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JOEL LEWIS: Hi.

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Welcome back to recitation.

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You've been learning in
lecture about matrices

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and their various
applications, and one of them

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is to solving systems
of linear equations.

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So I have here a system of
three linear equations for you.

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2x plus c*z equals 4, x
minus y plus 2z equals pi,

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and x minus 2y plus
2z equals minus 12.

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So what I'd like you
to do is the following.

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Find the value of c--
or all values of c--

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for which, first of all,
there's a unique solution

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to this system.

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Second of all, for which the
corresponding homogeneous

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system has a unique solution.

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So remember that the
corresponding homogeneous

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system is the system where you
just replace these constants

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on the right by 0.

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So it's a very
similar-looking system.

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The left-hand sides
are all the same,

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but the right-hand sides
are replaced with 0.

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So you want to find the value
of c for which this system has

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a unique solution,
the value of c

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for which the corresponding
homogeneous system has

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a unique solution, and
also the values of c

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for which the corresponding
homogeneous system has

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infinitely many solutions.

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Note that I'm not asking you to
solve this system of equations,

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although you're welcome
to do so if you like.

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Although, of course,
whether you can or not

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might depend on the value of c.

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So why don't you pause the
video, take a little while

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to work out the solutions
to these three questions,

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come back, and we can
work it out together.

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So hopefully you have some luck
working out these problems.

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Let's start working
through them together.

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So I'm actually going
to take parts a and b

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together at the same time.

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And the reason that
I'm going to do

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that is that one
thing you've learned

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is that a system has
a unique solution for,

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on the right-hand side-- sorry--
a system has a unique solution,

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like this, a square
system of linear equations

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has a unique
solution if and only

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if it has a unique
solution regardless

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of what the right-hand side is.

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So in particular, the answer
to a and the answer to b

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are exactly the same.

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So values of c for which this
system has a unique solution

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are exactly the same as values
of c for which the homogeneous

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system has a unique solution.

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Now the solutions will
be different, of course.

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But the value of
c-- or the values

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of c-- that make it
solvable uniquely,

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make it solvable uniquely
for all right-hand sides.

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And so which values
of c are those?

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Well, those are the
values of c for which

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the coefficient matrix on the
left-hand side is invertible.

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So if the coefficient
matrix on the left-hand side

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is invertible, then we
can solve this system

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and we get a unique solution.

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If it's not invertible,
then either we

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can't solve this system--
like, there are no solutions--

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or we can solve this system,
but there are infinitely

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many solutions.

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So in both questions
a and b, we're

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asking for the value of c for
which the coefficient matrix

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of the left-hand
side is invertible,

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and that will be when we
have a unique solution.

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So how do we know when
a matrix is invertible?

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Well, let's write down what
the matrix is first of all.

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So this matrix M that we're
after is equal to the matrix

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2, 0, c; 1, minus
1, 2; 1, minus 2, 2.

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So this is the coefficient
matrix M of that system,

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and we want to know for which
values of c is it invertible.

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Well, when is a
matrix invertible?

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A matrix is invertible--
square matrix is invertible--

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precisely when it has
non-zero determinant.

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So we just need to look at the
determinant of this matrix.

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So you've learned how to compute
determinants of matrices,

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I think.

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So let's, in this case,
we have the det M.

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So it's a sum or difference
of six different terms,

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and you could get
it, for example,

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by the Laplace expansion
if you wanted to.

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So I'm just going to write
out what the six terms are.

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So it's 2 times minus 1 times
2, plus 0 times 2 times 1,

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plus c times 1 times minus
2, minus c times minus 1

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times 1, minus 2 times minus 2
times 2, minus 0 times 1 times

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2.

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So this is the determinant
of this matrix.

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You can get it either
just by remembering

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which terms are which
and which get a plus sign

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and which get a minus sign, or
by doing the Laplace expansion,

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or by whatever other tricks
you might happen to know.

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So now we need to know whether
or not this determinant is 0.

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So let's work out what this is.

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So this is-- let me
start simplifying it.

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So this is minus 4
plus 0 minus 2c--

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this is minus minus
c, so plus c--

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this is minus minus 8, so plus
8, which is equal to 4 minus c.

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So the determinant--
right, two of those terms

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are 0, and so I just
get to leave them out.

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So the determinant of
this matrix is 4 minus c.

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And what we're
interested in is when

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this determinant is non-zero.

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So in particular, for c
not equal to 0-- sorry,

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for c not equal to 4-- when c
is not 4, the determinant of M

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is not 0.

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So when c is not 4,
determinant of M is not 0,

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so both systems-- both
the original system

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and the corresponding
homogeneous system--

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have a unique solution.

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So when c is not 4-- so
for most values of c--

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the determinant is
not 0, and the system

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has a unique solution.

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So when c is equal
to 4, what happens?

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Well, when c is equal to 4,
we're in the bottom case.

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We're in the case where
the homogeneous system has

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infinitely many solutions.

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OK?

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So let me write that over here.

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When c equals 4-- I'm
going to abbreviate

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again-- the homogeneous
system has-- I'm

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going to use this symbol-- this
sort of sideways eight symbol

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means infinity, so
I'm going to use it

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for infinitely many solutions.

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So when c is 4, the
homogeneous system

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has infinitely many solutions.

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And you might be
curious-- well, so let

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me say one more
thing about that.

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We know when the coefficient
matrix isn't invertible

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that the system either has zero
or infinitely many solutions.

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But the homogeneous system
always has a solution.

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It always has the solution
where everything is all 0.

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Right?

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So that's why we know that
it's infinitely many here.

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And one thing you might ask
is can you find any others?

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Can you find any solutions
that aren't just [0, 0, 0]?

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And the answer is yes.

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So this is now going beyond
when I asked you to do,

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but I think it's, you know,
an interesting thing to see.

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So if you wanted to find another
solution, what do you know?

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Well, let's go back to
the equations that we had.

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So when we're dealing
with a homogeneous system,

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the right-hand sides are 0.

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So I'm just going to cross
out these right-hand sides

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and replace them with 0
so we don't get confused.

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So this is 0, 0, and 0.

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So we're dealing with this
system: 2x plus c*z equals 0,

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x minus y plus 2z equals 0, and
x minus 2y plus 2z equals 0.

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OK, so if you want
a solution [x, y, z]

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to this system,
what do you know?

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Well, from the second equation,
you know that the vector

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[x, y, z] is orthogonal to
the vector 1, minus 1, 2.

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How do you know that?

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Because this left-hand
side, x minus y plus 2z,

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is equal to [x, y, z]
dot 1, minus 1, 2.

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And similarly from
the third equation,

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you know that the
vector [x, y,  z]

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is orthogonal to the
vector 1, minus 2, 2,

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because this left-hand side
is equal to [x, y, z] dot 1,

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minus 2, 2.

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Yeah?

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And that's equal to 0.

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So from the second
and third equations,

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you know that you're
looking for a vector that's

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orthogonal to both
x-- or sorry-- both 1,

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minus 1, 2, and 1, minus 2, 2.

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How do you get a
vector perpendicular

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to two known vectors?

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Well, you just take
their cross product.

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So let's go back over here.

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So to find one, you
take a cross product

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of two rows of the
coefficient matrix.

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So in this case, for example,
we can take these rows, 1,

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minus 1, 2; and 1, minus 2, 2.

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So, for example, the
vector 1, minus 1,

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2-- OK-- cross the
vector 1, minus 2, 2.

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Now I've kind of run
out of board space,

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so I'm not going to work out
precisely what this vector is

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for you.

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But if you like, you
can certainly check.

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You can compute this
cross product out

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with our nice formula
for the cross product.

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It will give you some
vector, and then you

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can check that that
vector is indeed

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a solution of the
homogeneous system.

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So that will give
us a second solution

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of the homogeneous system.

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Nontrivial we say, because
it's not just the 0 solution.

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So to quickly recap, we had
a system of linear equations.

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I've now crossed out what the
original right-hand side was.

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We had a system of
linear equations,

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and we were looking for a choice
of c for which that system had

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a unique solution and for which
the corresponding homogeneous

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system had a unique solution.

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And the values of c that
make that work are precisely

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the values of c such that
the coefficient matrix

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has a non-zero determinant.

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So that's true for
both parts a and b.

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And for part c, when we were
looking for what values of c

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give the homogeneous system
infinitely many solutions,

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the answer is any
other value of c.

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Any value of c for which
the coefficient matrix does

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have 0 determinant will give
you infinitely many solutions

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in the homogeneous case,
and in non-homogeneous cases

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will either give you 0 solutions
or infinitely many solutions.

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And then we also at
the end, we briefly

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discussed one way to
find nontrivial solutions

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in the homogeneous case
when there are infinitely

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many solutions.

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So I'll end there.