We changed the first nonzero entry beneath the pivot to a zero and found the following augmented matrix. -1 1 1 45 0 33 90 1 010 Next we will change the other entry beneath the pivot to a zero. Formulate the Type 2 row operation that wll change the first entry in R, to a zero. Applying the row operation results in the following matrix. -1 1 1 45 0 3 90 2

Linear Algebra: A Modern Introduction
4th Edition
ISBN:9781285463247
Author:David Poole
Publisher:David Poole
Chapter2: Systems Of Linear Equations
Section2.2: Direct Methods For Solving Linear Systems
Problem 3CEXP
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We changed the first nonzero entry beneath the pivot to a zero and found the following augmented matrix.
-1 1 1 45
0 33 90
1 010
Next we will change the other entry beneath the pivot to a zero. Formulate the Type 2 row operation that will
change the first entry in R, to a zero.
R. ?vR,
Applying the row operation results in the following matrix.
-1 1
1
45
0 3
90
2
Transcribed Image Text:We changed the first nonzero entry beneath the pivot to a zero and found the following augmented matrix. -1 1 1 45 0 33 90 1 010 Next we will change the other entry beneath the pivot to a zero. Formulate the Type 2 row operation that will change the first entry in R, to a zero. R. ?vR, Applying the row operation results in the following matrix. -1 1 1 45 0 3 90 2
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