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We've done a lot of work on multiplying, adding,

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subtracting and inverting matrices.

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So now let's delve a little into what a matrix is

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actually good for.

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And remember, all a matrix is is, a way of

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representing data.

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And all of those rules we learned, you can kind of view

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them as human-created rules.

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There's no fundamental thing in nature that says matrices

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have to be multiplied the way we learned.

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But I think you'll see as we progress into applications,

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that the way that matrix operations have been defined

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are actually quite useful.

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So let's go back to our Algebra 1 or Algebra 2.

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I forget when you tend to learn it.

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But let's go back to linear equations.

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So what were linear equations?

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Systems of linear equations.

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Well you had two lines, and you essentially had to figure

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out where the two lines intersected.

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So you might have had something like-- let me think

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of something-- 3x plus 2y.

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Is equal to 7.

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And then you might have, minus 6x plus 6y is equal to-- I

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need to do this in my head just to make sure that I get

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numbers that work out well-- equal to 6.

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I think this will work out well.

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And what was this problem essentially?

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Well this is a line, and this is a line.

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So you had to figure out where they intersected.

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And if you were to draw those two lines--

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Actually let's draw them.

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Just because this is all about getting intuition, and seeing

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how it maps over into the matrix world.

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And the word 'matrix world' has a whole new

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meaning after 1999.

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Let's see, so if that's my coordinate axes, what is this?

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I always have to put everything into y equals mx

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plus b form for me to-- So this equation is what?

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It's y is equal to 3/2 x plus 7/2.

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So 7/2 is what?

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It's like 3 1/2 or something?

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So if that's 7/2, that's going to have a slope of 3 1/2.

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So it's a little bit steeper than a slope of 1.

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So it's going to look something like that.

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That's that line.

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And then this line is going to look like what?

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I'll do a different color.

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It's going to look like-- It's the same thing is as--

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Oh, you know what?

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I did that wrong.

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Because that line, I just realized, is equal to

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minus 3x plus 7/2.

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Because when you take this to the other side, it becomes

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minus 3x divided by 2, so it's actually going

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to be downward sloping.

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So it will look something like this.

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It's going to be a little bit steeper than something that

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has a slope of negative 1, so I'm just approximating.

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So that line will look something like that.

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And this line, this will be y-- I'm just rewriting this--

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y is equal to x plus 1, if I'm right.

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Yeah.

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Because this go to the other side.

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Divide everything by 6.

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y is equal to x plus 1, so its y intercept will be-- We said

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this was 3 and 1/2, so maybe if this is 1.

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And it has a slope of 1.

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Then it'll just look like something like this.

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And so when you solve a system of equations, you're

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essentially looking for the x and the y values that satisfy

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both of these equations.

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This magenta line shows us all the x and y values that

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satisfy this first linear equation.

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And this green line shows all of the x's and y's that

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satisfy the second equation.

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And of course where they intersect shows us the

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particular x and y that satisfies both equations.

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So that's what we did in Algebra 1.

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We'd solve both of these equations for that.

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And we'd either do it by substitution, or we'd scale

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them and add them together, et cetera, et cetera.

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As you'll see, that's essentially just what we

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learned in the Gauss-Jordan elimination.

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It's the exact same thing.

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It's just when we did the Gauss-Jordan elimination, we

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just represented it a little different.

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But I think you know this much.

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But let's now do it in the matrix world.

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So how can we represent this problem as a matrix?

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We could write it like this, and we'll take out a little

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time to prove to you that it really is the same

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representation.

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If you define matrices the way we have defined them in their

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multiplication.

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You can define this problem as 3, minus 6, 2, 6.

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I just took the coefficients, 3, minus 6, 2, 6.

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And if I were to multiply that by soon. column

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vector matrix xy.

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And if I were to set that equal to another column vector

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matrix, 7, 6.

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Now you might want to pause it and actually just try to

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multiply this out, the way that we have learned to

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multiply matrices.

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And you will see that you get the same thing.

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But I will do it now, in case you don't

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want to do it yourself.

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So let's just multiply these two matrices.

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Let's multiply this matrix out and see what happens.

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So what do you do?

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You get your row information from the first matrix, column

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information from the second matrix.

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And this is of course the product matrix.

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So this is saying, 3 times x plus 2 times y is equal to 7.

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Well that's exactly what we wrote up here.

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3 times x plus 2 times y is equal to 7.

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And similarly when you multiply the bottom row, you

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get minus 6 times x plus 6 times y is equal to 6.

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So if that was a little confusing to you, go review

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how we multiply matrices.

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But if you just multiply this out, you'll get these exact

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same equations.

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So hopefully you understand that this is just another way

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of representing this problem.

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Although we've gotten rid of the plus signs

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and the equals signs.

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But of course you have to know the representation.

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But why is this useful?

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Why is this representation useful?

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Well let's call this matrix a.

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Let's call this vector x.

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It's not a variable.

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It's a vector.

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So maybe we'll bold it, or we'll put a little vector sign

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there or something.

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Whatever.

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But you'll see it in your textbook.

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It's bolded real heavy.

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And then we call this vector b.

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And the general notation-- if I remember it correctly-- is

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that anything that's a matrix or a vector is bolded.

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And matrices that are not vectors, that have more than

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one dimension in either of the dimensions,

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they're capital letters.

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While lower-case letters represent vectors.

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So these are matrices, but they're also vectors.

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So that's why they got the lowercase letters.

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And that's why this one got the uppercase letters.

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That's just convention.

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So this equation has the form ax equals b, where a is this

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matrix, x is this vector-- or this matrix, same thing-- and

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b is this column vector.

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So what does that do for us?

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Well, what happens if we knew a inverse?

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Well actually, let me take a step back.

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If these were numbers, what would we do?

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If I just gave you an algebra equation, ax is equal to b.

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How do you solve that?

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Well you would divide both sides of this equation by a.

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Or another way of saying it, you would multiply both sides

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of this equation by the inverse of a.

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So you would essentially say, 1/a times ax is equal

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to 1/a times b.

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And then these would cancel out, and you would get x is

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equal to b/a.

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That's how we would do it in a traditional

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simple, linear equation.

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So how would you do it here?

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Well what's the matrix analogy to division?

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And I'm going to give you the answer now.

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What's the analogy to multiplying by your inverse?

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Well, it's multiplying by your inverse.

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So what if we knew the matrix a inverse?

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We could just multiply both sides of this

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equation by a inverse.

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And remember, order matters.

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So it's not like when you're doing a linear equation you

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could multiply 1/a on this side.

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But then you can do it on the right side here.

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But no.

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Notice, I put it in front of the numbers in both cases.

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So you have to do it in front of the numbers in both cases.

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But if we know a inverse, and if a inverse exists, then we

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can multiply both sides-- you can say the left side of both

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sides of this equation by a inverse.

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a inverse times a, times the vector x is equal to a

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inverse times b.

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All I did is I took this expression, and I multiplied

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both sides by a inverse.

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And what's a inverse times a?

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Well that's just the identity matrix.

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That's the identity matrix, times x is equal

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to a inverse b.

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And of course that's just x.

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The identity matrix times any other matrix

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is just that matrix.

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So that's just the matrix x, or the vector x

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times a inverse b.

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So, if you're given a linear equation, and if you know the

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inverse of this matrix, to solve for x and y, we just

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have to multiply this number times the inverse.

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And you might say, Sal, that's such a pain.

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Because this is such a simple linear equation to solve.

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Why would I go through all the trouble of getting an inverse,

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and then multiplying the inverse times this number.

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And I would agree with you to some degree.

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That for a 2 by 2 system of equations, it is easier to

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solve it the way that you did it in Algebra 1 or Algebra 2.

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But if you're doing it for a 3 by 3, well, finding a matrix

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is still pretty difficult for a 3 by 3,

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so it's still difficult.

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But as you get to larger and larger numbers, it's

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sometimes-- well, finding a matrix can be difficult too--

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But actually the real place where it you really, really

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pays off is, let's say that you have a bunch of linear

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equations to solve.

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And the left hand side stays the same.

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But you keep changing the right hand side.

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So let's say you have ax equals b.

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And then you have another one that says, ax equals c,

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and ax equals d.

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And these numbers keep changing.

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And these numbers are the same.

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Then it really pays off to solve for the inverse.

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And then every time you need to find a new solution, you

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just multiply your new right-hand side times your

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inverse, and you just get the answer.

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And that'll really pay off when we view

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this in another way.

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But anyway, I wanted to show you that

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this is the same thing.

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And so let's solve it using what our

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knowledge is of matrices.

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Let me erase this here, and I know I'm running over time,

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but hopefully I'm not completely boring you.

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So I'm going to keep that there, just because I think

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it's nice to have that visual representation

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of what we're doing.

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Always to remember what's going on.

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So, what's a inverse?

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So first of all, the a inverse is equal to 1 over the

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determinant of a times the adjoint of this matrix.

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I don't want to get fancy with terminology and all that, but

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what was that?

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2 by 2 is fairly easy.

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You swap these two terms. You get a 6 and a 3.

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And then you make these two terms negative.

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So a minus 6 becomes a 6.

262
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And a 2 becomes a minus 2.

263
00:12:27,779 --> 00:12:30,500


264
00:12:30,500 --> 00:12:32,870
And what's the determinant of a?

265
00:12:32,870 --> 00:12:36,610
The determinant of a is equal to this times this minus this

266
00:12:36,610 --> 00:12:37,070
times this.

267
00:12:37,070 --> 00:12:38,830
So 3 times 6.

268
00:12:38,830 --> 00:12:45,840
3 times 6 is 18 minus this times this.

269
00:12:45,840 --> 00:12:47,540
So 6 times 2 is 12.

270
00:12:47,539 --> 00:12:48,429
That's a minus 6.

271
00:12:48,429 --> 00:12:49,319
That's minus 12.

272
00:12:49,320 --> 00:12:52,460
So minus minus 12.

273
00:12:52,460 --> 00:12:56,280
It's equal to plus.

274
00:12:56,279 --> 00:12:58,689
So 18 plus 12 is equal to 30.

275
00:12:58,690 --> 00:13:00,350
So what does a inverse equal?

276
00:13:00,350 --> 00:13:02,090
1 over 30 times this thing.

277
00:13:02,090 --> 00:13:10,870
So a inverse is equal to-- we could even keep the 1/30 on

278
00:13:10,870 --> 00:13:12,120
the outside.

279
00:13:12,120 --> 00:13:13,830


280
00:13:13,830 --> 00:13:15,490
That might simplify things.

281
00:13:15,490 --> 00:13:16,519
Well actually I'll put it--

282
00:13:16,519 --> 00:13:19,889
So a inverse is equal to what?

283
00:13:19,889 --> 00:13:21,120
This divided by 30.

284
00:13:21,120 --> 00:13:33,299
So that's 1/5, minus-- Actually I do want to keep it

285
00:13:33,299 --> 00:13:36,179
on the outside, because that's going to make the later

286
00:13:36,179 --> 00:13:38,079
multiplications easier.

287
00:13:38,080 --> 00:13:46,570
So anyway, a is equal to 1/30 times 6, minus 2, 6, 3.

288
00:13:46,570 --> 00:13:47,680
That's a inverse.

289
00:13:47,679 --> 00:13:49,370
So now let's solve for x and y.

290
00:13:49,370 --> 00:13:53,340
So we said x and y is equal to a inverse times b.

291
00:13:53,340 --> 00:13:58,990
So we could say x-- another way to write x is like this.

292
00:13:58,990 --> 00:14:00,820
x is just this vector.

293
00:14:00,820 --> 00:14:02,090
x and y.

294
00:14:02,090 --> 00:14:04,785
Not to get confused, this x is different than that x, even

295
00:14:04,784 --> 00:14:05,569
though I've written them the same.

296
00:14:05,570 --> 00:14:08,430
If I was a typographer, I would make this really bold,

297
00:14:08,429 --> 00:14:10,039
so that you know that this is a vector.

298
00:14:10,039 --> 00:14:11,889
Maybe I should put a vector notation.

299
00:14:11,889 --> 00:14:12,149
I don't know.

300
00:14:12,149 --> 00:14:14,480
You could do a bunch of things with it.

301
00:14:14,480 --> 00:14:17,889
It's equal to a inverse times this.

302
00:14:17,889 --> 00:14:20,659
So that's 1/30.

303
00:14:20,659 --> 00:14:22,490
I did that just for the matrix addition.

304
00:14:22,490 --> 00:14:26,919
I didn't divide everything by 30, just so the matrix

305
00:14:26,919 --> 00:14:28,579
multiplication's a little easier.

306
00:14:28,580 --> 00:14:32,509
Minus 2, 3, times 7/6.

307
00:14:32,509 --> 00:14:37,929


308
00:14:37,929 --> 00:14:40,615
And so what is this equal to?

309
00:14:40,615 --> 00:14:45,699
It's equal to 1/30 times-- I know I'm crowding this down

310
00:14:45,700 --> 00:14:46,810
here-- let's see.

311
00:14:46,809 --> 00:14:52,439
6 times 7 minus 2 times 6.

312
00:14:52,440 --> 00:14:59,240
So 6 times 7 is 42.

313
00:14:59,240 --> 00:15:01,690
Minus 2 times 6, so minus 12.

314
00:15:01,690 --> 00:15:04,830
So that's equal to the 30.

315
00:15:04,830 --> 00:15:07,220
And then 6 times 7 plus 2 times 6.

316
00:15:07,220 --> 00:15:11,090
So 6 times 7, once again is 42.

317
00:15:11,090 --> 00:15:12,820
Plus 2 times 6.

318
00:15:12,820 --> 00:15:17,420
So 42 plus 12 is 50.

319
00:15:17,419 --> 00:15:19,620
Is that right?

320
00:15:19,620 --> 00:15:22,990
6 times 7-- oh I'm sorry.

321
00:15:22,990 --> 00:15:23,700
This is a 3.

322
00:15:23,700 --> 00:15:25,320
That's why I was getting confused.

323
00:15:25,320 --> 00:15:27,700
See, it's important to have good penmanship.

324
00:15:27,700 --> 00:15:33,565
So it's 6 times 7 is 42, plus 3 times 6.

325
00:15:33,565 --> 00:15:38,120
So it's 42 plus 18, which is 60.

326
00:15:38,120 --> 00:15:40,480
And of course you divide both of them by 30.

327
00:15:40,480 --> 00:15:43,769
So you get the final xy.

328
00:15:43,769 --> 00:15:45,449
I'll write it here.

329
00:15:45,450 --> 00:15:46,480
I don't want to erase anything.

330
00:15:46,480 --> 00:15:52,850
So we get xy is equal to-- divide both of those by 30--

331
00:15:52,850 --> 00:15:58,409
is equal to 1 and 2.

332
00:15:58,409 --> 00:16:02,419
And so that tells us that these two linear equations

333
00:16:02,419 --> 00:16:10,159
intersect at the point x is equal to 1, y is equal to 2.

334
00:16:10,159 --> 00:16:12,610
That might seem a little bit like a lot of work, but that's

335
00:16:12,610 --> 00:16:14,480
just because I took the time to explain it and all that.

336
00:16:14,480 --> 00:16:17,000
But if you just immediately took that, represented it this

337
00:16:17,000 --> 00:16:19,240
way, found the inverse, and multiplied, it wouldn't have

338
00:16:19,240 --> 00:16:20,049
taken you that much time.

339
00:16:20,049 --> 00:16:22,099
And I encourage you to do that as an exercise.

340
00:16:22,100 --> 00:16:23,529
Anyway, I'll see you in the next video.

341
00:16:23,529 --> 00:16:25,610
And in the next video, we're going to do this exact same

342
00:16:25,610 --> 00:16:29,009
problem, but we're going to see that this data represents

343
00:16:29,009 --> 00:16:30,370
a different problem.

344
00:16:30,370 --> 00:16:31,620
See

345
00:16:31,620 --> 00:16:31,899