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Cramer's Rule - Problem 2
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I want to show you how Cramer’s Rule works on a system of three equations with three unknowns. I have a system like that right here. Now it’s going to work a lot like the two equations with two unknowns system.

First thing I have to do is look at the determinant of the coefficient matrix. Now the coefficient matrix is going to look like this; 2 3 -1, 3 -1 2, 7 -4 3. And the determinant of that, is that. I'm going to call that value d. That’s the determinant of the coefficient matrix.

Now I’m going to need to calculate three other determinants. The next one is a, I’ll call it a. Now basically it’s the same determinant, only I’m going to replace this first column, which is the x column, with these constants here. So let me just write the other constants; 3 -1 -4, -1 2 3, and I’ll substitute in the 14, 4 and -4.

And then b is going to be the same determinant, only now I’m going to put the constants in this column. So 2 3 7, -1 2 3, and now I put my 14 4 -4 in the second column. And finally c equals 2 3 7, and I’ve got 3 -1 -4, and I put my constants in the last column, 14 4 -4.

How does this work? So now I’m going to calculate the x, y and z values this way. It’s going to be a over d, this determinant over this determinant. For the y values it's going to be b over d. Notice, the one I’m choosing for the top is going to be the one where I replaced the y coefficients with the constants. So that goes with the y equals part of the solution; b over d. And this is going to be c over d. C is the one where I replaced the z coefficients with the constants. Now I’ve got 4 determinants here.

The 3 by 3 determinants, they are not easy to evaluate so I’m going to use the TI to evaluate all four of them. Let’s take a look at that now.

The first thing we have to do, in order to evaluate determinants is, learn how to enter a matrix so that we can evaluate its determinant. I go into the matrix mode. I’ve already taken the liberty of entering a, b and c, so we’ll enter d.

We have to go to the edit menu and then down to d, enter. So you need to change this to a 3 by 3 and now we just enter the values. 2 3 -1, 3 -1 2, 7 -4 3. Now we’re ready to take the determinant.

So we quit and we go second matrix, and now on the Math menu under matrices, it’s the very first thing, the determinant. I’ll calculate the determinant of and now I need to go into the matrix menu again. You want to choose the name d. Let’s do d first.

So I just hit the number 4 to get d. Let’s calculate that determinant, it’s 30. So we’ll remember that and we’ll use it later. Now let’s do the same thing. I’m going to use the second entry so I don’t have to keep pressing buttons. Let me get the determinant of a. I need to go into names again, and get a, I'll just hit enter. I get 30. Second entry again, let’s do the same thing for b. Second matrix, I choose number 2, hit enter, 150. This is very quick if you use second entry a lot, one more.

Now we have all four determinants. D has a determinant of 30, 'a' has a determinant of 30, b is 150, c is 90. Let’s go back to the board and let’s finish our problem. We’re back. Let’s fill in the numbers that we just found.

We found that d was 30, a was 30, b was 150 and c was 90. So according to this, x is going to be a over d, that’s 30 over 30, that’s 1. Y is going to be b over d, that’s 150 over 30, that’s 5. And z is going to be c over d, that’s 90 over 30, that’s 3. And that’s our solution; 1, 5, 3.

Cramer’s Rule works very similarly for a 3 by 3 system as it does for a 2 by 2 system. Only this time, because you have three coordinates to your solution, you’re going to have 4 determinants that you have to evaluate. The first one is the determinant of the coefficient matrix, and the rest of them, you take one column of the coefficient matrix, and replace it with the constants in your equation, and then you calculate. You always calculate x, y and z as a ratio of two determinants. This one in the bottom is always the determinant of the coefficient matrix.

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