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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 and

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their various applications, and
one of them is to solving

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systems of linear equations.

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So I have here a system
of three linear

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equations for you.

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2x plus cz equals 4, x minus y
plus 2z equals pi, and x minus

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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 to

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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, but the right-hand

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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 for which the

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corresponding homogeneous system
has a unique solution,

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and also the values of c for
which the corresponding

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homogeneous system has
infinitely many solutions.

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

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

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Although, of course, whether you
can or not might depend on

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the value of c.

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

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work out the solutions to these
three questions, come

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back, and we can work
it out together.

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So hopefully you have some
luck working out these

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problems. 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 that is that one thing

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you've learned is that a system
has a unique solution

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for, on the right-hand
side-- sorry--

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a system has a unique solution,
like this, a square

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system of linear equations has
a unique solution if and only

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if it has a unique solution
regardless of what the

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right-hand side is.

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

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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 value of c--

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that make it solvable uniquely,
make it solvable

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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 the coefficient

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matrix on the left-hand
side is invertible.

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So if the coefficient matrix
on the left-hand side is

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invertible, then we can solve
this system and we get a

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unique solution.

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If it's not invertible, then
either we can't solve this

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system-- like there
are no solutions--

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or we can solve this system,
but there are

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infinitely many solutions.

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So in both questions a and b,
we're asking for the value of

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c for which the coefficient
matrix of the left-hand side

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is invertible, and that
will be when we

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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

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matrix 2, 0, c.

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1, minus 1, 2.

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1, minus 2, 2.

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So this is the coefficient
matrix M of that system, and

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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

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matrices, I think.

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

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or difference of six different
terms, and you could get it,

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for example, 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 times 1,

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minus 2 times minus 2 times 2,
minus 0 times 1 times 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 which terms are

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which and which get a plus sign
and which get a minus

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sign, or by doing the Laplace
expansion, or by whatever

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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--

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so plus c--

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this is minus minus 8--

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

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So the determinant-- right, two
of those terms are 0, and

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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 this

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determinant is non-zero.

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

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equal to 4, when c is not 4--

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the determinant of M is not 0.

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

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both systems-- both the original
system and the

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corresponding homogeneous
system--

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have a unique solution.

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So when c is not 4--

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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--

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I'm going to abbreviate
again--

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the homogeneous system has--

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I'm going to use this symbol--

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this sort of sideways eight
symbol means infinity, so I'm

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going to use it for infinitely
many solutions.

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So when c is 4, the homogeneous
system has

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infinitely many solutions.

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And you might be curious-- well,
so let me say one more

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thing about that.

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We know when the coefficient
matrix isn't invertible that

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the system either has 0 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, but I

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

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So if you wanted
to find another

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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, the

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right-hand sides are 0.

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

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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 cz equals 0, x

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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--

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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

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vector xyz 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, is equal to

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xyz dot 1, minus 1, 2.

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And similarly from the third
equation, you know that the

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

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because this left-hand
side is equal to xyz

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dot 1, 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, you know that

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you're looking for a vector
that's orthogonal to both x--

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or sorry-- both 1, 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 of two rows of

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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, 2--

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OK--

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cross the vector
1, minus 2, 2.

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Now I've kind of run out of
board space, so I'm not going

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to work out precisely what
this vector is 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 with our nice

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formula for the cross product.

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It will give you some vector,
and then you can check that

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that vector is indeed
a solution of

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the homogeneous system.

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So that will give us a
second solution of

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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, and we were looking

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for a choice of c for which
that system had a unique

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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 the

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values of c such that the
coefficient matrix has a

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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, the

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answer is any other
value of c.

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Any value of c for which the
coefficient matrix does have 0

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determinant will give you
infinitely many solutions in

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the homogeneous case, and in
non-homogeneous cases will

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either give you 0 solutions or
infinitely many solutions.

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And then we also at the end, we
briefly discussed one way

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to find nontrivial solutions in
the homogeneous case when

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there are infinitely
many solutions.

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So I'll end there.

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