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PROFESSOR: Hi.
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Welcome to the second
special recitation
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on exam problem solving.
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As you may have experienced,
when you do homework problems,
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you can give every
problem or even every step
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a careful thought.
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You may even try different
methods in solving one problem,
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just to check your answers and
also to find the optimal way.
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But in the real exam, these
may not be available to you
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anymore, because time
is the main issue.
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So you want to do things
fast but accurately
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at the same time.
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Today's recitation is
going to focus on that.
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So we're going to look
at a real exam problem.
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So this is the problem from a
15-minute linear algebra exam.
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OK.
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By now, you've developed
enough background
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to completely
solve this problem.
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So because this is
a 15-minute exam,
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a good suggestion on time
spending on this problem
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would be no more than
15 minutes, which
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means within 15
minutes, you have
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to read through the
questions, understand
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what they're asking for,
and completely solve
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all three parts.
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Why don't you hit Pause now, and
try to complete this problem as
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if you were in an exam.
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And time yourself
for 15 minutes.
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If you finish early, don't
forget to check your answer.
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You want to get all the
credits you deserve in an exam.
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OK.
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I'll come back later
and show you how
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I would speed up in the exam.
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OK.
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Have you finished?
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Well, let's solve
this problem together.
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We are looking at a
4 by 4 matrix A here.
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As you can see, this
matrix is made up
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by the examiner in
a rather casual way,
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because you have
numbers from 1 to 12
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as the entries of this matrix.
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And in the first part, we want
to find all the nonzero terms
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in this big formula to
compute determinant of A.
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So the determinant of
A-- so that's Part 1--
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that's equal to a big summation
of plus or minus a_(1,alpha),
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a_(2,beta), a_(3,gamma),
and a_(4,delta).
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So what I'm doing here is I
choose one entry from each row
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with the columns
being all different.
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So in other words, if I take
this column numbers down--
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let me write it here-- alpha,
beta, gamma, and delta,
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I want this to be a permutation
of numbers 1, 2, 3, 4.
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OK.
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So how would you do it?
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Well, if you follow
this order, you
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may want to start
with the first row.
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And you go through
all the possibilities
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of this alpha, beta, gamma,
and delta, and at the end,
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you drop the terms which are 0.
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But if you do it in
that way, how many terms
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do you have to compute?
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This is a 4 by 4 matrix.
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So in general, this sum
will contain 4 factorial,
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which is 24, terms.
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That sounds time consuming.
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So can you do it
in a faster way?
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Well, since we only
care about nonzero terms
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in this sum, let's look at
where 0 occurs in matrix A.
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They're here.
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You have zero entries
in these four spots.
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OK.
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So they all occur in the third
row and the fourth row, which
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means when you make your
choice of the last two entries,
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in order not to get
zero, you can only
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choose within this red box.
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Because you want
to avoid these 0's.
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Which means the choice
of the last two entries
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can only be either
9, 12 or 10, 11.
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OK.
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Now, if that's the case, what
will happen to the first two
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entries?
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Since they all have to come
from different columns, which
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means when you choose
the first two entries,
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they can only be
from this red box.
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Which means the choice
of the first two entries
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can only be either 1, 6 or 2, 5.
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So how many terms
am I talking about?
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Two possibilities here and
two possibilities here,
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which comes to four.
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So in other words, instead
of computing 24 terms,
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you only need four terms here.
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Let's put them down.
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OK.
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So we start with this
choice: 1, 6, 9, 12.
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So that's 1 times
6 times 9 times 12.
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Well, these are from
diagonal, right?
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So of course, this coordinate
numbers would just be a_(1,1),
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a_(2,2), a_(3,3), a_(4,4).
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This is the perfect
alignment of 1 2,
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3, 4, so the sign of this
term is just a plus sign.
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Let's continue.
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Now I'm going to look
at 1, 6 but 10, 11.
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So 1 times 6 times 10 times 11.
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What is the coordinate
number-- what
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are the coordinate
numbers of this choice?
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a_(1,1), a_(2,2), 10
comes from a_(3,4).
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So here I have 4 in the front.
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The last one is
a_(4,3), so right here.
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This is the permutation
of 1, 2, 3, 4.
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In order to get back to this,
I have to exchange 3 and 4.
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Just do it once.
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OK.
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Which means I need a
negative sign in front.
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A negative sign in front
of the entire product.
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Well, let's continue.
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I'm going to write it down here.
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As you can see the blue
part indicate the signature
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of columns.
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So next term, I'm
going to put-- I've
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exhausted the first possibility
of choosing 1, 6, right?
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So let's look at 2, 5.
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2 times 5 times 9 times 12.
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What is the column?
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What are the column numbers?
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2 comes from (1, 2) entry,
so I have a 2 in the front.
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5 is (2, 1) entry, so
2, 1, and then 3, 4.
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3, 4.
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Again, I need one
exchange to get back
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to 1, 2, 3, 4,
which means I have
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a negative sign in the front.
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The last one would be 2, 5.
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2 times 5 times 10, 11,
so times 10 times 11.
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Columns are a_(1,2),
a_(2,1), a_(3,4), a_(4,3).
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So you exchange the first two
spots, and the last two spots.
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You get back to 1, 2, 3, 4.
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But you have to do twice,
two exchanges, which means
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you need a plus sign in front.
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That's it.
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This big summation formula comes
down to the sum of four terms.
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And you can compute it.
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If you don't make
any algebra mistake,
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the result should be 8.
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Did you get the right answer?
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All right, time is
really precious.
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Let's move on to
the second part.
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In the second part, we need
to find cofactors c_(1,1),
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c_(1,2), c_(1,3), and c_(1,4).
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Let me put the second part here.
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So we're looking
for the cofactors
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of the first row of
matrix A. Let's just
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write everything down.
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c_(1,1).
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c_(1,1) is the cofactor
of this entry here.
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So you're looking at
the determinant of this
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left over 3 by 3 matrix.
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So that's the determinant
of-- copy it down--
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[6, 7, 8; 0, 9, 10; 0, 11, 12].
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How would you compute this?
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This is a 3 by 3 matrix.
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Of course, you can use the
big summation formula again.
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In other words,
you can write down
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this determinant-- this
formula, but for the specific 3
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by 3 matrix.
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But that will involve three
factorial terms, right?
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Which is six terms.
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All right.
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Is there a way that
you can do it faster?
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Just notice that the first
column of this matrix
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has only one nonzero entry,
which is this (1, 1) entry 6.
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So why do we just expand
this along the first column
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and use cofactors?
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Let's do it.
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This determinant is equal
to 6 times its cofactor.
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And its cofactor comes to
the determinant of this 2
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by 2 matrix, which is easy.
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That becomes 9 times
12 minus 10 times 11.
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And if you compute this
correctly, that should be -12.
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That's not too bad.
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Let's look at the
second one, c_(1,2).
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c_(1,2) is the determinant
of this 3 by 3 matrix.
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So I have to delete the first
row and the second column.
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And I read what is left
over and I put it here.
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So that's
[5, 7, 8; 0, 9, 10; 0, 11, 12].
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Same thing.
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The first column has
only one nonzero entry.
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You expand it along the
first column, use cofactors,
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the result will be 10.
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There's one thing that
I've forgotten here.
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Because we are looking
at (1, 2) entry.
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So for the cofactor, I have
to put an extra negative sign
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here.
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So this is actually -1 times
the determinant of this 3
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by 3 matrix.
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And the result will be 10.
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OK, let's continue.
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c_(1,3).
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Let's hope the computation is
going to get easier and easier.
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So c_(1,3) is the determinant
of the leftover 3 by 3 matrix.
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So I'll directly read from that.
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That's going to be 5, 6,
8; 0, 0, 10; and 0, 0, 12.
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What is the determinant
of this matrix?
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00:12:24,080 --> 00:12:27,110
Again, you can use
the same method
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as we did for the first two,
because the first column has
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only one nonzero entry.
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And if you do that,
the result should be 0.
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But you should be
able to tell it
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without any direct computation.
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Why is that?
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Well, clearly the
first two columns
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are linearly dependent, because
the second column is 6 over 5
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times the first column.
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So this is a singular matrix.
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Right away, the
determinant is 0.
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00:13:00,150 --> 00:13:01,010
So what is c_(1,4)?
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c_(1,4) is the determinant of
the matrix of [5, 6, 7; 0, 0,
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9; 0, 0, 11].
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Same thing.
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00:13:17,310 --> 00:13:21,150
It's singular, so
its determinant is 0.
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00:13:21,150 --> 00:13:21,870
All right.
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00:13:21,870 --> 00:13:25,050
So that completes
the second part.
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00:13:25,050 --> 00:13:27,580
You can move on to
the third part now,
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00:13:27,580 --> 00:13:30,890
but before you do
that, just notice
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this may be a good point
to check your answer
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from the first part.
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Because you have
all the cofactors
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of the first row of A, and
if you use the other formula
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to compute the
determinant of A, you
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can see the determinant of A
is equal to the dot product
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of the first row
with its cofactors.
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So because the last
two cofactors is 0,
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so I only have two
terms in the sum.
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So determinant A is equal to
a_(1,1) entry, which is 1,
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times c_(1,1), which is -12,
plus a_(1,2) entry, which is 2,
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times c_(1,2), which is 10.
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What is that?
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-12 plus 20, that give you 8.
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OK, at least the answers
from the first two parts
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are consistent.
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So by now, you should be
more confident to move on
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00:14:43,510 --> 00:14:45,560
to the third part.
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I have a problem of space.
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00:14:47,330 --> 00:14:50,510
So please allow me to
put the third part here.
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I'm going to come all
the way back here.
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So that's my third part.
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00:14:55,970 --> 00:14:57,570
What is the third part?
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00:14:57,570 --> 00:15:03,130
Third part asks you to find
the first column of A inverse.
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Well, it seems that involves
more computation of cofactors,
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00:15:07,300 --> 00:15:11,900
but as many well
designed exam problems,
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00:15:11,900 --> 00:15:15,100
the answer from third part
can be directly derived
249
00:15:15,100 --> 00:15:19,030
from the first part
and the second part.
250
00:15:19,030 --> 00:15:21,800
That's how you should do it.
251
00:15:21,800 --> 00:15:25,050
So what is A inverse?
252
00:15:25,050 --> 00:15:31,570
The formula for A inverse is
equal to 1 over determinant
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00:15:31,570 --> 00:15:40,050
of A times the transpose of
a matrix C. This matrix C
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00:15:40,050 --> 00:15:43,690
is composed by
cofactors of matrix A.
255
00:15:43,690 --> 00:15:47,840
We want to find the first
column of A inverse.
256
00:15:47,840 --> 00:15:58,260
So the first column
of A inverse should
257
00:15:58,260 --> 00:16:02,470
be one of our determinants
of A, this constant,
258
00:16:02,470 --> 00:16:07,280
times the first row of
C, but transpose, right?
259
00:16:07,280 --> 00:16:15,030
So I have to put here, first
row of C, but transpose.
260
00:16:17,700 --> 00:16:21,620
Determinant of A comes
from the first part,
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00:16:21,620 --> 00:16:25,700
and the first row of C
comes from the second part.
262
00:16:25,700 --> 00:16:29,480
So I just copy what I have
from the first two parts down.
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00:16:29,480 --> 00:16:31,750
That's 1 over 8.
264
00:16:31,750 --> 00:16:38,840
This column vector will
become [-12; 10; 0; 0].
265
00:16:41,540 --> 00:16:42,930
That's it.
266
00:16:42,930 --> 00:16:45,370
That completes this problem.
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00:16:45,370 --> 00:16:46,770
Have you got your answer right?
268
00:16:49,590 --> 00:16:52,580
OK, before I finish,
there are two things
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00:16:52,580 --> 00:16:54,730
that I like to remind you.
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00:16:54,730 --> 00:16:58,960
First, as you can see,
this is an exercise
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00:16:58,960 --> 00:17:03,690
on the big summation formula
to compute determinant of A.
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00:17:03,690 --> 00:17:07,960
In previous recitation, we
practice using the combination
273
00:17:07,960 --> 00:17:11,020
of elimination and the
method by cofactors
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00:17:11,020 --> 00:17:12,730
to compute determinants.
275
00:17:12,730 --> 00:17:16,000
But we should never
forget this formula.
276
00:17:16,000 --> 00:17:17,819
Because it always works.
277
00:17:17,819 --> 00:17:20,579
And in a lot of cases,
this will turn out
278
00:17:20,579 --> 00:17:23,480
to be an easy way to
compute determinant.
279
00:17:23,480 --> 00:17:27,260
For example, for
this matrix A here.
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00:17:27,260 --> 00:17:30,940
And second, in
your real exam, it
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00:17:30,940 --> 00:17:34,930
would be really helpful
if you can put down
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00:17:34,930 --> 00:17:39,900
your work like this, because it
helps you to check your work.
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00:17:39,900 --> 00:17:43,990
And also, even if you don't
get your final answer correct,
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00:17:43,990 --> 00:17:47,130
this may get you
some partial credits.
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00:17:47,130 --> 00:17:51,180
OK, I'm going to stop here,
and thank you for watching.