WEBVTT
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MARTINA BALAGOVIC: Hi.
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Welcome to recitation.
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Today's problem is
about change of basis.
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It says the vector
space of polynomials
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in x of degree up to 2 has
a basis 1, x, and x squared.
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That's the obvious
basis that you would
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write for that vector space.
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But today we're going to
consider another basis, w_1,
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w_2, and w_3.
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And we don't know what w_1,
w_2, and w_3 are explicitly.
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What we know is that
their values at x
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equals minus 1, 0, and 1 are
given by this table here.
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So they are 1, 0, 0;
0, 1, 0; and 0, 0, 1.
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We're asked to do the following.
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We're asked to express
this polynomial-- so y
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of x is minus x plus 5-- in
this basis, w_1, w_2, w_3.
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We're asked to find
the change of basis
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matrices between these two
bases, 1, x, x squared,
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and w_1, w_2, w_3.
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And finally, we're asked to
find the matrix of taking
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derivatives, which is a
linear map on this space,
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in both of these basis.
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And let me give you an extra
level of challenge, which
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is to try to do as much of this
as possible without explicitly
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finding w_1, w_2, and w_3.
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I'll let you think
about the problem
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and then you can come back
and compare your solution
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with mine.
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Hi.
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Welcome back.
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So to start with
the problem a, we
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need to find coefficients alpha,
beta, and gamma so that y of x
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is expressed through,
with this coefficient,
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in this new basis,
w_1, w_2, and w_3.
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now, one way to do
that would be to look
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at this table of
values, explicitly
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find w_1, w_2, and w_3, so
a quadratic polynomial is--
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all the information
we need about it
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is in values at three points.
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So say w_1 is a*1 plus b times
x plus c times x squared.
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Find a, b, and c.
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Find w_1, w_2, w_3 explicitly,
and then go back to this system
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and try to find
alpha, beta and gamma.
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However, there's a trick.
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Let's try to see if we
can do it without finding
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w_1, w_2, and w_3 explicitly.
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So let me try to see
what are the values of y
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at these points.
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So y is minus x plus 5.
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So the values are 6, 5, and 4.
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And let me, instead of
considering this equation,
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let me evaluate it at x is
minus 1, x is 0, and x is 1.
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What I get through this is
that w at minus 1-- actually,
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let me write this.
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I get w at minus 1,
which is a number,
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equals alpha times
w_1 at minus 1,
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which is a number, plus
beta times w_2 at minus 1,
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plus gamma times w_3 at minus 1.
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And similarly at 0,
and similarly at 1.
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And now, let me think of
this as a linear system that
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has unknowns alpha,
beta, and gamma
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coefficients, these values
here at minus 1, 0, and 1.
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And the right-hand
side, well, what's
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written here on the left-hand
side. y at minus 1, y at 0,
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and y at 1.
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If you write this in a matrix
and read off coefficients
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from there, you get
the following system.
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So this is the matrix of the
system read off from here.
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These are the unknowns.
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And these are the values
of the right-hand side.
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And I hope you'll agree that
this is a very easy system
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to solve.
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We just get alpha is 6,
beta is 5, and gamma is 4.
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So the solution to the first
part is y equals 6*w_1 plus
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5*w_2 plus 4*w_3.
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And let's notice another thing.
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No matter what
values we put here,
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this matrix is always
going to stay the same.
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It's only the right-hand
side that's going to change.
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So if we're given
any other polynomial
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now to express in a
basis w_1, w_2, and w_3,
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we don't have to
do any thinking.
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We don't have to do
any computations.
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What we do is go back to
our table at the beginning
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and just read off-- let's
go back to the table,
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and just read these values.
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So in this case, y is 6
times w_1, 5 times w_2,
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and 3 times w_3.
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And that's already
a hint to solving
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the b part, which is
find the change of basis
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matrices between 1, x, x
squared and w_1, w_2, w_3.
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Change of basis matrices
means expressing
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one basis in terms of another.
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So as a part of the
problem, we will
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have to express 1,
x, and x squared
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in terms of w_1, w_2, w_3.
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So let's just find their
values at these three points.
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One is a constant, it just
takes value 1 everywhere.
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x takes value minus 1 at
minus 1, 0 at 0, and 1 at 1.
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And x squared takes values 1,
0, and 1 at minus 1, 0, and 1.
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And from this we can already
conclude the part b here,
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we can conclude that 1
equals w_1 plus w_2 plus w_3.
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That x equals
minus w_1 plus w_3.
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And that x squared
equals w_1 plus w_3.
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And from this, we
can immediately
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write one change
of basis matrix.
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Namely, since we know
how to express 1, x,
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and x squared in terms
of w_1, w_2, and w_3,
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we can just copy
this information over
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to this matrix, getting 1, 1,
1; minus 1, 0, 1; and 1, 0, 1.
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So which matrix is this?
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This is a matrix--
so we have 1, x,
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and x squared expressed in
terms of w_1, w_2, and w_3.
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So if we feed this matrix
something expressed
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in the basis 1, x, and
x squared, say a, b,
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and c, what it's
going to throw out
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is the same polynomial expressed
in this basis here, w_1, w_2,
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and w_3.
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So I'm going to just write that
this is a matrix of this basis
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change.
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How do we get the other one?
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Well, very easy.
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We know it's just
the inverse of A.
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So this is going to
be the matrix that
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takes something
written in this basis
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and transfers it to this basis.
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I'm not going to
calculate the inverse
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of a matrix in front of you.
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Instead I'm going to
consult my oracle.
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Sorry about that.
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And my oracle says
that the inverse
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should be 0, 1, 0; minus 1/2,
0, 1/2; and 1/2, minus 1, 1/2.
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And that solves the b part.
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Let's go into the c part.
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The c part required us to find
a matrix of taking derivatives,
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which is a linear map in
the space of polynomials,
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in both of these basis.
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So let's first do the
1, x, x squared basis
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because that one's easier.
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I'm going to call it D_x.
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So I'm going to work in
basis 1, x, x squared.
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And what I want to express
is the transformation
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of taking derivatives.
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So here I'm going
to write the vector
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to which taking derivatives maps
the polynomial 1, which is 0.
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And that's this expressed in
the basis 1, x, x squared.
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In the second column I'm
going to write x prime,
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the vector to which
D_x sends to vector
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x, and that's equal to 1,
which expressed in this basis
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is 1, 0, 0.
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And here I'm going
to write x squared
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prime, which is 2x, which,
expressed in this basis,
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is just this.
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That one was easy.
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For the other one,
well we could calculate
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w_1, w_2, w_3 explicitly,
take the derivatives,
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go back to the table
and repeat the procedure
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that we did already.
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So expressing these derivatives
in terms of w_1, w_2, w_3,
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and that's a lot of work.
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But we pretty much already
did most of this work.
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So we know how to take
derivatives in this basis,
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and we know how to go
between these two basis.
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So if we want to take a
derivative of something written
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in the basis w_1, w_2,
w_3, well, let's first
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write this something in
basis 1, x, x squared.
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Then let's take a
derivative of it.
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And then let's write it
back in the original basis
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that we want.
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So it's multiplication
of three matrices.
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We have all three-- matrix
multiplication is easy.
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And my oracle, again, says
that this should be minus 3
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over 2, 2, minus 1 over 2;
minus 1 over 2, 0, 1 over 2;
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and 1 over 2, minus 2, 3 over 2.
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And that solves the problem.
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Now, one thing that I would
want to discuss in the end
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is how did you do with
respect to my challenge, which
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was let's try to do as
much of it as possible
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without finding w_1,
w_2, and w_3 explicitly.
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And it seems like
we did really well.
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There's nowhere on the
board written w_1 equals,
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w_2 equals, w_3 equals.
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But is it really so?
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It's not.
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We calculated the matrix
of A inverse here.
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And what this really means is
that w_1, w_2, and w_3, written
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in the basis 1, x, and x
squared, are as follows.
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w_1 is minus 1/2 x
plus 1/2 x squared.
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w_2 is 1 minus x squared.
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And w_3 is 1/2 x
plus 1/2 x squared.
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So you can check your work
with the help of this matrix
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in case you did find w_1,
w_2, and w_3 explicitly.
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And that's all I
wanted to say today.