Stitz-Zeager_College_Algebra_e-book

# The identity matrix in the super sized augmented

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Unformatted text preview: lace R2 with 7 R3 + R2 0 0 1 −1 0 0 1 −1 Finally, we take care of the 2 in R1 above the leading 1 in R2. 3 3 100 120 Replace R1 with −2R2 + R1 0 1 0 0 −−−−−−−−− 0 −−−−−−−−→ 0 1 0 0 0 1 −1 0 0 1 −1 To our surprise and delight, when we decode this matrix, we obtain having to deal with any back-substitution at all. 3 100 x Decode from the matrix 0 1 0 0 −−−−−−−→ y −−−−−−− 0 0 1 −1 z the solution instantly without = 3 = 0 = −1 Note that in the previous discussion, we could have started with R2 and used it to get a zero above its leading 1 and then done the same for the leading 1 in R3. By starting with R3, however, we get more zeros ﬁrst, and the more zeros there are, the faster the remaining calculations will be.2 It is also worth noting that while a matrix has several3 row echelon forms, it has only one reduced row echelon form. The process by which we have put a matrix into reduced row echelon form is called Gauss-Jordan Elimi...
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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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