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Unformatted text preview: 1.2. Matrices and Gaussian Elimination Suppose you need to solve the following system of linear equations: x y + 2 z = 3 2 x 2 y + 8 z = 22 x 2 y = 2 We’re going to be doing elimination over and over, which means writing equations over and over, which is a real pain, since only the numbers will be changing. So we are going to convert the system into a matrix and work with the matrix instead: x y + 2 z = 3 2 x 2 y + 8 z = 22 x 2 y = 2 = ⇒ 1 1 2 2 2 8 1 2 3 22 2 (A matrix in general is a rectangular arrangement of numbers.) Note that if a variable is missing from an equation, we put a 0 in the matrix. Also, I’ve drawn a vertical bar separating the coefficients of the variables from the numbers on the righthand side. The rightmost column is treated differently, and this is a reminder of that. What would be the best kind of matrix to have? What would be the best kind of matrix to have? Answer: 1 1 1 A B C What would be the best kind of matrix to have? Answer: 1 1 1 A B C Because it represents the system x = A y = B z = C which says that the solution is ( A,B,C ) . Now, how do we get from 1 1 2 2 2 8 1 2 3 22 2 to 1 1 1 A B C without changing the solution? There are three types of “operations” (things we do to a matrix) which will not change the solutions when we do them, and these three types will ALWAYS allow us to put our matrix in the desired form! 1. Swap two rows of the matrix. For instance, if I swap the second and third rows of the matrix, I do the following: 1 1 2 2 2 8 1 2 3 22 2 = ⇒ 1 1 2 1 2 2 2 8 3 2 22 The book’s notation for this row operation is SWAP ( R 2 ,R 3 ) ....
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 Spring '10
 heckman
 Linear Equations, Equations, Gaussian Elimination, Matrices, Row echelon form, Row, other applications

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