Cornell University
PhD. in Mathematics
Norm was 4th at the 2004 USA Weightlifting Nationals! He still trains and competes occasionally, despite his busy schedule.
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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 a harder problem. I want to take the inverse of a 3 by 3 square matrix. Here is my matrix A, and I've written my matrix A here, and then I've augmented it by adding the order three identity matrix here. The object here is to use row operations to switch this around till I get the identity matrix on the left, and whatever is left on the right, is going to be my inverse matrix.
So row operations. I need to get the identity matrix on the left. So what I'm going to try to do is get rid of this 2 first. One way to do that is to multiply the top row by -2, and add it to the bottom row.
So that would look like -2 times row one plus row 3. Now that's not going to affect rows 1, and 2, so I can just copy those over. 1 6 4, 1 0 0. 0 -2 -1, 0 1 0. Now multiplying 1 by -2, I get -2 plus 2 is 0, -12 plus -1 is -13, -8 plus 1 is -7, -2 plus 0 is -2. 0 plus 0 is 0. 0 plus 1 is 1. So now I have a zero here, all I need to do is get zeros here, here, here, and ones on the diagonal. Easy so what next?
Let's try to get rid of this -13. I want to get a zero here. So what I'm going to do is I'm going to take this row, and multiply this -2 by -13/2, and add it to this row. That is going to create a lot of fractions, and it's going to be messy, but I've got to do it.
So -13/2 times row 2 plus row 3. It looks a little funny there you go. Now that's not going to affect rows 1, and 2, so I can just copy those down 1 6 4, 1 0 0. 0 -2 -1, 0 1 0. Then nothing is going to happen here because -13/2 times 0, 0 plus 0 is 0. When I multiply this by -13/2, I get 13 minus 13 is 0, so that's what I wanted to happen. Then -1 times -13/2 is +13/2 minus 7. -7 is -14/2, so I have 13/2 minus 14/2, -½. I get 0 minus 2, I get -13/2 plus 0 -13/2, and I get 0 plus 1, 1. So now I'm very close. I'm going to multiply this row by -2, and this one by -½, and I'll get the one that I need here, and the one that I need here.
So -2 times row 3, and -½ times row 2, so I'm doing two at a time here. First row stays the same 1 6 4, 1 0 0, and then I get this row is being multiplied by -½, I get 1. So 0, 1 times -½ is ½ times -½ is 0, -½, and 0.
This row is getting multiplied by -2 which will get rid of my fractions. So I get 1 times -2 is 4 times -2 is 1 times -2 is -2, very close. I need to get rid of this ½, so I'm going to add -½ times this row to this row. -½ row 3 plus row 2. So that's going to leave the first row the same. I haven't done anything to the first row yet.
The bottom row remains unchanged 0 0 1, 4 13 -2. So -½ times row 3, this is going to be 0 plus 0 is 0. 0 plus 1 is 1. -½ plus ½ is 0. -2 plus 0 is -2. -13/2 minus a ½ is -14/2, -7. -½ times this one is 1 plus 0 is 1. That was tough.
Now I want to get a 0 here, and here. I'll do that in two steps. First I'm going to add -6 times row 2 to row 1. That will leave row 2, and row 3 unchanged. Notice from there down it looks like the identity matrix I'm getting close. 0 1 0, 0 0 1. -2 -7 1, and 4 13 -2. So I multiply this row by -6, and add it to the top. So I'll get 0 plus 1 is 1. I get -6 plus 6 is 0. I get 0 plus 4 is 4. I get 12 plus 1 is 13. I get 42 plus 0 is 42. I get -6 plus 0, -6.
Now I have to add -4 times this row to this row, and that will get rid of this guy, and that will be my last step, I'll have the identity matrix. So unfortunately I'm going to have to take this up here. So I've got -4 times row 3 plus row 1.
Now that means that rows 2, and 3 are going to remain the same, so I can copy those over; 0 1 0, -2 -7 1. 0 0 1, 4 13 -2. Now -4 times 0 plus 1 is 1. -4 times 0 plus 0 is 0. -4 times 1 is -4 plus 4, 0 perfect. -4 times 4 is -16 plus 13 is -3. -4 times 13 is -52 plus 42 is -10. -4 times -2 is 8, plus -6 is 2, and I made it.
Finally I've got the identity matrix on the left, so this must be the inverse of matrix A. So I can finally write A inverse equals -3 -10 2, -2 -7 1, 4 13 -2. That was exhausting, but I feel like I'm a better person for having done this.
Unit
Systems of Linear Equations and Matrices