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.