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Solving Linear Systems Using Matrix Algebra - Problem 3 2,268 views

Teacher/Instructor Norm Prokup
Norm Prokup

Cornell University
PhD. in Mathematics

Norm was 4th at the 2004 USA Weightlifting Nationals! He still trains and competes occasionally, despite his busy schedule.

Let’s try another problem. This time I’m going to use my calculator to help me out. I have the system 5x plus 4y minus z equals 3 2x minus y equals -1, -2x minus 5y equals plus z equals -2.

Now the first step in solving a system using matrices is to convert this into a matrix equation. So I start by looking at the left hand aside. I want to make a coefficient matrix out of this. So I look at the coefficient is 5 4 -1, 2 -1 0, remember to represent missing term as a 0. -2 -5 1, so that’s my coefficient matrix and I’ll call that a.

And then I want my variable matrix x y z. You can tell that this is a right variable matrix because you multiply you get 5x plus 4y minus z, that’s the first line. And of course that has to equals the constant matrix, 3 -1 -2. So this I’m calling capital X and this one I’ll call b, a times X equals b.

That’s our matrix equation. I want to use the TI-84 to find the inverse of this coefficient matrix, so let’s take a look at the TI-84 quickly. So here we are on the TI-84 we get the home screen right now. I want to show you how to enter matrices because that’s how we are going to have to deal with our finding our inverse matrix.

So you go second, and then the reciprocal. So -1 and that the matrix menu on TI-84. And we want to edit, that’s the third menu, so I need to hit arrow, right arrow and I’m going to edit matrix A. So I hit enter.

Now the first thing you have to do is enter the dimensions. You see this two by two, we need to make ours a three by three. So I hit 3 enter 3 enter, and now I've got a three by three matrix. Now all I have to do is enter the entries and it goes across and down. So 5 enter, 4 enter, -1 enter. Oops! A little mistake there.

Watch out for the difference between the minus sign and the negative. And then 2 enter, minus 1 enter, 0 enter, minus 2 enter, minus 5 enter, and 1 enter. Just double check, that looks good. So that’s our matrix A.

Now let me quit. Now to get the inverse, what I want to do is in the home screen is I want to call up the matrix menu again. And I want to just get the name matrix of A. You can’t just type this in, you have to actually go into the matrix menu and hit matrix A. And the first thing it gives you is matrix A. I want to take the reciprocal of that and actually not the reciprocal, I’m sorry, the inverse. We can’t take the reciprocal of matrices. But if you bit the x to the -1, button you get that same symbol, that means inverse of the matrix.

So hit enter, Bam! There it is. So that’s our inverse matrix and we'll go back to the board and use this to solve our system. So we have our inverse matrix, let’s write that down. I've got a inverse equals 1 -1 1, 2 -3 2 and 12 -17 13. Recall how we solve a system of equations using the matrices.

We start with the matrix equation ax equals b and the solution is going to be x equals a inverse b. Remember a inverse on the left. Now a inverse is 1 -1 1, 2 -3 2 and 12 -17 13 and then b is the constant matrix here; 3 -1 -2. So it’s just a question of, multiplying. We have 3 plus 1, 4, minus 2 is 2. Then we have 6 plus 3 is 9, minus 4 is 5 and we have 36 plus 17 that’s 53 minus 26 is 27.

So according to this we have x equals 2, 2, y equals 5 and z equals 27. Now let’s double check this on the TI-84.

We are back. So if I want to find out if my solution on the board was right, I’m going to need to enter the matrix b the constant matrix. So I’m going to go into the matrix mode, second matrix and I need to edit. 2 over and I’ll make it matrix b. So number 2 and I need to change this to a 3 by 1. And the entries are 3 -1 and -2.

So all that’s left to do, is to multiply a-inverse, which is actually the answer that I have there. Second answer times matrix b, so I just enter 2 for matrix b and there it is. X equals 2, y equals 5 and z equals 27.

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