College Algebra: Using Cramer's Rule in 3x3 Matrix

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So we saw how Cramer's Rule can be used to solve a system of equations where you have two linear equations and two variables. But actually that exact same method works no matter how many equations you have, no matter how many variables, as long as they're the same.
So let me just show you that really fast by taking a look at the following three by three example. And look how easy it is to solve it. No more elimination of variables, now it's just direct, x + y - z = -2. There's the first equation. 2x - y + z = 0, that's the second equation. And the last equation is x - 2y + 3z = 1. There's the third equation. And I want to solve this system, find the x, y or z, where they all meet up, where they all intersect, using Cramer's Rule.
So the first thing I do is compute--now in this case, how many determinants will there be? There will be actually four determinants. There's going to be the D determinant, then there'll be the D[x], D[y] and D[z]. So what's D sub nothing, naked D? That's going to be the determinant of all the coefficients. So I just rip off the coefficients right here and write that in, 1, 1, -1; 2, -1, 1; 1, -2, 3. I just took off the coefficients and now I want to compute the determinant. So what does the determinant equal? Well let's think about and see. This is a three by three baby. So what I'm going to do is have to expand around one of the rows or columns. I'm going to expand around here. So the first thing I do is take a look at this 1 and remember the signs, plus, minus, plus. So I take a look at that 1, and I write it down. So there's a 1. And I'm going to multiply it by what I get when I cross these people out and take the determinant of what's left over. So that's -3 minus -2. So -3 minus -2 is -3 + 2, which is just -1. This is a -1 right in here. Then I subtract a 1. So I've got a 1 right here times, and then I get rid of this and get rid of that, and I take a look at the two by two matrix that remains, and take its determinant. So it would be 6 - 1, which is just 5. And then the sign tells me to add. And what do I add? I add -1, so add that last term, -1. See how I'm expanding around that particular row? So I have -1. And I multiply that by what I get when I get rid of the row and column containing that. So I see this little matrix here, take the determinant, -4 minus -1. So -4 + 1 = -3. So I have a -3. And so what do I see? I see this equals, -1, -5, and this is a +3. And so I see a -6 + 3 = -3. So this determinant D is actually equal to -3. Let me write that down. So D = -3. Notice it's not 0, so if this method will work, if D were to equal 0, I'd have to stop this method right away.
So anyway, that takes care of D. Now I've got to compute these other guys. Let me compute now D[x]. Remember what D[x] is. D[x] is exactly D, but I get rid of the x column and replace it by these. So I just put in a -2, 0, 1 in place of this, and then copy everything else. So 1, -1, -2; -1, 1, 3, and I have to compute this determinant. So this is sort of like determinant central here, folks, with all these determinants, so many determinants, so little time. Let's see what this equals. Again I'll expand around here just for fun. So remember it's plus, minus, plus. I'm going to start to be a little bit faster now. So I take the -2, and I multiply it by that determinant, which is going to be -3 minus -2. So -3 + 2, which is -1. Then I minus, I subtract, the 1 times what I get when I cross that out. And what do I see there? I see a 0 x 3, which is 0, and then -1. I subtract, so I just see a -1. And then I see a plus, a -1 times, and what I see here is what? Well this is a 0. And then I have minus -1 becomes just a +1. So here I see 2 + 1 - 1, which is just 2. So I see D[x] = 2. Now I bet that these are taking a teeny bit of time, but they're not hard at all once you get the hang of it.
Now let's find D[y]. So how do I find D[y]? Well I come back to D, and I get rid of the y column and replace it by these, and compute that determinant. No big deal, so D[y] equals the determinant of--I keep the x column, but now the y column gets replaced by this -2, 0, 1. And then the z column remains the same. And now I have to compute that. So what would that give me? Well let's see. Again, I'll expand around here. Remember the signs, plus, minus, plus. So I have a 1. And then I've got to take that determinant. That's a 0 - 1. So that's -1. Then I've got a minus -2 times what? Well times that determinant, which is, notice, 6 - 1, which is 5. And then I have a plus -1 times that determinant, which is 2 - 0, or 2. So what I see is -1 + 10 - 2. So that's going to be 10 - 3, which equals 7. Great, notice all I'm doing is adding and subtracting and multiplying integers. That's all I'm doing here, no division even, no division. So D[y] = 7.
Okay, one more, I've got to find D[z]. What's D[z]? I take the z column and replace it by this, and compute that determinant. So for D[z], I keep the x column. I keep the y column, but now in the z column I put this in. Let's compute that. Well, not a problem, because what do I see? I see a 1 times that determinant, which is just -1 - 0. So it's -1. And then I see a minus, remember it's a minus sign now because we're in that row, a 1 and then its determinant here, which is 2 - 0. That's just 2 + -2 times that determinant, which is -4 minus -1, which is -3. So I see -1, -2 and +6. So that gives me a net gain of 3. So the determinant of this matrix is 3. So D[z] = 3.
So what's the solution to the system? Well x will just equal D[x] over D. So that's going to be . Y is going to be D[y] over D, which is going to be 7 . And z is going to be D[z] over D, which is going to be . And so I see the solution is simply and -1. And you can check that. You can check those answers by just plugging in these values into x, y and z in each of these three equations, and see they're all satisfied.
It's absolutely amazing. I was able to solve this with not a lot of complicated solving, and the x's drop out, and the z's drop out... No one dropped. In fact, no one dropped at all. This was just adding, subtracting and multiplying the coefficients in a very clever way, making these numbers, getting answers, a really cool way of solving these kind of systems. Try these and you should really enjoy them, because it's a lot easier than solving it the other way. Have fun.
Systems of Equations
Cramer's Rule
Using Cramer's Rule in a 3x3 Matrix Page [2 of 2]