Stitz-Zeager_College_Algebra_e-book

The variables x y and z are the dimensions of the box

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Unformatted text preview: trix and then guide you through some exercises using a 3 × 3 matrix. Consider the matrix 6 15 14 35 C= from Exercise 1 above. We know that det(C ) = 0 which means that CX = 02×2 does not have a unique solution. So there is a nonzero matrix Y such that CY = 02×2 . In fact, every matrix of the form −5t 2 Y= t is a solution to CX = 02×2 , so there are infinitely many matrices such that CX = 02×2 . But consider the matrix 3 X41 = 7 It is NOT a solution to CX = 02×2 , but rather, CX41 = 3 7 6 15 14 35 = 123 287 3 7 = 41 In fact, if Z is of the form Z= then CZ = 6 15 14 35 3 7t t = 3 7t t 123 7t 41t = 41 3 7t t = 41Z for all t. The big question is “How did we know to use 41?” We need a number λ such that CX = λX has nonzero solutions. We have demonstrated that λ = 0 and λ = 41 both worked. Are there others? If we look at the matrix equation more closely, what we really wanted was a nonzero solution to (C − λI2 )X = 02×2 which we know exists if and only if the determinant of C...
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This note was uploaded on 05/03/2013 for the course MATH Algebra taught by Professor Wong during the Fall '13 term at Chicago Academy High School.

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