Inverse Matrix
Inverse matrix is a mathematical tool used in linear algebra, which finds the inverse of a given matrix. The inverse matrix of a square matrix A is defined to be another square matrix B such that AB = BA = I, where I denotes the identity matrix. The inverse of matrix A, by definition, is a matrix M whose property is that, when multiplied by A, it yields the identity matrix. If I multiply M by A and get identity, then it follows that the two properties are, in fact, equivalent. So if you apply for transformation A first, then M, you actually undo the transformation A, and vice versa. These two transformations are opposites of each other, inverse of each other. So, to make sense of this, we need A to be a square matrix-that is. it must have the same number of rows as columns (size n by n). Linear algebra is a general fact. You will see more in detail if you study linear algebra. For now you should just admit that. We denote the matrix M by A inverse, like this. OK? And then what is it good for? Well, for example, the solution to a linear system–so what’s a linear system in our new language–it’s a matrix times some unknown vector X equals some known vector B. So how do we solve that? We can solve for X by computing A inverse B. Why does that work? From there. we ask how we can get from one point on our graph to another. Let’s be cautious. OK, well, I’m going to re-use this matrix, but I’m going to erase it and rewrite it anyway. So, AX is equal to B. We multiplied both sides of the equation by the inverse of A. A inverse times AX is A inverse B. After that, A inverse times A is identity, so we have that X equals A inverse B. To solve a system of equations, first multiply both sides of each equation by its inverse and then subtract the two equations to get rid of any constants. If you have a calculator that can invert matrices, then it can solve linear systems quickly. Now, we should still learn how to compute these things by hand. Notice, if we know that A inverse will be on the left of B and not after it, we can confirm this by reproducing the calculation. It’s exactly in this derivation. I mean, to get from here to here, multiplied the elements on the left by the inverse. The result was that those were simplified. IfI had put A inverse on the right, the equation would be AXA inverse, which may not make sense, and even if it does make sense, it doesn’t simplify. The basic rule is that if there’s an A on both sides of the equation, you have to multiply by A inverse on one side so that it cancels with the A on the other side. So, just remember that if you have a square matrix times a column vector, the product makes sense when the matrix is on the left and the vector is on the right. The other one does not work. It is not sized correctly. If A is a square matrix and X is a column vector, then this product makes sense. The other one does not make sense because it is not sized correctly. Nice. Inverse matrix: ________________ Inverse of A matrix M such that AM=I MA=I Nead: A square matrix M=A^-1 n*n _____________________ Solution to AX=B is X=A^-1 B
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