Stitz-Zeager_College_Algebra_e-book

To that end we note that a x0 a ax0 a in a a in

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Unformatted text preview: −−−−−→ 0 3 0 −1 3 0 4 We proceed to get a leading 1 in R2. 2 1 1 −1 1 −1 0 −3 −3 3 3 3 Replace R2 with 2 R2 2 2 19 0 4 1 −− − − − − − − − − − − −→ 0 3 3 3 0 4 0 −1 3 0 4 1 0 −3 −2 3 19 2 4 3 3 3 0 −1 1 0 −3 −2 3 6 1 19 2 0 −1 3 We now zero out the entry below the leading 1 in R2. 1 1 2 0 −3 −3 0 −1 −2 1 −3 1 −1 3 3 3 Replace R3 with −4R2 + R3 19 1 6 1 19 − − − − − − − − − 1 6 1 −−−−−−−−→ 0 0 2 2 0 4 0 −1 0 0 −24 −5 −35 3 Next, it’s time for a leading 1 in R3. 1 1 −3 0 −1 −2 1 −1 3 3 3 1 Replace R3 with − 24 R3 19 1 6 1 1 − − − − − − −→ 0 0 −− − − − − − − 2 0 0 −24 −5 −35 0 0 The matrix is now in row echelon form. To get the reduced last leading 1 we produced and work to get 0’s above it. 1 2 1 −3 0 −1 −3 3 Replace R2 with −6R3 + R2 1 6 1 19 − − − − − − − − − −−−−−−−−→ 0 2 5 35 0 0 1 24 24 row echelon form, we start with the 1 −1 3 1 0 0 0 Lastly, we get a 0 above the leading 1 of R2. 1 2 1 −3 1 0 −1 −3 3 Replace R1 wi...
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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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