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MARTINA BALAGOVIC: Welcome.
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Today's problem actually
appeared in a quiz.
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It appeared in quiz one in
fall of 1999 as question four.
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The problem puts the usual solve
the following system upside
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down by saying we have some
matrix and we know that all
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the solutions to A*x equals
this vector here, [1, 4, 1, 1],
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all the solutions to this
problem are given by x equals
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[0, 1, 1] plus any
number c times [1, 2, 1].
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And we're asked
to say everything
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that we can about the
columns of the matrix A.
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So I'm going to let you pretend
that you are on an exam,
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try to solve it yourself,
and then come back
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and compare your
solution with mine.
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OK, welcome back.
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So the first thing
that you should
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think about in this
sort of situation
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is what is the size of A?
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Well, we want to multiply A with
an x that has three entries,
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so A should have three columns.
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Let me call those columns
c_1, c_2, and c_3.
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And when I take some linear
combinations of c_1, c_2
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and c_3, I'm going to get this
vector here, [1, 4, 1, 1].
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So all the c_i's, c_1, c_2,
and c_3 are vectors in R_4.
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Now, if you know
about, if you learned
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about particular solutions
and special solutions, then
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my notation here
shouldn't surprise you.
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I'm going to call
this vector here x_p,
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and this vector here x_s.
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And I'm going to
use the fact that
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x_p plus c times x_s
satisfies A times this
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equals b-- I will call this
vector b-- for any number c.
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In particular, what
I'm going to conclude
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is that when c equals 0 we
get A times x_p equals b.
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But also that when
c equals 1, we
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get A times x_p plus
A times x_s equals b.
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Replacing this by b, we
get that this implies
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that A times x_s equals 0.
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So in trying to find
what are the columns
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c_1, c_2, and c_3 of the matrix
A, let's look at these two
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equations.
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x_p satisfies A
times x_p equals b,
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and x_s satisfies A
times x_s equals 0.
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Again, if you know what
particular and special
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solutions are this
shouldn't surprise you.
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But we also know
what x_p and x_s are,
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so let's use them to try to
calculate c_1, c_2, and c_3.
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A times x_p equals b means that
the linear combination of c_1,
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c_2, and c_3 encoded in the
vector x_p, which is [0, 1, 1],
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gives the vector b.
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So c_1, c_2, c_3 times
[0, 1, 1] gives us [1, 4, 1, 1].
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In other words, c_2
plus c_3 equal b.
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Let's turn our attention
to A times x_s equals 0.
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This says that c_1,
c_2, c_3 times--
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x_s was defined to be
[0, 2, 1]-- equals 0.
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In other words, 2 times
c_2 plus c_3 equals 0.
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Now solving this system where
the unknowns are vectors
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but it's still just a
linear system, we can see,
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for example, from the second
equation that c_3 equals minus
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2*c_2.
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And plugging it back into
the original equation,
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getting c_2 minus
2*c_2 equals b,
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from which it follows that
c_2 is equal to minus b,
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and that c3 is
equal to 2 times b.
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So from this tiny
amount of information--
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we just knew the solutions to
this one particular equation
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involving A-- we got
the second column of A
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and the third column of
A completely explicitly
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calculated.
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Now, what can we say
about the first column?
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I said before that all the
solutions of A*x equals b are
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of the form a particular
solution plus some number times
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a special solution.
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And the information that we have
is that there's just one number
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here.
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So we said everything, once
we remove this vector here,
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everything that we get here
will satisfy A times x equals 0.
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And the fact that
everything that
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satisfies A times x equals 0 is
a multiple of this one vector
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that was given to us means
that the null space of A
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has dimension one.
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There is just one
special solution.
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So dimension of the
null space of A is 1.
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So rank of A is the
number of columns
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minus this dimension of null
space, and it's equal to 2.
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As rank of A is equal
to 2, the number
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of linearly independent
columns needs to be 2 as well.
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So the only thing that
we can say at this point
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is if the first column
was also a multiple of b,
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as the second and the third are,
then the rank would be smaller
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than 2.
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So that is the only
thing that cannot happen.
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So c_1 is not a multiple of b.
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Not any multiple, including
not a zero multiple.
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And that's pretty much
everything we can say.
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Yes, if it was some
other multiple of it,
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then we would be able to
find some other vector here
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and we would have
two parameters.
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But it's not, and
this is everything
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that we can say about it.