CHAPTER 05.11: SYSTEM OF EQUATIONS: Distinguishing between consistent and inconsistent system of equations based on rank of matrices Example 3

 

 

In this segment we will talk about how we can distinguish between consistent and inconsistent system of equations based on rank of a matrix. So I am going to do an example. The problem statement is this set of equations. 25 5 1 64 8 1 89 13 2. X1 X2 and X3. That is the unknown vector. It is equal to 106.8 177.2 and 280.0 Consistent or inconsistent? So to be able to figure out whether this system of equations is consistent or inconsistent, what we are to do is: Let’s fist write this in the symbolic form. What we have to show is that: What is the rank of A and what is the rank of the (A:C) matrix. If the rank of the coefficient matrix of the (A:C) matrix, then we have a consistent system of equations. If the rank of the coefficient matrix is less than the rank of the (A:C) matrix, then it is an inconsistent system of equations.

 

So let’s concentrate finding the rank of A. So here we have A, which is a 3 by 3 matrix, so that is the largest order of the sub-matrix which I can find there. And the determinate of A is turning out to be 0. If the determinate of A is turning out to be 0 and that is the only 3 by 3 sub-matrix that I can find of A, then that itself tells me that the rank of A is less than or equal to 2. It can’t be 3 because the only square matrix by the order of 3 by 3 of that determinate is 0, so the rank of a is less than or equal to 2. So let’s see if it is 2 1 or 0. So let’s suppose I take a sub matrix like this one a 2 by 2. Now I’m going to look at 2 by 2 sub matrices. The determinate of this sub matrices, which is 5 1 8 1, is turning out -3. That is not equal to 0. So what I am able to find out here is that it is a 2 by 2 matrix for which the determinate is not equal to 0, that tells me the rank of A is equal to 2.

 

Now what I have to do is, hey, what is the rank of the (A:C) matrix? The augmented matrix is the  coefficient matrix. And then we have an additional column, which is the right hand side vector. So we have: 25 5 1 64 8 1 89 13 2. Then we have this vector right here, our forth column. And what we want to be able to show or what we want to be able to see is: What is the rank of this augmented matrix? So this is 3 rows and 4 columns so I know the largest square sub-matrix that I can get is what? It’s the order 3 by 3. So if I look at the 3 by 3 sub matrix, which is the largest square matrix I can get out of this-Let’s look at this particular matrix right here- the determinate of 25 5 1 64 8 1 and 89 13 2- that is giving me 0. That tells me, hey, I still have to look at the other sub-matrices which are 3 by 3 size to see whether I am getting a determinate which is not equal to 0 or do I keep on getting 0.

 

So let me take another sub-matrix right here. Let’s suppose I take-I get rid of the 1st column, for example. Let me see what the determinate of that is. The determinate of this augmented matrix without the 1st column. 5 1 106.8 and I’m still looking at 3 by 3 matrices- 3 by 3 sub-matrices. 177.2 13 2 and 280.0. The determinate of this matrix is turning out to be 12. That is not equal to 0. So the determent of this matrix is equal to 12 and its not equal to 0, what  that means is that hey, I found a sub-matrix, which is by order 3 by3 with the determinate is not equal to 0 so that tells me that the rank of the augmented matrix is 3. So what I have found out is the rank of the A matrix which is the coefficient matrix, came out to be 2. The rank of the augmented matrix came out to be 3. And I’m finding out the rank of the coefficient matrix is not the same as the rank of the augmented matrix, which tells me the rank of this system of equations, is inconsistent. When we talk about a system being inconsistent that means no solution exists. And that is the end of this segment.