Select the first nonzero entry in the second row as the pivot. In this case we designate the entry -8 in the second row and second column as the pivot. [1 2 o -8 -11 0 -22 -9 1 Formulate the row operation that will change the first nonzero entry in the third row to zero. First, we will multiply each entry by the absolute value of the entry in the other row. The second entry in R, is ). which has absolute ]. which has absolute value 8. The second entry in R, is [ value 22. Therefore, the row operation will be of one of the following forms. BR, + 22R, BR, - 22R, Next, we will choose whether to add or subtract the rows. The pivot and the entry below it in R, have --Select- V so the following row operation will change the second entry in R, to zero. BR, V 22R, Using technology to apply the row operation, we arrive at the following result. 2 1 -8 -11
Select the first nonzero entry in the second row as the pivot. In this case we designate the entry -8 in the second row and second column as the pivot. [1 2 o -8 -11 0 -22 -9 1 Formulate the row operation that will change the first nonzero entry in the third row to zero. First, we will multiply each entry by the absolute value of the entry in the other row. The second entry in R, is ). which has absolute ]. which has absolute value 8. The second entry in R, is [ value 22. Therefore, the row operation will be of one of the following forms. BR, + 22R, BR, - 22R, Next, we will choose whether to add or subtract the rows. The pivot and the entry below it in R, have --Select- V so the following row operation will change the second entry in R, to zero. BR, V 22R, Using technology to apply the row operation, we arrive at the following result. 2 1 -8 -11
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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