Introduction
This book begins with the central problem of linear algebra: solving linear equations. The most important ease, and the simplest, is when the number of unknowns equals the number of equations. We have n equations in n unknowns, starting with n = 2: Two equations 1x + 2y = 3 Two unknowns 4x + 5y = 6. (1) The unknowns are x and y. I want to describe two ways, elimination and determinants, to solve these equations. Certainly x and y are determined by the numbers 1, 2, 3, 4, 5, 6. The question is how to use those six numbers to solve the system. 1. Elimination Subtract 4 times the first equation from the second equation. This eliminates x from the second equation. and it leaves one equation for y: (equation 2)−4(equation 1) −3y = −6. (2) Immediately we know y = 2. Then x comes from the first equation 1x+2y = 3: Back-substitution 1x+2(2) = 3 gives x = −1. (3) Proceeding carefully, we cheek that x and y also solve the second equation. This should work and it does: 4 times (x = −1) plus 5 times (y = 2) equals 6. 2. Determinants The solution y = 2 depends completely on those six numbers in the equations. There most be a formula for y (and also x) It is a “ratio of determinants” and I hope you will allow me to write it
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