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PROFESSOR: Hello, I'm Linan.

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Welcome to the recitation
of Linear Algebra.

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It's my great
pleasure to guide you

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through the first recitation.

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In the first lecture, we
learned some important concepts.

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We discussed how to view a
linear system of equations

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from different points.

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And we discussed the concepts
such as row picture, column

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picture, and the form in matrix.

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Some of them may be new to you.

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So today, we're going to
use this simple example

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to review those concepts.

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We're going to work with
this simple system of two

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equations with two unknowns.

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So I would like you
to first solve it,

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and then to find out the
associated row picture

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and column picture
with this system.

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After you're done,
we're also going

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to discuss the matrix form
of this linear system.

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Why don't you pause
the video now and try

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to work them out on your own.

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A good suggestion
would be, you can

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try to sketch your answer in
an xy coordinate like this.

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OK, I'll see you in a while.

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I hope you have just
had some fun with it.

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Let's work on it together.

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Well, we're going to
solve these equations.

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As you can see, we have
two unknowns, x and y.

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And they have to satisfy these
two equations at the same time.

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How would you solve it?

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A very simple way would
be you substitute x by y,

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in terms of y.

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So let's do it this way.

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So we use the second equation
to rewrite x as 2y minus 1.

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So 2y minus 1.

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Then you plug this into
the first equation.

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This implies twice-- x would
be replaced by this-- 2 times

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2y minus 1 plus y is equal to 3.

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When you simplify
this equation here,

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you will arrive at that
5y minus 2 is equal to 3.

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That simply tells
you y is equal to 1.

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If y is equal to 1, then
we go back to this formula.

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We see that x is
also equal to 1.

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That's it.

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This is the answer to
this linear system.

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And it's easy enough.

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Both x and y are 1.

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Now, let's try to find out its
row picture and column picture.

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So I'm going to work on
this xy coordinate here.

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First, let's look
at row picture.

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Please review what
a row picture is.

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So by row picture,
I mean you have

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to look at this linear
system according to each row.

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So what is the first row?

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Well, the first row is an
equation with two unknowns.

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So 2 times x plus
y is equal to 3.

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What is this equation, exactly?

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As you may remember,
this equation

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actually gives you a
straight line in xy-plane.

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So now let's put the line here.

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I want a line that
satisfies 2x plus y is 3.

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Let's first set x to be 0.

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If x is 0, by the
first equation,

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y should be 3, which is here.

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Now, let's set x to be 1.

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If x is 1, then y is also 1.

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So what I have now are
two points on the line.

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And this is enough,
because on the xy-plane,

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two points will uniquely
determine a straight line.

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And that's the line
we're looking for.

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So all we need to do is to
connect these two points.

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Let me try to draw
this line straight.

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So this is the
line that is given

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by 2x plus y is equal to 3.

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This is the easiest way
to determine a line.

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You just need to pick two
points that are on the line.

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Then you connect them, and here
you have the straight line.

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So that's the first row.

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Now let's look at
the second row.

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The second row is x
minus 2y is equal to -1.

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Same thing.

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We're going to locate two
points on the second line.

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So again, let's set
x to be 0 first.

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If x is 0, then y
should be 1/2, right?

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So this line has to
cross this point.

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That's x, 0; y, 1/2.

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Then let's put x to be 1 again.

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So if x is 1, then y is also 1.

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So we're going to use
the same point here.

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Now we have two points in the
second line to connect them.

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So by connecting them, I will
have my second straight line.

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So this is the line
corresponding to x minus 2y

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is -1.

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This is the row
picture, because we

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have separated
the two equations,

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and we look at
them respectively.

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The first equation
gives me this line,

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and the second equation
gives me this line.

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Then what do I mean by
solving this linear system?

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Well, we are putting
these two lines together,

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and we are looking
at a point which

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is on the first line and
second line at the same time.

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Then clearly, that's the point
where they intersect, right?

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We can see from the
answer over there,

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the coordinate of this
point should be (1, 1).

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This is also clear from the
construction of the two lines.

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We have noticed
that this point is

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on the first line and the
second line at the same time.

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Two lines meeting
at the point (1, 1).

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That's the row picture
of this linear system.

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Now, let's move on to
the column picture.

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Again, I will need
an xy coordinate.

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So, where can I find my columns?

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If you look at
the two equations,

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then you focus on the
coefficient in front

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of x in both equations.

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What would that be?

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Well, in the first equation,
I have a 2 in front of x.

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In the second equation,
I have a 1 in front of x.

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I want to put them together
as a column vector.

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Let me call it v_1.

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And I'm going to do the same
thing to the coefficients

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in front of y.

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In the first equation, I
have a 1 in front of y.

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In the second equation, I
have a -2 in front of y.

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Put them together.

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Call it a column vector v_2.

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These are the columns
I'm looking for.

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Then what does that
linear system say?

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Well, I have extracted the
coefficients in front of x.

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Now I can consider x to be the
coefficient of this vector.

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And same thing,
I'm going to view y

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as the coefficient
of this vector.

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Then what that linear system is
doing is just to sum them up.

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That gives you the left-hand
side of the linear system.

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What is the right-hand side?

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Again, you put the two
constants as a column

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vector, which is [3, -1].

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That is the right-hand side.

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So what I'm doing here is I'm
taking the linear combination

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of v_1 and v_2.

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And the coefficients are given
by x and y, respectively.

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And I want the result of this
combination to be [3, -1].

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Now let's incorporate
this into this picture.

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I have a v_1, so I'm going
to draw a vector v_1.

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x is 2, y is 1, so v_1 is here.

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That's v_1.

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And I need a v_2.

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x is 1.

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y is -2.

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So that's my v_2.

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I want to take the sum
of x multiple of v_1

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and y multiple of v_2.

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And I want the result
to be [3,  -1].

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Well, taking a hint from
the previous consideration,

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we know that both x
and y should be 1.

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So I'm actually summing one
copy of v_1 and one copy of v_2.

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So how do you indicate the
sum of these two vectors?

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You complete the parallelogram
spanned by these two vectors.

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Then the vector
given by the diagonal

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is the sum of v_1 and v_2.

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Is this vector going
to be [3,  -1]?

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Well, we can check.

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The x-coordinate will
be 2 plus 1, which is 3.

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That's 2 plus 1, 3.

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And the y-coordinate
will be 1 minus 2.

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So 1 minus 2,
which should be -1.

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That's it.

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That's one multiple of v_1
and one multiple of v_2.

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The sum will be [3, -1].

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And that's the row picture.

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Where does that that "x is equal
to y is equal to 1" come from?

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It comes from solving
the linear system.

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It comes from the row picture.

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So here, we have found out
the row picture and the column

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picture of this linear system.

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What I would like to
mention is the matrix form

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of this linear system.

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So what is the matrix form?

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What if I put these two
column vectors together?

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So I want to put them
back to back, v_1 and v_2.

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And I call this matrix to be
A. So if you write it out,

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A should be given
by 2, 1; 1, -2.

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Matrix A has v_1 and v_2
as its column vectors,

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and it's a 2-by-2 matrix.

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If I consider-- if I take
this into account, then

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what will be the left-hand
side of the linear system?

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In other words, what would
be the left-hand side

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of this equation?

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This is actually matrix
A multiplying a vector

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given by x, y.

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So that's [2, 1; 1, -2]
multiplying x and y.

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So you put both unknowns
together as a column vector.

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That's the left-hand
side of the equation.

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And again, the
right-hand side is given

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by this column vector, [3; -1].

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This is the matrix form
of this linear system.

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We can actually
solve this directly.

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In other words, we can get
this unknown vector at once,

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both x-coordinate
and y-coordinate.

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Let's recall, if you have a
scalar equation like this--

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let's say a is some
constant times x is unknown

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is equal to b.

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If a is non-zero,
what would be x?

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So clearly, x
should be b over a.

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I can also write it
as a inverse times b.

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That's what we do
when we have numbers.

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So here, what we
want to apply is

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a similar idea but to matrix.

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What you want to
find is a matrix

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A inverse such that
A inverse times A

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is equal to an identity
matrix, which is [1, 0; 0, 1].

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This may be new to you,
but as you go further

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into this course, this idea will
become more and more natural.

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If such an inverse
matrix exists, then

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what would be this vector here?

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Then [x; y] will simply be
A inverse times [3; -1].

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That will give you the answer.

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Here, I'm not going
to go into detail,

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but we will return to
this later in this course.

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I hope this simple
example is helpful to you

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in reviewing what you've
learned in the lecture.

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Thank you for watching,
and I'm looking forward

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to seeing you again.