Find the rank of the augmented matrix of the following system of equations: negative nine 𝑥 minus three 𝑦 equals two and negative 45𝑥 minus 15𝑦 equals 10.
Now, the first step in this problem is to find the augmented matrix for our system of equations. And to set up our augmented matrix, we’re gonna have the first column is our 𝑥 coefficients, the second column is our 𝑦 coefficients, and the third column is going to be our constants or the solutions to each of the equations. So we’re gonna have the matrix which reads negative nine, negative three, two, negative 45, negative 15, 10. And sometimes in an augmented matrix, you may also see a dotted line or a line between the 𝑦 terms and the constants. And this is just to show the partition between the two, so where the equal signs would go.
I’m just gonna leave it as one matrix just for what we’re doing today. Okay, so we’ve now found our augmented matrix. So now, what we want to do is find the rank of this augmented matrix and how we’re going to do this. So first of all, before we try and calculate the rank, we have a couple of rules that can help us think about what the rank might be. And these two rules are if 𝑟 is less than 𝑐, then the max rank is 𝑟. And that’s where 𝑟 is the rows and 𝑐 is the number of columns. And if 𝑟 is greater than 𝑐, the max rank is 𝑐. Well, if we apply it to our matrix, we can see that we’ve got two rows and three columns. So 𝑟 is equal to two and 𝑐 is equal to three. So, therefore, 𝑟 is less than 𝑐. So that means that our max rank is going to be two.
So we now know the maximum rank of our augmented matrix. But let’s work out what the rank is going to be. So now, if we want to calculate the rank of our augmented matrix, what we need to see is how many linear independent rows there are because this will help us determine what the rank is. So what we mean by a linearly independent vector row is where a row cannot be made up by being a multiple of a scalar of one of the other rows. Or it can’t be made with a linear combination of the other rows, for instance, by adding two rows together.
Well, if we take a look at our matrix, we can see that we haven’t got two linearly independent rows. And that’s because row two is equal to five multiplied by row one, so is the scalar multiple of row one. And that’s because if you do negative nine by five, you get negative 45. Or negative three multiplied by five gives you negative 15. Or two multiplied by five gives you 10.
So, therefore, as we’ve shown that both the rows cannot be linearly independent, that means that our matrix only has one linearly independent row. So, therefore, we can say that the rank of the augmented matrix of the system of equations, negative nine 𝑥 minus three 𝑦 equals two and negative 45𝑥 minus 15𝑦 equals 10, is one. And we can just check that it satisfies what we found earlier. And that is that the max rank is gonna be two. Well, one is less than two. So it satisfies that. So we can see that definitely the rank is equal to one.