CHAPTER 05.10: SYSTEM OF
EQUATIONS: Distinguishing between consistent and inconsistent system of
equations based on rank Example 2 In
this segment we will talk about whether through an example whether we can
distinguish if a system of equations is consistent or inconsistent. And let’s
pose the problem, this is the problem statement: is this particular set of
equations consistent or inconsistent? Is it consistent or inconsistent? So
that’s what we have to figure out. So the first thing which we have to do is
we have to find the – let’s write it down – in the symbolic form. This is A X
is equal to C, where A is the coefficient matrix, X is the solution vector,
and C is the right hand side vector. And see that whether this particular
system of equations is consistent or inconsistent. So
the first thing which I have to do is to find out the rank of A. So I have to
find the rank of the coefficient matrix and then to rank of the augmented
matrix which will be the C vector as the fourth column attached to the A
matrix and see if the two ranks are the same. If they are the same, then it
is consistent. If the rank of A is less than the rank of the augmented matrix,
then it is inconsistent. Now in this case, in order to find the rank of A,
what I have to do is I have to look for the largest order of the square
submatrices. This is a three by three matrix, so the largest square submatrix
is the matrix itself. SO determinant of A is in this case 0. So it’s a three
by three, three by three is the largest order of the square submatrix which I
can get. So I know that rank of A is less than or equal to three of course.
The determinant of A is equal to 0. If the determinant of A is equal to 0,
that means that the rank of the A matrix cannot be three. Because there are
no other three by three submatrices out of this three by three matrix. SO we
know that the rank of A is going to be less than or equal to two. Let’s
go and see that – hey – is it two. So it is two, well let’s suppose we take a
two by two matrix which is like this one right here and in this case, so we
have the determinant of the submatrix which I choose; 5, 1, 8, 1 let’s
suppose. What do I get? I get 5 times 1 minus 8 times 1 which is -3. That’s
not equal to 0. So I do have a two by two submatrix for which the determinant
is not equal to 0. So that tells me that the rank of A – that the rank of A
in fact is two. Let’s go and see whether we can show – what we can show about
the rank of the augmented matrix? The augmented matrix for the example which
we have is that we take the A matrix, which is 25, 5, 1, 64, 8, 1, 89, 13, 2,
106.8, 177.2, 284.0. So what we want to be able to do – this is the augmented
matrix which I can denote like this, and we want to find out – hey what is
the largest square submatrix – largest order of square submatrix – for which
the determinant is not equal to 0. So here I have three rows and four columns
and I know that the largest square submatrix is going to be of the order of
three by three. So I need to look at three by three matrices. So if I look at
– let’s suppose if I take the first three rows and first three columns again,
25, 5, 1, 64, 8, 1, 89, 13, 2, 106.8, 177.2, 284.0 – and I find the
determinant of that, I get zero. Which I had in the previous case as well
when I was looking at the rank of A matrix. But that doesn’t make that the
rank of the augmented matrix is not 3, because there are other three by three
square submatrices which are possible. Let’s
look at – let’s say hey – I’m not going to account for the first row, first
column. I get 5, 1, 106.8, 8, 1, 17.2, 13, 2, 284.0.
By taking out the first column I get these three rows and three columns
submatrix of my augmented matrix and I find the determinant of the and that also turns out to be 0. But I still cannot
give up because there are other three by three square submatrices. So let me
get rid of the second column. So if I get rid of the second column here, I
get 25, 1, 106.8, 64, 1, 177.2, 89, 2, 289.0. That
is three by three submatrix, which I get by getting rid of the second column.
The determinant of this is also equal to 0. So I’m not having any good luck
in trying to figure out the three by three matrix for which the determinant is
equal to 0. But I’ve got to look at all the possibilities till I find out –
hey either all of the possible three by three submatrices have a determinant
equal to zero, or one of them start to have a determinant which is not equal
to 0. So
let’s find out the determinant of the other submatrix, the three by three
submatrix which I can find by getting rid of the third column. So then I have
25, 5, 64, 8, 89, 13, 106.8, 177.2, 284.0 and if you
do this, you’ll find the determinant of this is also equal to 0. So what you
are finding out is that the determinant of all the possible three by three
submatrices is 0. There are no more three by three - other three by three - submatrices
available. But of all the three by three submatrices the determinant is equal
to 0. So what that basically tells me is that the rank of the augmented
matrix has to be less than or equal to 2. It cannot be three submatrix because
I cannot find a three by three submatrix for which the determinant is
non-zero. Let’s
go and see what’s the rank of the augmented matrix.
Is it 2, is it 1, is it zero? So what I’ll have to do for that is that I’ll
have to take a two by two submatrix and see whether the determinant is not
equal to zero. So I can do – is I can maybe take this right here. I could
take this one right here for example and if I find the determinant of that
which is the determinant of 5, 1, 8, 1, it is one of the two by two
submatrices which turns out to be -3. So I have been able to find a two by
two matrix for which the determinant is nonzero. So the rank of the augmented
matrix is equal to 2. So what I had was that the rank of A was 2. The rank of
the augmented matrix is 2. So since the two ranks are the same that means
this is still a consistent system of equations. Because the consistency of a
system of equations is dependent on the rank of the coefficient matrix and
the rank augmented matrix, and those being equal. So it is a consistent system
of equations. And that’s the end of this example. |