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LINAN CHEN: Hi everyone.

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I'm Linan.

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Welcome back to recitation.

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In recent lectures, we
have studied the properties

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of the determinant.

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And we also derived the
formula to compute it.

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Today we're going to
put what we learned

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into practice by considering
these two examples.

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So we want to find out the
determinants of these two 5

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by 5 matrices.

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And as you can
see, matrix A has x

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along this diagonal, and in the
first four rows, y to the right

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of x, except for the last row.

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And zero entries
everywhere else.

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And matrix B also has
x along this diagonal

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and y everywhere else.

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Before starting, let me
help you review what you

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can do to compute determinants.

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Of course, you can
carry out elimination

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to transform your original
matrix into upper triangular

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matrix.

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Or you can use this
big summation formula.

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Another choice would be
you can do it by cofactors.

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Namely, you can expand
your original matrix

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along any row or any column,
and then the determinant

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is simply given
by the dot product

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of that row or that
column with its cofactors.

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Why don't you pause the video
now and try to work on them

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yourself.

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Whenever you're ready, I'll
come back and show you my way.

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I hope you just had some
fun with these two problems.

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Now let's look at them together.

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Let's look at matrix A first.

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As you can see, there are a lot
of zero entries in matrix A.

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So perhaps you don't need
elimination to introduce more

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0's.

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Furthermore, we observe
this pattern of A,

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and you notice that if
I cover the last row

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and the first column, so
if this column and this row

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are not here, what is left
over is simply a 4 by 4

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lower triangular matrix.

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And similarly, if you
cover the first column

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and the first row, what is
left over here is simply a 4

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by 4 upper triangular matrix.

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This is telling
us that we should

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calculate the determinant
of A by the third method.

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So we should expand along
the first column of A,

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and we calculate the cofactors.

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Let's do it now.

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So determinant of A,
is equal to the (1, 1)

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entry of A, which is
x, times the cofactor

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of that spot, which is the
determinant of the leftover 4

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by 4 matrix.

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And it's upper triangular,
so its determinant is simply

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given by x to the power 4.

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Plus, the only other
nonzero entry in that column

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is the y at the very bottom.

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So you put y here.

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And you multiply
y by the cofactor

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of that spot, which is the
determinant of the leftover 4

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by 4 matrix again.

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In this case, it's
lower triangular

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and its determinant
is y to the power 4.

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So I have a y to the power 4.

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But not quite.

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In principle, there
should be another factor

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here indicating the sign.

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And the sign in this
case, well because the y

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is the entry in the fifth
row and the first column,

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so this factor
should be negative 1

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to the power 5 plus 1.

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And of course, it's just 1.

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So the determinant
of A is simply

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equal to x to the fifth power
plus y to the fifth power.

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Did you get the correct answer?

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Well, the determinant
of A is not too bad,

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because A has a lot
of zero entries.

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Now let's look at the
determinant of matrix B.

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I have another copy of B here.

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So B also has a very
clear structure.

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It has x along its diagonal,
and y everywhere else.

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But in general, B does
not have any zero entry.

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So perhaps our first step should
be carrying out elimination

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to introduce zero
entries into matrix B.

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Of course, you can do it
by the regular routine.

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You start with the
first row, find a pivot,

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and eliminate the second
row and the third row

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and so on and so forth.

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But in this case,
there's a shortcut.

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If you compare two rows
that are next to each other,

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for example, if we compare the
fourth row and the fifth row,

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you notice that they have
a lot of entries in common.

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And they're only different
at these two spots.

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So imagine if I subtract the
fourth row from the fifth row.

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So if I do the
following operation--

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so I subtract this
row from the last row.

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Then the new fifth row
should become the following.

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So this row will become 0,
0, 0, y minus x, x minus y.

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You see, just by this
simple operation,

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I have introduced three
zero entries at once.

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And it's similar with the
fourth row and the third row.

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They have common entries
here, here, and here.

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So you subtract the third
row from the fourth row.

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You update the fourth row to
0, 0, y minus x, x minus y, 0.

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Again, three zero entries.

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And same thing happened to the
second row and the third row.

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So you subtract the second
row from the third row

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and your new third
row will become 0,

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y minus x, x minus y, 0, 0.

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Finally, you subtract the
first row from the second row,

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and then you update
the second row

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to y minus x, x
minus y, 0, 0, 0.

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Let's keep the
first row unchanged.

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So I'm going to copy here.

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All right.

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By row elimination, we have
introduced many zero entries

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to matrix B. Is
there anything else

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that I can take advantage of?

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Let's observe the pattern
of this new matrix.

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As you can see, in each row,
you have two nonzero entries,

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except for the first row.

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And they're only
different by a sign.

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So if, somehow, you can figure
out a way to sum them up,

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you will get even
more zero entries.

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So let's do it.

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That's going to involve
the operations on column.

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So here is how I do it.

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I'm going to keep the
last column unchanged.

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So the last column is
y, 0, 0, 0, x minus y.

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What I will do is I will add
a copy of the last column

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to the fourth column.

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So this is what I'm going to do.

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Add one copy of the last
column to the fourth column.

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Now the new fourth column will
become 2y, 0, 0, x minus y, 0.

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As you can see, by doing
this, I have killed this spot.

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So I have introduced
one more zero entry

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into my fourth column.

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If you continue, you may
want to add the fourth column

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to the third column.

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Let's see what will
happen if you do that.

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So if you add the fourth
column to the third column,

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now what should appear here is
2y, 0, x minus y, 0, y minus x.

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But in this new third
column, you still

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have two zero entries,
which is the same

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as the original third column.

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So although you've
killed this spot,

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but you've introduced
a new nonzero entry.

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So is there a way that we
can kill this spot too?

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You may have noticed that if you
add a copy of the fifth column

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to this column again, then that
spot will have been killed.

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So let's do it.

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If I add this
column to this one--

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I'm going to just
update it here-- then

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the first entry
should become 3y.

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These are unchanged.

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And the last spot becomes 0.

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It reflects here
as you are adding

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a copy of the fourth column
and a copy of the fifth column

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to the third column.

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Now you've got the idea.

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And you continue.

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What do you do with
the second column?

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This time you will have to add
a copy of the third column,

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a copy of the fourth column, and
the copy of the fifth column.

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So you update the second column
to be 4y, x minus y, 0, 0, 0.

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Eventually, what you will
do to the first column

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would be you add everything
to the first column.

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So a copy of each.

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Then the first
column will become--

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so I don't have
enough spot here,

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so make it here-- x plus 4y.

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Then everything else is 0.

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This is fun.

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And the result's really nice.

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This is wonderful
because this is simply

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upper triangular matrix.

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Now you tell me: what
is the determinant of B?

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The determinant of
B is the determinant

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of this upper triangular matrix.

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So you simply multiply
everything together,

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that's x plus 4y times x
minus y to the fourth power.

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So I hope you enjoyed
these two examples.

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Maybe your method is
different from mine,

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but at least these
two examples teach us

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that you can be flexible
in combining methods

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in calculating determinants.

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Thanks for watching,
and I'm looking forward

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to see you soon.