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A = Consider the matrix 0 0 6 181 -6-3010-60 15-200 and suppose that we have the following row reduction to its RREF B. 0 0 6 181 15-200 -6-3010-60 -6-3010-60 A= 15-200 006 181 15-200 15-200 00-2-60 001 30 00 6 181 006 181 150 6 0 001 30 15060 00130 = B 006181 00001 The row reduction gives us that B = E5E4E3E2E₁ A where 0 0 1 1 0 0 Ει = 0 1 0 |E₂ = E2 6 1 0 1 0 0 0 0 1 1 0 0 1 2 0 E3 = 0 -1/2 0 |E₁ = 0 1 O 0 0 0 1 0 0 1 1 0 0 E5 = 0 1 0 0 -6 1 We therefore have an invertible matrix U so that B = UA U =

A = Consider the matrix 0 0 6 181 -6-3010-60 15-200 and suppose that we have the following row reduction to its RREF B. 0 0 6 181 15-200 -6-3010-60 -6-3010-60 A= 15-200 006 181 15-200 15-200 00-2-60 001 30 00 6 181 006 181 150 6 0 001 30 15060 00130 = B 006181 00001 The row reduction gives us that B = E5E4E3E2E₁ A where 0 0 1 1 0 0 Ει = 0 1 0 |E₂ = E2 6 1 0 1 0 0 0 0 1 1 0 0 1 2 0 E3 = 0 -1/2 0 |E₁ = 0 1 O 0 0 0 1 0 0 1 1 0 0 E5 = 0 1 0 0 -6 1 We therefore have an invertible matrix U so that B = UA U =

Intermediate Algebra
Intermediate Algebra
19th Edition
ISBN:9780998625720
Author:Lynn Marecek
Publisher:Lynn Marecek
Chapter4: Systems Of Linear Equations
Section4.5: Solve Systems Of Equations Using Matrices
Problem 4.80TI: Perform the needed row operation that will get the first entry in row 2 to be zero in the augmented...
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A = Consider the matrix 0 0 6 181 -6-3010-60 15-200 and suppose that we have the following row reduction to its RREF B. 0 0 6 181 15-200 -6-3010-60 -6-3010-60 A= 15-200 006 181 15-200 15-200 00-2-60 001 30 00 6 181 006 181 150 6 0 001 30 15060 00130 = B 006181 00001 The row reduction gives us that B = E5E4E3E2E₁ A where 0 0 1 1 0 0 Ει = 0 1 0 |E₂ = E2 6 1 0 1 0 0 0 0 1 1 0 0 1 2 0 E3 = 0 -1/2 0 |E₁ = 0 1 O 0 0 0 1 0 0 1 1 0 0 E5 = 0 1 0 0 -6 1 We therefore have an invertible matrix U so that B = UA U =
Transcribed Image Text:A = Consider the matrix 0 0 6 181 -6-3010-60 15-200 and suppose that we have the following row reduction to its RREF B. 0 0 6 181 15-200 -6-3010-60 -6-3010-60 A= 15-200 006 181 15-200 15-200 00-2-60 001 30 00 6 181 006 181 150 6 0 001 30 15060 00130 = B 006181 00001 The row reduction gives us that B = E5E4E3E2E₁ A where 0 0 1 1 0 0 Ει = 0 1 0 |E₂ = E2 6 1 0 1 0 0 0 0 1 1 0 0 1 2 0 E3 = 0 -1/2 0 |E₁ = 0 1 O 0 0 0 1 0 0 1 1 0 0 E5 = 0 1 0 0 -6 1 We therefore have an invertible matrix U so that B = UA U =
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