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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.