Questions about MatricesDate: 24 Feb 1995 18:58:12 -0500 From: Steve Toub Subject: A matrix and its inverse Dear Dr. Math- In using my calculator to figure out matrices, I use the formula, |A|^-1|B|. I know that I am multiplying the inverse of matrix |A| by the matrix |B|, but I don't UNDERSTAND why I am doing this. What is the inverse of a matrix? How would you find this? How do you multiply two matrices together? I am fairly knowledgeable in mathematics, but could you try to explain this so that a pre-calculus student can understand. Thanks, Steve Date: 27 Feb 1995 19:28:40 -0500 From: Dr. Ken Subject: Re: A matrix and its inverse Hello there! Well, here's the story. Remember back when you were learning about multiplication of numbers? You learned that if you multiply any number by 1, you'll get that number. And you learned that if you multiply 6 by 1/6, or 3/4 by 4/3, you'll get 1. Well, that's essentially the situation that's happening with the matrices. You may know that when you multiply any n-by-n matrix by the identity matrix (the n-by-n matrix that has all ones down the main diagonal and zeroes everywhere else), you get that same matrix back again. So if we let the letter, I, represent the n-by-n identity matrix, and A represent any other n-by-n matrix, then we have A * I = A and I * A = A. This is much like the situation in the real numbers: x * 1 = x and 1 * x = x. Where do inverses come in? Well, the inverse of an n-by-n matrix, A, is the matrix you can multiply A by to get I, just like the inverse (reciprocal) of a real number, x is the number you can multiply x by (1/x) to get 1. Remember that for a second. Now I'll see if I can explain matrix multiplication to you. Let's say we have two matrices A and B: 1 2 2 3 4 5 4 7 6 5 9 9 5 6 0 2 7 3 A B To multiply them together, we take a row from A and a column from B and multiply each entry in the row and the column together, and then add up what we get. The result will be a new 3x3 matrix: 1*3 + 2*5 + 2*2 1*4 + 2*9 + 2*7 1*5 + 2*9 + 2*3 17 36 29 4*3 + 7*5 + 6*2 4*4 + 7*9 + 6*7 4*5 + 7*9 + 6*3 = 59 121 101 5*3 + 6*5 + 0*2 5*4 + 6*9 + 0*7 5*5 + 6*9 + 0*3 45 74 79 A*B A*B Notice that multiplication of matrices isn't commutative: if you multiply B*A you'll get something completely different, so A*B doesn't equal B*A. Also notice that you can't multiply a 3x4 matrix by a 5x2 matrix. If you have a something-by-n matrix, you can only multiply it on the right by an n-by-something-else matrix. By the way, I think multiplication of matrices is something you'd understand a lot better if you got someone to actually show it to you, instead of trying to just tell you like I did. Now back to the question of inverses. Here is an example of a matrix and its inverse (in fact, it's A and its inverse): 1 2 2 -18 6 -1 1 0 0 4 7 6 15 -5 1 0 1 0 5 6 0 -5.5 2 -.5 0 0 1 A A Inverse A * A Inverse Why would we ever want to study inverses? For a lot of the reasons we studied reciprocals. If we had the equation x * y = 5, then we could mutiply both sides on the left by 1/x, and we'd get y = 5/x. Well, it's the same with matrices: if we have the equation A * B = C, we can multiply by A^-1 and we'll get B = C * A^-1. So that's what inverses are. Keep in mind that this is a tricky subject: the whole college course Linear Algebra is essentially the study of matrices. So don't give up, and try experimenting with them some more! Date: 27 Feb 1995 21:54:27 -0500 From: Steve Toub Subject: Re: Re: A matrix and its inverse Ken- Thank you so much for your help with matrices. I totally understand the multiplication part now, and I understand the idea of an inverse and what it is, but not how to find it. Is there some sort of formula or equation to finding an inverse? The reason that I am asking these questions is that I am in the process of writing a shareware math stack for High School aged kids. The programming part is easy, it is the math that I have a little knowledge problem with. By the way, if I ever get the program done, I will be sure to include your name under the credits. Thanks, Steve Date: 3 Mar 1995 12:37:19 -0500 From: Dr. Ethan Subject: Re: Re: A matrix and its inverse Hey, this isn't Ken but I should be able to help you. Do you know how to row reduce a matrix? That is the process that we will be using. In row reducing there are three things that you can legally do: 1. You can switch two rows. 2. Multiply a row by a constant 3. Add one row plus a constant times another and replace one by the sum. Let's see an example. | 2 4 3 | = r1 So we could use r3 | 1 1 0 | | 1 3 2 | = r2 move one to get r2 | 1 3 2 | | 1 1 0 | = r3 r1 | 2 4 3 | Move three r3 | 1 1 0 | r2 - r3 | 0 2 2 | r1 - 2r3 | 0 2 3 | Now we get to how to find inverses. If we want the inverse of the matrix that we have been dealing with we write it like this | 2 4 3 | 1 0 0 | | 1 3 2 | 0 1 0 | | 1 1 0 | 0 0 1 | Then we use the techniques of row reduction to get the left side to look like the right side. Whatever remains on the right side is the inverse. Let's see how it works. We'll start with the steps I already showed you: r3 | 1 1 0 | 0 0 1 | r3 | 1 1 0 | 0 0 1 | relabel a1 r2 | 1 3 2 | 0 1 0 | then r2 - r3 | 0 2 2 | 0 1 -1 | a2 r1 | 2 4 3 | 1 0 0 | r1 - 2r3 | 0 2 3 | 1 0 -2 | a3 next a1 - .5 a2 | 1 0 -1 | 0 -.5 1.5 | relabel w1 .5 a2 | 0 1 1 | 0 .5 -.5 | w2 a3 - a2 | 0 0 1 | 1 -1 -1 | w3 then w1 + w3 | 1 0 0 | 1 -1.5 .5 | w2 - w3 | 0 1 0 | -1 -.5 .5 | w3 | 0 0 1 | 1 -1 -1 | Okay, so now the matrix on the right should be the inverse. You can check this by multiplying it by the original matrix using the method taught to you by Ken. There is another method for finding inverses involving things called determinants. However I think that it would be most useful for you to get a linear algebra book to see it explained as it is long and involves more linear algebra. Hope that helps. Ethan Doctor On Call |
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