MARTINA BALAGOVIC: Welcome.
Today's problem actually appeared in a quiz.
It appeared in quiz one in fall of 1999 as question four.
The problem puts the usual solve the following system upside
down by saying we have some matrix and we know that all
the solutions to A*x equals this vector here, [1, 4, 1, 1],
all the solutions to this problem are given by x equals
[0, 1, 1] plus any number c times [1, 2, 1].
And we're asked to say everything
that we can about the columns of the matrix A.
So I'm going to let you pretend that you are on an exam,
try to solve it yourself, and then come back
and compare your solution with mine.
OK, welcome back.
So the first thing that you should
think about in this sort of situation
is what is the size of A?
Well, we want to multiply A with an x that has three entries,
so A should have three columns.
Let me call those columns c_1, c_2, and c_3.
And when I take some linear combinations of c_1, c_2
and c_3, I'm going to get this vector here, [1, 4, 1, 1].
So all the c_i's, c_1, c_2, and c_3 are vectors in R_4.
Now, if you know about, if you learned
about particular solutions and special solutions, then
my notation here shouldn't surprise you.
I'm going to call this vector here x_p,
and this vector here x_s.
And I'm going to use the fact that
x_p plus c times x_s satisfies A times this
equals b-- I will call this vector b-- for any number c.
In particular, what I'm going to conclude
is that when c equals 0 we get A times x_p equals b.
But also that when c equals 1, we
get A times x_p plus A times x_s equals b.
Replacing this by b, we get that this implies
that A times x_s equals 0.
So in trying to find what are the columns
c_1, c_2, and c_3 of the matrix A, let's look at these two
equations.
x_p satisfies A times x_p equals b,
and x_s satisfies A times x_s equals 0.
Again, if you know what particular and special
solutions are this shouldn't surprise you.
But we also know what x_p and x_s are,
so let's use them to try to calculate c_1, c_2, and c_3.
A times x_p equals b means that the linear combination of c_1,
c_2, and c_3 encoded in the vector x_p, which is [0, 1, 1],
gives the vector b.
So c_1, c_2, c_3 times [0, 1, 1] gives us [1, 4, 1, 1].
In other words, c_2 plus c_3 equal b.
Let's turn our attention to A times x_s equals 0.
This says that c_1, c_2, c_3 times--
x_s was defined to be [0, 2, 1]-- equals 0.
In other words, 2 times c_2 plus c_3 equals 0.
Now solving this system where the unknowns are vectors
but it's still just a linear system, we can see,
for example, from the second equation that c_3 equals minus
2*c_2.
And plugging it back into the original equation,
getting c_2 minus 2*c_2 equals b,
from which it follows that c_2 is equal to minus b,
and that c3 is equal to 2 times b.
So from this tiny amount of information--
we just knew the solutions to this one particular equation
involving A-- we got the second column of A
and the third column of A completely explicitly
calculated.
Now, what can we say about the first column?
I said before that all the solutions of A*x equals b are
of the form a particular solution plus some number times
a special solution.
And the information that we have is that there's just one number
here.
So we said everything, once we remove this vector here,
everything that we get here will satisfy A times x equals 0.
And the fact that everything that
satisfies A times x equals 0 is a multiple of this one vector
that was given to us means that the null space of A
has dimension one.
There is just one special solution.
So dimension of the null space of A is 1.
So rank of A is the number of columns
minus this dimension of null space, and it's equal to 2.
As rank of A is equal to 2, the number
of linearly independent columns needs to be 2 as well.
So the only thing that we can say at this point
is if the first column was also a multiple of b,
as the second and the third are, then the rank would be smaller
than 2.
So that is the only thing that cannot happen.
So c_1 is not a multiple of b.
Not any multiple, including not a zero multiple.
And that's pretty much everything we can say.
Yes, if it was some other multiple of it,
then we would be able to find some other vector here
and we would have two parameters.
But it's not, and this is everything
that we can say about it.