and then just taking the determinant by calculating the product of the primary diagonal elements. Turned out that I copied a number wrong and I kept wondering if I have done anything wrong since my determinant always became 0. 😀

I also once used Laplace to calculate the determinants in some homework. Turned out I've forgotten about basic determinant rules and so I calculated the easiest stuff with the most difficult way.

sal why dont you teach me the math behind electromagnetics. could sure use your help there haha

Cool, I was making a Class Library in C# for matrices and needed a method for determining the determinant. It was much easier than I originally thought it would be to implement (because of recursion). Sal, you should do some Computer Science videos. I'm tired of everyone asking me to tutor them.

Thanks a lot, very helpfull indeed. Keep up the good work

sal u should post videos for computer science major !!!!!!!!

thank you so much, this is brilliant, it made such an impact!

<3

Multiplying by 0 is the shiznit.

Sal, you rock as usual.

the cramer method is fastest way of getting a 3×3

great for nontraditional students who didn't do this stuff in high school. I need these things "dumbed down" (I mean that in the most respectful way). Thanks for the help!

damn my brain wanna explode!!

I know this is just the natural function of numbers and has no sentient quality to it whatsoever but somehow this feels evil…

u rock!

have an exam tommorow ..

Dosent matter you there ! my savior thank you !

"We are afraid of what we don't understand." … Well I guess you dont get this then ;P Haha naah just kidding :))) Cheeers maan!

frickin awesome dude

Finally!!! I swear my linear algebra teacher sucks. 20 min on YouTube is better than an 1hr of lecture.

I'm glad I found your channel. It helps a lot. Thanks! and the subtitles are appreciated as well

But why does this work? I mean, why does when the det=0 when calculated this way makes the matrix not invertable?

Yes, but only because 1/0 is undefined.

How dare you insult your linear algebra teacher!

yea, QQ some more like you know his algebra teacher

Thank you so much.

Quiz tomorrow, thanks for your help!

nice

final exam in the next 3 hours, thanks 😀

you get, you get…..YOU GET

sorry i had to make a snarky comment, thanks for the help!

Hoooly shit, you just gifted me with the ability of enjoying the beauty of math! Thanks dude!

still informative. thankyou

Thank you! Very helpful and straightforward.

I love you man!!!!!!

Why is the sound gone both for videos I watch of Khan and PatrickJMT? All other youtube videos have functional sound! Why must it target the two math gurus when I am in the most of need?? 🙁

Wow, I actually get it! First math teacher who actually taught me anything! 😛

thank you, Khan Academy, you saved my life!

This video was really helpful for my comp sci class. Thanks so much!

you are an angel from heaven

would this work the same if there were negative coefficients?

gauss-jordan elimination determinant anyone?

i keep getting -7 and cant find my error, anyone get the same thing?

16:07 how is 2×3 = -6 ?

it's supposed to be 6. Then -2×6 = -12 and -2x(-12) = 24 is the final result of that part.

which software do u use in this lesson and other lessons? +Khan Academy

thank you so much, finally I understood matrices

Thanks! I wish my professors were as fluent in teaching as you are.

thank you so much

Great video! Thank you. I'm just wondering why you switch sides with minus, so you suddenly do an addition?

sir.Khan at the yellow color 0-9 why u didn't multiply (-2 | -9)

that was awesome

I noticed while I was watching that the recursive rule also applies to a 2×2 matrix. therefore I would argue that even more fundamental than the definition of the determinant for a 2×2 matrix (7:55) is the determinant for a 1×1 matrix, which is itself. (and the recursive rule can also be applied to the 2×2 matrix in this way) It's the same thing really, just thought I'd point it out for people like myself, for whom looking at it like this makes it easier to understand/remember.

How is it exactly that after a few minutes of watching this I understand how to get a determinant but the professor at my school couldn't teach this in a week without fumbling the explanation???? You're the light at the end of the tunnel, Saul. Keep up the awesome work!

what about that formula?

Great, how would you code this on matlab using recursion.

This is Cofactor expansion?

How about the determinant of a 2 by n matrix which won't have a basic (ad-bc) when you ignore the first row.Then, how do you do it?

Great,

Correct me if wrong. It's n-1 because it takes at least two factors to make a coefficient? Successive applications of multiplications (sign rules also) until a 2×2 matrix is isolated.. But the iterations are progressively going to multiply as well. A formula and help from a computer can lessen the burden of the signed arithmetic as well as the assure that a simple mistake won't bring the entire house of cards down..Plus in either case little is to be learned of the mechanics of matrices. If a 4×4 is under your belt then relax. Go back and ponder How this insane mess ever got started — Derive the reasoning or application at the origin. This is a fair attempt at running you through the game in a tiny amount of time. Although it's a good video it means reviewing recursively. :0)

… and may I add this must be considered a Magic Act. Keep your eyes directed to every change (..what was that for..) and listed to every word !..

… also the biggest mistake a teacher makes (#1) is assuming what's obvious to them is obvious to you. .. It takes a intuitive person with patience and time to make what appears overwhelming just a matter of the course..:0)

… Finally this little dip in the bottomless depths of mathematics should make all but the insane shy away unless having the type Brain geared for tedium and not allergic to chalk dust from a revolving blackboard..

.. personally I like computers because there's only '1's & '0's. math is no problemo.. :0)

Science, Mathematics together with even the statisticians can't explain the fact that a Pascal triangle has a marble which either balances on a peg or rockets out of the array. go figure ..

No no ! if this is dumbfounding just don't even consider Non-Linear Algebra.. If there really is a Hell, I'm sure it's non-linear. .. Arg

But why are we changing signs???)

I see a previous exam is computing a 6×6 without a calculator, is it really that bad? Seems really time consuming, even if you understand it.

u r not going to done 5×5 and 6×6 matrix question in 20 mins!

I was investigating n by n determinant, finally I have found and understood this type of determinant.

Actually there is another way to express determinant of arbitrary square matrix, that is using the Leibniz formula. This way we use permutation function to pair any indices of matrix elements. For details see Wikipedia about determinant of n x n matrix

Okay… The introduction to matrices and multiplation is enough for me for today…

Thanks!

you save my life

OMG！！You save me

I was about ready to drop out but now it all makes sense <3

but why is it that the determinant results in the same value regardless of the row or column from which it is expanded

Slow death

Very useful

O'boy, I can't wait to do these computations!

THANK YOU SO MUCH!!!!!