WEBVTT
00:00:05.176 --> 00:00:06.300
MARTINA BALAGOVIC: Welcome.
00:00:06.300 --> 00:00:10.840
Today's problem actually
appeared in a quiz.
00:00:10.840 --> 00:00:16.440
It appeared in quiz one in
fall of 1999 as question four.
00:00:16.440 --> 00:00:20.420
The problem puts the usual solve
the following system upside
00:00:20.420 --> 00:00:24.130
down by saying we have some
matrix and we know that all
00:00:24.130 --> 00:00:29.500
the solutions to A*x equals
this vector here, [1, 4, 1, 1],
00:00:29.500 --> 00:00:33.750
all the solutions to this
problem are given by x equals
00:00:33.750 --> 00:00:38.820
[0, 1, 1] plus any
number c times [1, 2, 1].
00:00:38.820 --> 00:00:40.400
And we're asked
to say everything
00:00:40.400 --> 00:00:43.180
that we can about the
columns of the matrix A.
00:00:43.180 --> 00:00:46.067
So I'm going to let you pretend
that you are on an exam,
00:00:46.067 --> 00:00:47.900
try to solve it yourself,
and then come back
00:00:47.900 --> 00:00:49.400
and compare your
solution with mine.
00:00:57.210 --> 00:00:58.610
OK, welcome back.
00:00:58.610 --> 00:01:00.350
So the first thing
that you should
00:01:00.350 --> 00:01:02.350
think about in this
sort of situation
00:01:02.350 --> 00:01:04.140
is what is the size of A?
00:01:04.140 --> 00:01:10.710
Well, we want to multiply A with
an x that has three entries,
00:01:10.710 --> 00:01:14.734
so A should have three columns.
00:01:14.734 --> 00:01:25.300
Let me call those columns
c_1, c_2, and c_3.
00:01:25.300 --> 00:01:27.700
And when I take some linear
combinations of c_1, c_2
00:01:27.700 --> 00:01:32.420
and c_3, I'm going to get this
vector here, [1, 4, 1, 1].
00:01:32.420 --> 00:01:43.380
So all the c_i's, c_1, c_2,
and c_3 are vectors in R_4.
00:01:46.150 --> 00:01:49.130
Now, if you know
about, if you learned
00:01:49.130 --> 00:01:55.064
about particular solutions
and special solutions, then
00:01:55.064 --> 00:01:56.730
my notation here
shouldn't surprise you.
00:01:56.730 --> 00:01:59.410
I'm going to call
this vector here x_p,
00:01:59.410 --> 00:02:01.990
and this vector here x_s.
00:02:01.990 --> 00:02:05.210
And I'm going to
use the fact that
00:02:05.210 --> 00:02:15.990
x_p plus c times x_s
satisfies A times this
00:02:15.990 --> 00:02:22.020
equals b-- I will call this
vector b-- for any number c.
00:02:26.580 --> 00:02:28.850
In particular, what
I'm going to conclude
00:02:28.850 --> 00:02:40.030
is that when c equals 0 we
get A times x_p equals b.
00:02:40.030 --> 00:02:42.680
But also that when
c equals 1, we
00:02:42.680 --> 00:02:53.750
get A times x_p plus
A times x_s equals b.
00:02:53.750 --> 00:02:57.830
Replacing this by b, we
get that this implies
00:02:57.830 --> 00:03:04.820
that A times x_s equals 0.
00:03:04.820 --> 00:03:07.230
So in trying to find
what are the columns
00:03:07.230 --> 00:03:10.840
c_1, c_2, and c_3 of the matrix
A, let's look at these two
00:03:10.840 --> 00:03:11.770
equations.
00:03:11.770 --> 00:03:15.370
x_p satisfies A
times x_p equals b,
00:03:15.370 --> 00:03:19.330
and x_s satisfies A
times x_s equals 0.
00:03:19.330 --> 00:03:21.440
Again, if you know what
particular and special
00:03:21.440 --> 00:03:23.610
solutions are this
shouldn't surprise you.
00:03:23.610 --> 00:03:26.660
But we also know
what x_p and x_s are,
00:03:26.660 --> 00:03:32.340
so let's use them to try to
calculate c_1, c_2, and c_3.
00:03:32.340 --> 00:03:38.630
A times x_p equals b means that
the linear combination of c_1,
00:03:38.630 --> 00:03:44.990
c_2, and c_3 encoded in the
vector x_p, which is [0, 1, 1],
00:03:44.990 --> 00:03:46.730
gives the vector b.
00:03:46.730 --> 00:04:05.850
So c_1, c_2, c_3 times
[0, 1, 1] gives us [1, 4, 1, 1].
00:04:08.890 --> 00:04:15.110
In other words, c_2
plus c_3 equal b.
00:04:17.649 --> 00:04:20.959
Let's turn our attention
to A times x_s equals 0.
00:04:20.959 --> 00:04:30.040
This says that c_1,
c_2, c_3 times--
00:04:30.040 --> 00:04:37.960
x_s was defined to be
[0, 2, 1]-- equals 0.
00:04:37.960 --> 00:04:47.550
In other words, 2 times
c_2 plus c_3 equals 0.
00:04:47.550 --> 00:04:51.170
Now solving this system where
the unknowns are vectors
00:04:51.170 --> 00:04:54.100
but it's still just a
linear system, we can see,
00:04:54.100 --> 00:04:58.490
for example, from the second
equation that c_3 equals minus
00:04:58.490 --> 00:05:00.350
2*c_2.
00:05:00.350 --> 00:05:02.870
And plugging it back into
the original equation,
00:05:02.870 --> 00:05:11.310
getting c_2 minus
2*c_2 equals b,
00:05:11.310 --> 00:05:16.120
from which it follows that
c_2 is equal to minus b,
00:05:16.120 --> 00:05:20.280
and that c3 is
equal to 2 times b.
00:05:23.290 --> 00:05:25.620
So from this tiny
amount of information--
00:05:25.620 --> 00:05:29.870
we just knew the solutions to
this one particular equation
00:05:29.870 --> 00:05:32.980
involving A-- we got
the second column of A
00:05:32.980 --> 00:05:35.310
and the third column of
A completely explicitly
00:05:35.310 --> 00:05:36.540
calculated.
00:05:36.540 --> 00:05:39.980
Now, what can we say
about the first column?
00:05:39.980 --> 00:05:46.760
I said before that all the
solutions of A*x equals b are
00:05:46.760 --> 00:05:51.880
of the form a particular
solution plus some number times
00:05:51.880 --> 00:05:54.660
a special solution.
00:05:54.660 --> 00:05:58.340
And the information that we have
is that there's just one number
00:05:58.340 --> 00:05:58.840
here.
00:05:58.840 --> 00:06:03.640
So we said everything, once
we remove this vector here,
00:06:03.640 --> 00:06:09.380
everything that we get here
will satisfy A times x equals 0.
00:06:09.380 --> 00:06:12.010
And the fact that
everything that
00:06:12.010 --> 00:06:16.050
satisfies A times x equals 0 is
a multiple of this one vector
00:06:16.050 --> 00:06:21.360
that was given to us means
that the null space of A
00:06:21.360 --> 00:06:23.390
has dimension one.
00:06:23.390 --> 00:06:26.000
There is just one
special solution.
00:06:26.000 --> 00:06:34.080
So dimension of the
null space of A is 1.
00:06:37.700 --> 00:06:43.880
So rank of A is the
number of columns
00:06:43.880 --> 00:06:49.850
minus this dimension of null
space, and it's equal to 2.
00:06:49.850 --> 00:06:52.560
As rank of A is equal
to 2, the number
00:06:52.560 --> 00:06:57.350
of linearly independent
columns needs to be 2 as well.
00:06:57.350 --> 00:06:59.740
So the only thing that
we can say at this point
00:06:59.740 --> 00:07:03.090
is if the first column
was also a multiple of b,
00:07:03.090 --> 00:07:07.860
as the second and the third are,
then the rank would be smaller
00:07:07.860 --> 00:07:08.990
than 2.
00:07:08.990 --> 00:07:11.240
So that is the only
thing that cannot happen.
00:07:11.240 --> 00:07:22.400
So c_1 is not a multiple of b.
00:07:22.400 --> 00:07:26.000
Not any multiple, including
not a zero multiple.
00:07:26.000 --> 00:07:29.160
And that's pretty much
everything we can say.
00:07:29.160 --> 00:07:30.999
Yes, if it was some
other multiple of it,
00:07:30.999 --> 00:07:33.165
then we would be able to
find some other vector here
00:07:33.165 --> 00:07:34.760
and we would have
two parameters.
00:07:34.760 --> 00:07:38.040
But it's not, and
this is everything
00:07:38.040 --> 00:07:40.190
that we can say about it.