WEBVTT

00:00:06.570 --> 00:00:07.290
PROFESSOR: Hi.

00:00:07.290 --> 00:00:09.720
Welcome to the second
special recitation

00:00:09.720 --> 00:00:12.000
on exam problem solving.

00:00:12.000 --> 00:00:15.570
As you may have experienced,
when you do homework problems,

00:00:15.570 --> 00:00:18.820
you can give every
problem or even every step

00:00:18.820 --> 00:00:20.320
a careful thought.

00:00:20.320 --> 00:00:23.620
You may even try different
methods in solving one problem,

00:00:23.620 --> 00:00:27.920
just to check your answers and
also to find the optimal way.

00:00:27.920 --> 00:00:31.400
But in the real exam, these
may not be available to you

00:00:31.400 --> 00:00:35.590
anymore, because time
is the main issue.

00:00:35.590 --> 00:00:38.790
So you want to do things
fast but accurately

00:00:38.790 --> 00:00:40.160
at the same time.

00:00:40.160 --> 00:00:43.230
Today's recitation is
going to focus on that.

00:00:43.230 --> 00:00:46.660
So we're going to look
at a real exam problem.

00:00:46.660 --> 00:00:54.500
So this is the problem from a
15-minute linear algebra exam.

00:00:54.500 --> 00:00:56.000
OK.

00:00:56.000 --> 00:00:58.640
By now, you've developed
enough background

00:00:58.640 --> 00:01:01.190
to completely
solve this problem.

00:01:01.190 --> 00:01:04.819
So because this is
a 15-minute exam,

00:01:04.819 --> 00:01:08.050
a good suggestion on time
spending on this problem

00:01:08.050 --> 00:01:10.900
would be no more than
15 minutes, which

00:01:10.900 --> 00:01:13.440
means within 15
minutes, you have

00:01:13.440 --> 00:01:17.050
to read through the
questions, understand

00:01:17.050 --> 00:01:20.410
what they're asking for,
and completely solve

00:01:20.410 --> 00:01:22.360
all three parts.

00:01:22.360 --> 00:01:26.510
Why don't you hit Pause now, and
try to complete this problem as

00:01:26.510 --> 00:01:28.280
if you were in an exam.

00:01:28.280 --> 00:01:31.070
And time yourself
for 15 minutes.

00:01:31.070 --> 00:01:34.620
If you finish early, don't
forget to check your answer.

00:01:34.620 --> 00:01:38.330
You want to get all the
credits you deserve in an exam.

00:01:38.330 --> 00:01:39.180
OK.

00:01:39.180 --> 00:01:41.210
I'll come back later
and show you how

00:01:41.210 --> 00:01:42.780
I would speed up in the exam.

00:01:52.880 --> 00:01:53.700
OK.

00:01:53.700 --> 00:01:55.370
Have you finished?

00:01:55.370 --> 00:01:58.320
Well, let's solve
this problem together.

00:01:58.320 --> 00:02:02.750
We are looking at a
4 by 4 matrix A here.

00:02:02.750 --> 00:02:06.100
As you can see, this
matrix is made up

00:02:06.100 --> 00:02:09.150
by the examiner in
a rather casual way,

00:02:09.150 --> 00:02:11.960
because you have
numbers from 1 to 12

00:02:11.960 --> 00:02:14.380
as the entries of this matrix.

00:02:14.380 --> 00:02:18.920
And in the first part, we want
to find all the nonzero terms

00:02:18.920 --> 00:02:22.900
in this big formula to
compute determinant of A.

00:02:22.900 --> 00:02:32.170
So the determinant of
A-- so that's Part 1--

00:02:32.170 --> 00:02:39.020
that's equal to a big summation
of plus or minus a_(1,alpha),

00:02:39.020 --> 00:02:45.720
a_(2,beta), a_(3,gamma),
and a_(4,delta).

00:02:45.720 --> 00:02:49.790
So what I'm doing here is I
choose one entry from each row

00:02:49.790 --> 00:02:52.830
with the columns
being all different.

00:02:52.830 --> 00:02:57.410
So in other words, if I take
this column numbers down--

00:02:57.410 --> 00:03:04.500
let me write it here-- alpha,
beta, gamma, and delta,

00:03:04.500 --> 00:03:09.920
I want this to be a permutation
of numbers 1, 2, 3, 4.

00:03:09.920 --> 00:03:10.720
OK.

00:03:10.720 --> 00:03:12.530
So how would you do it?

00:03:12.530 --> 00:03:15.190
Well, if you follow
this order, you

00:03:15.190 --> 00:03:17.570
may want to start
with the first row.

00:03:17.570 --> 00:03:20.170
And you go through
all the possibilities

00:03:20.170 --> 00:03:24.280
of this alpha, beta, gamma,
and delta, and at the end,

00:03:24.280 --> 00:03:26.860
you drop the terms which are 0.

00:03:26.860 --> 00:03:30.660
But if you do it in
that way, how many terms

00:03:30.660 --> 00:03:32.770
do you have to compute?

00:03:32.770 --> 00:03:35.130
This is a 4 by 4 matrix.

00:03:35.130 --> 00:03:39.730
So in general, this sum
will contain 4 factorial,

00:03:39.730 --> 00:03:41.750
which is 24, terms.

00:03:41.750 --> 00:03:43.860
That sounds time consuming.

00:03:43.860 --> 00:03:46.740
So can you do it
in a faster way?

00:03:46.740 --> 00:03:50.310
Well, since we only
care about nonzero terms

00:03:50.310 --> 00:03:55.400
in this sum, let's look at
where 0 occurs in matrix A.

00:03:55.400 --> 00:03:57.110
They're here.

00:03:57.110 --> 00:04:01.120
You have zero entries
in these four spots.

00:04:01.120 --> 00:04:01.750
OK.

00:04:01.750 --> 00:04:06.320
So they all occur in the third
row and the fourth row, which

00:04:06.320 --> 00:04:11.330
means when you make your
choice of the last two entries,

00:04:11.330 --> 00:04:14.500
in order not to get
zero, you can only

00:04:14.500 --> 00:04:17.820
choose within this red box.

00:04:22.079 --> 00:04:24.750
Because you want
to avoid these 0's.

00:04:24.750 --> 00:04:28.230
Which means the choice
of the last two entries

00:04:28.230 --> 00:04:34.430
can only be either
9, 12 or 10, 11.

00:04:34.430 --> 00:04:35.240
OK.

00:04:35.240 --> 00:04:38.710
Now, if that's the case, what
will happen to the first two

00:04:38.710 --> 00:04:40.110
entries?

00:04:40.110 --> 00:04:44.440
Since they all have to come
from different columns, which

00:04:44.440 --> 00:04:46.470
means when you choose
the first two entries,

00:04:46.470 --> 00:04:50.690
they can only be
from this red box.

00:04:54.260 --> 00:04:57.630
Which means the choice
of the first two entries

00:04:57.630 --> 00:05:04.140
can only be either 1, 6 or 2, 5.

00:05:04.140 --> 00:05:07.810
So how many terms
am I talking about?

00:05:07.810 --> 00:05:11.840
Two possibilities here and
two possibilities here,

00:05:11.840 --> 00:05:14.360
which comes to four.

00:05:14.360 --> 00:05:18.600
So in other words, instead
of computing 24 terms,

00:05:18.600 --> 00:05:21.520
you only need four terms here.

00:05:21.520 --> 00:05:24.060
Let's put them down.

00:05:24.060 --> 00:05:24.670
OK.

00:05:24.670 --> 00:05:31.530
So we start with this
choice: 1, 6, 9, 12.

00:05:31.530 --> 00:05:37.330
So that's 1 times
6 times 9 times 12.

00:05:37.330 --> 00:05:40.380
Well, these are from
diagonal, right?

00:05:40.380 --> 00:05:46.350
So of course, this coordinate
numbers would just be a_(1,1),

00:05:46.350 --> 00:05:50.700
a_(2,2), a_(3,3), a_(4,4).

00:05:50.700 --> 00:05:52.560
This is the perfect
alignment of 1 2,

00:05:52.560 --> 00:05:57.420
3, 4, so the sign of this
term is just a plus sign.

00:05:57.420 --> 00:05:58.680
Let's continue.

00:05:58.680 --> 00:06:04.900
Now I'm going to look
at 1, 6 but 10, 11.

00:06:04.900 --> 00:06:11.590
So 1 times 6 times 10 times 11.

00:06:11.590 --> 00:06:14.630
What is the coordinate
number-- what

00:06:14.630 --> 00:06:17.480
are the coordinate
numbers of this choice?

00:06:17.480 --> 00:06:23.170
a_(1,1), a_(2,2), 10
comes from a_(3,4).

00:06:23.170 --> 00:06:26.110
So here I have 4 in the front.

00:06:26.110 --> 00:06:31.470
The last one is
a_(4,3), so right here.

00:06:31.470 --> 00:06:34.040
This is the permutation
of 1, 2, 3, 4.

00:06:34.040 --> 00:06:39.130
In order to get back to this,
I have to exchange 3 and 4.

00:06:39.130 --> 00:06:40.600
Just do it once.

00:06:40.600 --> 00:06:41.360
OK.

00:06:41.360 --> 00:06:45.400
Which means I need a
negative sign in front.

00:06:48.130 --> 00:06:51.620
A negative sign in front
of the entire product.

00:06:51.620 --> 00:06:52.495
Well, let's continue.

00:06:55.150 --> 00:06:56.540
I'm going to write it down here.

00:06:56.540 --> 00:07:00.300
As you can see the blue
part indicate the signature

00:07:00.300 --> 00:07:01.910
of columns.

00:07:01.910 --> 00:07:05.530
So next term, I'm
going to put-- I've

00:07:05.530 --> 00:07:10.100
exhausted the first possibility
of choosing 1, 6, right?

00:07:10.100 --> 00:07:12.840
So let's look at 2, 5.

00:07:12.840 --> 00:07:16.700
2 times 5 times 9 times 12.

00:07:20.680 --> 00:07:22.140
What is the column?

00:07:22.140 --> 00:07:24.340
What are the column numbers?

00:07:24.340 --> 00:07:28.740
2 comes from (1, 2) entry,
so I have a 2 in the front.

00:07:28.740 --> 00:07:34.200
5 is (2, 1) entry, so
2, 1, and then 3, 4.

00:07:34.200 --> 00:07:37.090
3, 4.

00:07:37.090 --> 00:07:41.530
Again, I need one
exchange to get back

00:07:41.530 --> 00:07:44.590
to 1, 2, 3, 4,
which means I have

00:07:44.590 --> 00:07:48.200
a negative sign in the front.

00:07:48.200 --> 00:07:52.070
The last one would be 2, 5.

00:07:52.070 --> 00:07:58.438
2 times 5 times 10, 11,
so times 10 times 11.

00:08:01.050 --> 00:08:09.900
Columns are a_(1,2),
a_(2,1), a_(3,4), a_(4,3).

00:08:09.900 --> 00:08:14.270
So you exchange the first two
spots, and the last two spots.

00:08:14.270 --> 00:08:16.250
You get back to 1, 2, 3, 4.

00:08:16.250 --> 00:08:20.550
But you have to do twice,
two exchanges, which means

00:08:20.550 --> 00:08:22.810
you need a plus sign in front.

00:08:22.810 --> 00:08:24.000
That's it.

00:08:24.000 --> 00:08:28.850
This big summation formula comes
down to the sum of four terms.

00:08:28.850 --> 00:08:30.940
And you can compute it.

00:08:30.940 --> 00:08:33.830
If you don't make
any algebra mistake,

00:08:33.830 --> 00:08:35.929
the result should be 8.

00:08:38.950 --> 00:08:40.940
Did you get the right answer?

00:08:40.940 --> 00:08:42.669
All right, time is
really precious.

00:08:42.669 --> 00:08:44.780
Let's move on to
the second part.

00:08:44.780 --> 00:08:48.890
In the second part, we need
to find cofactors c_(1,1),

00:08:48.890 --> 00:08:50.540
c_(1,2), c_(1,3), and c_(1,4).

00:08:54.950 --> 00:08:56.330
Let me put the second part here.

00:09:02.590 --> 00:09:04.730
So we're looking
for the cofactors

00:09:04.730 --> 00:09:08.850
of the first row of
matrix A. Let's just

00:09:08.850 --> 00:09:10.470
write everything down.

00:09:10.470 --> 00:09:11.640
c_(1,1).

00:09:11.640 --> 00:09:16.140
c_(1,1) is the cofactor
of this entry here.

00:09:16.140 --> 00:09:19.840
So you're looking at
the determinant of this

00:09:19.840 --> 00:09:22.280
left over 3 by 3 matrix.

00:09:22.280 --> 00:09:28.020
So that's the determinant
of-- copy it down--

00:09:28.020 --> 00:09:37.070
[6, 7, 8; 0, 9, 10; 0, 11, 12].

00:09:37.070 --> 00:09:38.880
How would you compute this?

00:09:38.880 --> 00:09:41.490
This is a 3 by 3 matrix.

00:09:41.490 --> 00:09:45.400
Of course, you can use the
big summation formula again.

00:09:45.400 --> 00:09:48.830
In other words,
you can write down

00:09:48.830 --> 00:09:52.980
this determinant-- this
formula, but for the specific 3

00:09:52.980 --> 00:09:54.850
by 3 matrix.

00:09:54.850 --> 00:09:58.550
But that will involve three
factorial terms, right?

00:09:58.550 --> 00:10:00.450
Which is six terms.

00:10:00.450 --> 00:10:01.550
All right.

00:10:01.550 --> 00:10:05.200
Is there a way that
you can do it faster?

00:10:05.200 --> 00:10:08.300
Just notice that the first
column of this matrix

00:10:08.300 --> 00:10:14.180
has only one nonzero entry,
which is this (1, 1) entry 6.

00:10:14.180 --> 00:10:18.070
So why do we just expand
this along the first column

00:10:18.070 --> 00:10:20.340
and use cofactors?

00:10:20.340 --> 00:10:22.340
Let's do it.

00:10:22.340 --> 00:10:27.080
This determinant is equal
to 6 times its cofactor.

00:10:27.080 --> 00:10:30.260
And its cofactor comes to
the determinant of this 2

00:10:30.260 --> 00:10:33.110
by 2 matrix, which is easy.

00:10:33.110 --> 00:10:40.360
That becomes 9 times
12 minus 10 times 11.

00:10:40.360 --> 00:10:43.770
And if you compute this
correctly, that should be -12.

00:10:47.710 --> 00:10:48.960
That's not too bad.

00:10:48.960 --> 00:10:53.280
Let's look at the
second one, c_(1,2).

00:10:53.280 --> 00:10:59.220
c_(1,2) is the determinant
of this 3 by 3 matrix.

00:10:59.220 --> 00:11:03.000
So I have to delete the first
row and the second column.

00:11:03.000 --> 00:11:07.100
And I read what is left
over and I put it here.

00:11:07.100 --> 00:11:19.940
So that's
[5, 7, 8; 0, 9, 10; 0,  11, 12].

00:11:19.940 --> 00:11:21.020
Same thing.

00:11:21.020 --> 00:11:24.490
The first column has
only one nonzero entry.

00:11:24.490 --> 00:11:28.770
You expand it along the
first column, use cofactors,

00:11:28.770 --> 00:11:33.080
the result will be 10.

00:11:33.080 --> 00:11:35.450
There's one thing that
I've forgotten here.

00:11:35.450 --> 00:11:38.360
Because we are looking
at (1, 2) entry.

00:11:38.360 --> 00:11:43.190
So for the cofactor, I have
to put an extra negative sign

00:11:43.190 --> 00:11:43.770
here.

00:11:43.770 --> 00:11:48.720
So this is actually -1 times
the determinant of this 3

00:11:48.720 --> 00:11:50.210
by 3 matrix.

00:11:50.210 --> 00:11:52.770
And the result will be 10.

00:11:52.770 --> 00:11:56.400
OK, let's continue.

00:11:56.400 --> 00:11:58.140
c_(1,3).

00:11:58.140 --> 00:12:01.460
Let's hope the computation is
going to get easier and easier.

00:12:01.460 --> 00:12:06.820
So c_(1,3) is the determinant
of the leftover 3 by 3 matrix.

00:12:06.820 --> 00:12:09.350
So I'll directly read from that.

00:12:09.350 --> 00:12:17.880
That's going to be 5, 6,
8; 0, 0, 10; and 0, 0, 12.

00:12:20.740 --> 00:12:24.080
What is the determinant
of this matrix?

00:12:24.080 --> 00:12:27.110
Again, you can use
the same method

00:12:27.110 --> 00:12:30.580
as we did for the first two,
because the first column has

00:12:30.580 --> 00:12:32.610
only one nonzero entry.

00:12:32.610 --> 00:12:36.450
And if you do that,
the result should be 0.

00:12:36.450 --> 00:12:39.520
But you should be
able to tell it

00:12:39.520 --> 00:12:41.990
without any direct computation.

00:12:41.990 --> 00:12:43.570
Why is that?

00:12:43.570 --> 00:12:46.950
Well, clearly the
first two columns

00:12:46.950 --> 00:12:53.040
are linearly dependent, because
the second column is 6 over 5

00:12:53.040 --> 00:12:54.720
times the first column.

00:12:54.720 --> 00:12:56.820
So this is a singular matrix.

00:12:56.820 --> 00:13:00.150
Right away, the
determinant is 0.

00:13:00.150 --> 00:13:01.010
So what is c_(1,4)?

00:13:04.870 --> 00:13:12.068
c_(1,4) is the determinant of
the matrix of [5, 6, 7; 0, 0,

00:13:12.068 --> 00:13:16.550
9; 0, 0, 11].

00:13:16.550 --> 00:13:17.310
Same thing.

00:13:17.310 --> 00:13:21.150
It's singular, so
its determinant is 0.

00:13:21.150 --> 00:13:21.870
All right.

00:13:21.870 --> 00:13:25.050
So that completes
the second part.

00:13:25.050 --> 00:13:27.580
You can move on to
the third part now,

00:13:27.580 --> 00:13:30.890
but before you do
that, just notice

00:13:30.890 --> 00:13:33.690
this may be a good point
to check your answer

00:13:33.690 --> 00:13:35.910
from the first part.

00:13:35.910 --> 00:13:38.800
Because you have
all the cofactors

00:13:38.800 --> 00:13:43.150
of the first row of A, and
if you use the other formula

00:13:43.150 --> 00:13:45.520
to compute the
determinant of A, you

00:13:45.520 --> 00:13:52.360
can see the determinant of A
is equal to the dot product

00:13:52.360 --> 00:13:56.000
of the first row
with its cofactors.

00:13:56.000 --> 00:13:58.990
So because the last
two cofactors is 0,

00:13:58.990 --> 00:14:02.080
so I only have two
terms in the sum.

00:14:02.080 --> 00:14:08.040
So determinant A is equal to
a_(1,1) entry, which is 1,

00:14:08.040 --> 00:14:19.930
times c_(1,1), which is -12,
plus a_(1,2) entry, which is 2,

00:14:19.930 --> 00:14:22.560
times c_(1,2), which is 10.

00:14:25.880 --> 00:14:27.510
What is that?

00:14:27.510 --> 00:14:33.300
-12 plus 20, that give you 8.

00:14:33.300 --> 00:14:37.270
OK, at least the answers
from the first two parts

00:14:37.270 --> 00:14:38.750
are consistent.

00:14:38.750 --> 00:14:43.510
So by now, you should be
more confident to move on

00:14:43.510 --> 00:14:45.560
to the third part.

00:14:45.560 --> 00:14:47.330
I have a problem of space.

00:14:47.330 --> 00:14:50.510
So please allow me to
put the third part here.

00:14:50.510 --> 00:14:53.030
I'm going to come all
the way back here.

00:14:53.030 --> 00:14:55.970
So that's my third part.

00:14:55.970 --> 00:14:57.570
What is the third part?

00:14:57.570 --> 00:15:03.130
Third part asks you to find
the first column of A inverse.

00:15:03.130 --> 00:15:07.300
Well, it seems that involves
more computation of cofactors,

00:15:07.300 --> 00:15:11.900
but as many well
designed exam problems,

00:15:11.900 --> 00:15:15.100
the answer from third part
can be directly derived

00:15:15.100 --> 00:15:19.030
from the first part
and the second part.

00:15:19.030 --> 00:15:21.800
That's how you should do it.

00:15:21.800 --> 00:15:25.050
So what is A inverse?

00:15:25.050 --> 00:15:31.570
The formula for A inverse is
equal to 1 over determinant

00:15:31.570 --> 00:15:40.050
of A times the transpose of
a matrix C. This matrix C

00:15:40.050 --> 00:15:43.690
is composed by
cofactors of matrix A.

00:15:43.690 --> 00:15:47.840
We want to find the first
column of A inverse.

00:15:47.840 --> 00:15:58.260
So the first column
of A inverse should

00:15:58.260 --> 00:16:02.470
be one of our determinants
of A, this constant,

00:16:02.470 --> 00:16:07.280
times the first row of
C, but transpose, right?

00:16:07.280 --> 00:16:15.030
So I have to put here, first
row of C, but transpose.

00:16:17.700 --> 00:16:21.620
Determinant of A comes
from the first part,

00:16:21.620 --> 00:16:25.700
and the first row of C
comes from the second part.

00:16:25.700 --> 00:16:29.480
So I just copy what I have
from the first two parts down.

00:16:29.480 --> 00:16:31.750
That's 1 over 8.

00:16:31.750 --> 00:16:38.840
This column vector will
become [-12; 10; 0; 0].

00:16:41.540 --> 00:16:42.930
That's it.

00:16:42.930 --> 00:16:45.370
That completes this problem.

00:16:45.370 --> 00:16:46.770
Have you got your answer right?

00:16:49.590 --> 00:16:52.580
OK, before I finish,
there are two things

00:16:52.580 --> 00:16:54.730
that I like to remind you.

00:16:54.730 --> 00:16:58.960
First, as you can see,
this is an exercise

00:16:58.960 --> 00:17:03.690
on the big summation formula
to compute determinant of A.

00:17:03.690 --> 00:17:07.960
In previous recitation, we
practice using the combination

00:17:07.960 --> 00:17:11.020
of elimination and the
method by cofactors

00:17:11.020 --> 00:17:12.730
to compute determinants.

00:17:12.730 --> 00:17:16.000
But we should never
forget this formula.

00:17:16.000 --> 00:17:17.819
Because it always works.

00:17:17.819 --> 00:17:20.579
And in a lot of cases,
this will turn out

00:17:20.579 --> 00:17:23.480
to be an easy way to
compute determinant.

00:17:23.480 --> 00:17:27.260
For example, for
this matrix A here.

00:17:27.260 --> 00:17:30.940
And second, in
your real exam, it

00:17:30.940 --> 00:17:34.930
would be really helpful
if you can put down

00:17:34.930 --> 00:17:39.900
your work like this, because it
helps you to check your work.

00:17:39.900 --> 00:17:43.990
And also, even if you don't
get your final answer correct,

00:17:43.990 --> 00:17:47.130
this may get you
some partial credits.

00:17:47.130 --> 00:17:51.180
OK, I'm going to stop here,
and thank you for watching.