Linear Algebra and Its Applications (5th Edition)

Published by Pearson
ISBN 10: 0-32198-238-X
ISBN 13: 978-0-32198-238-4

Chapter 1 - Linear Equations in Linear Algebra - 1.2 Exercises: 4

Answer

The initial matrix pivot positions are row 1, column 1 (value=1), row 2, column 2 (value=5), and row 3, column 3 (value=9). We go across pivot columns 1, 2, and 3 to find the new matrix, comprised of rows (1, 0, -1, 0), (0, 1, 2, 0), and (0, 0, 0, 1). This has pivot points at (R1, C1), (R2, C2), and (R3, C4).

Work Step by Step

This 3x4 matrix has row 1 values (1, 3, 5, 7), row 2 values (3, 5, 7, 9), and row 3 values (5, 7, 9, 1). We want to row reduce this matrix to reduced echelon form and label the pivot positions in the final and initial matrices. Additionally, we want to list the pivot columns. First, the initial matrix pivot positions are row 1, column 1 (value=1), row 2, column 2 (value=5), and row 3, column 3 (value=9). We go across pivot columns 1, 2, and 3. Add -3 times the first row to the second row for row 2=(0, -4, -8, -12). Next, add -5 times the first row to the third row for row 3=(0, -8, -16, -34). Then, multiply the second row by $\frac{-1}{4}$ for row 2=(0, 1, 2, 3). Add 8 times the 2nd row to the third row for row 3=(0, 0, 0, -10), then multiply the third row by $\frac{-1}{10}$ for row 3=(0, 0, 0, 1). Then, add -3 times the third row to the 2nd row for row 2=(0, 1, 2, 0), -7 times the third row to the first row for row 1=(1, 3, 5, 0), and -3 times the 2nd row to the first row for row 1=(1, 0, -1, 0). The pivot positions for the new matrix, comprised of rows (1, 0, -1, 0), (0, 1, 2, 0), and (0, 0, 0, 1) are the three initial values of 1: (R1, C1), (R2, C2), (R3, C4).
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