Matrix_Division

Matrix_Division - Matrix Division Je Stuart c 2008 High...

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Matrix Division Je/ Stuart c 2008 ax = b where a and b are real numbers. Why do we need division? In its earliest form, division must have arisen to answer questions such as, "If our foraging party of three gathered seven melons, 3 x = 7 where x which is to say, in solving problems of the form ax = b where a and b are known numbers. Of course, we all learned early in our educations that ax = b has a unique solution when a 6 = 0 , namely, x = b=a . Although this seems trivially ax = b . We have observed that when a 6 = 0 , there is a unique solution for every possible choice of b , namely x = b=a . What happens when a = 0 ? Probably, your ±rst thought is that there is no solution since you cannot divide by zero. Is this correct? Certainly, if b 6 = 0 ; we cannot ±nd a real number x such that 0 x 6 = 0 . In the language of linear systems, when a = 0 and b 6 = 0 , the system ax = b is inconsistent. But what happens when b = 0 ? In this case, the equation becomes 0 x = 0 , which holds for every real number x . That is, when a = 0 and b = 0 , the system ax = b has in±nitely many solutions. Summarizing: Theorem 1 Let a and b be real numbers. ± If a = 0 , then for all real numbers b with b 6 = 0 , ax = b is inconsistent; and for b = 0 , ax = b ± If a 6 = 0 , then for all real numbers b , ax = b is consistent and has a unique solution, x = b=a . What happens for linear systems Ax = b with more equations and more variables? From the Trichotomy Theorem, we know that exactly one of three outcomes occurs for the linear system Ax = b where A and b are known: the system is inconsistent, the system is consistent with a unique solution, or the system is consistent with in±nitely many solutions. Is there some analogous property to a 6 = 0 that distinguishes the situation in which there is a unique solution to Ax = b when Ax = b is solvable, or that distinguishes the situation in which Ax = b is solvable for all vectors b ? When do multiple solutions occur for Ax = b ? Recall that when we form the product Ax , we are creating a vector that is a linear combination of the columns of A using the scalars that are the entries of x . One consequence of this is that for any matrix A , we can always ±nd a vector b for which Ax = b is 1
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consistent; just choose any vector b that is a linear combination of the columns of A , or even simpler, just choose b to be one of the columns of the matrix A . Also recall that for a consistent system, multiple solutions occur exactly when
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This note was uploaded on 02/22/2012 for the course EE 441 taught by Professor Neely during the Spring '08 term at USC.

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Matrix_Division - Matrix Division Je Stuart c 2008 High...

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