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PROFESSOR: Hi.

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Welcome to recitation.

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My name is Martina, and I'll
be your recitation instructor

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for some of these
linear algebra videos.

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Today's problem is
a straightforward

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solve the following
linear system

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with four equations
and four unknowns,

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using the method of elimination.

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The system is x minus y minus
z plus u equals 0, 2x plus 2z

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equals 8, minus y
minus 2z equals -8,

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and 3x minus 3y minus
2z plus 4u equals 7.

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And although you might
know different ways

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to solve the system
at this point,

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the method of
elimination is going

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to show up a million
times during these videos,

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so it's really important
to get it right.

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So I suggest you try solving
this system now, using

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the method of elimination as
it was described in class.

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I'm going to leave you
alone with the problem.

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You should pause the
video, solve it yourself,

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and then come back and compare
your solution with mine.

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And we're back.

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So the method of elimination,
if you remember it from class,

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consisted of replacing this
system with an equivalent

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system-- equivalent meaning
they have the same solution--

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by a series of row operations.

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Row operations are not
supposed to change the solution

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to the system.

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And they're, for
example, exchange

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the order of two equations.

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Multiply an equation
with a nonzero number,

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and add a nonzero multiple
of one equation to the other.

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So let's do that.

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As we're going to do this
series of arithmetic operations,

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we don't really want
to copy the names

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of the variables and the
equality signs every time.

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So we're going to keep the
important information which

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are these numbers.

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these coefficients
here, we're going

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to keep that
information in a matrix.

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So let's write a matrix.

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Each row is going to
correspond to an equation,

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and each column is going to
correspond to an unknown.

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So the first row
is 1, -1, -1, 1.

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The second row, corresponding
to the second equation,

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is 2, 0, 2, 0.

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And you want to be very careful
to put 0's on the right spots

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here.

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The third equation is
0-- I haven't left myself

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enough room-- 0, -1, -2, 0.

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And the fourth
row, corresponding

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to the fourth equation,
is 3, -3, -2, 4.

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And as we care about the
right-hand side as well,

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we're going to copy this
information as well,

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and get the augmented
matrix of the system.

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0, 8, -8, 7.

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OK, and now let's try
reducing this matrix

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to an upper triangular one.

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We start with the
first column, and we're

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going to use this
number, called a pivot,

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to get rid of all
the numbers under it,

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so to get 0's here and here.

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A way to do it is-- well,
to get rid of this 2,

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I have to multiply
the first row by -2,

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and add it to the second one.

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Writing this here is
not strictly necessary,

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but I like to do a
bit of bookkeeping,

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because I'm prone to
make a lot of errors

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while doing this simple
arithmetic operations.

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And then if I get
to the end, figure

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out I made a mistake
somewhere, this bookkeeping

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makes it easier to
backtrack and find the place

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where I made this mistake.

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So we replace this matrix
with another matrix.

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The first row stays the same.

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1, -1, -1, 1, 0.

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The second row gets replaced
by the second row minus 2 times

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the first row.

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The aim of that is to get
a 0 here, so that's good.

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Next, on this position
here, we get -2 times -1,

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which is 2, plus 0 which is 2.

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-1 times -2, which is
2, plus 2, which is 4.

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-2 plus 0 which is -2.

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And 8 minus 2
times 0 which is 8.

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The third row
already has a 0 here,

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so I can just copy it over.

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0, -1, -2, 0, -8.

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And to get a 0 here, I'm going
to multiply the first row by -3

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and add it to the fourth row
and get 0, -3 times -1 is 3,

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minus 3 is 0.

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-3 minus 1-- oh, sorry-- 3 minus
2 is 1, and -3 plus 4 is 1.

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And there we go.

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The first column looks
like a first column

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of an upper triangular matrix.

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Now let's do the same
to the second column.

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This is going to be our
pivot, the number that we use

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to get rid of numbers under it.

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And we see that to
get rid of this number

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here, we will need to
multiply it with 1/2.

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So multiply the whole
second row with 1/2,

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and add it to the third row.

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The matrix that we get will
have the first row the same.

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It stays the same.

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1, -1, -1, 1, 0.

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The second row stays the same.

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0, 2, 4, -2, 8.

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The third row gets
replaced by the third row

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plus 1/2 times the second row
and becomes 0, 0, 2 minus 2

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which is 0, -1 plus 0 which is
-1, and 4 minus 8 which is -4.

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And the fourth row already has
a 0 here so I just copy it over.

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0, 0, 1, 1, 7.

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And now let's look
at this matrix.

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It has the first two columns
as they're supposed to be,

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0's under the diagonal.

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And now we want to get a 0 here.

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Normally what I would do is
to circle this number here,

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multiply it by something
so that I get a -1,

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and add it to this
row to get a 0 here.

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But that's not going to work.

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You might remember from lecture
that 0's can never be pivots.

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Or you can just try finding
a number such that 0

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times this number
equals -1, and seeing

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that such a number
doesn't exist,

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because you're always
going to get 0.

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So we can proceed
as we did until now.

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But is there another
way to get a 0 here?

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There is a very
simple row operation,

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which consists just of switching
the third and the fourth row.

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It certainly doesn't change
the solution of the system.

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So let's do that.

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And let's get the next matrix
which is 1, -1, -1, 1, 0.

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0, 2, 4, -2, 8.

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Then we put the fourth row here.

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0, 0, 1, 1, 7.

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And we put the third row
here, 0, 0, 0, -1, -4.

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And there it is.

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This is an upper
triangular matrix.

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So in the same way as at the
beginning, we had a system

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and then wrote the
matrix representing it,

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this matrix also
represents a system.

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And this system has the same
solutions as the initial system

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but is much easier to solve.

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Now let's write this
back as a system,

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and let me do that not starting
from the first equation,

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but starting from
the last equation.

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So the last equation
here reads -u

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equals -4, which
is, as equations go,

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a pretty easy one to solve.

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The solution is u equals 4.

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Now let's go back to
the third equation.

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The third equation
reads y-- no, I'm

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sorry-- z-- the third column
corresponds to z-- plus u

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equals 7.

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But we know what's u now.

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So, it reads z plus 4 equals 7,
which just becomes z equals 3.

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The second equation is 2
times y plus 4 times z,

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but z is 3, minus 2 times
u, but u is 4, equals 8.

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And from this, one can easily
compute that y is equal to 2.

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And finally, the first equation
reads x minus y minus z plus u

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equals 0, or x equals 1.

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And this is our solution.

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x, y, z, and u equal
1, 2, 3, and 4.

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This finishes the problem,
but I would very strongly

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encourage you now to
take this solution

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and plug it back into
the original system

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and check if it's
really a solution.

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And that's all I
wanted to say today.