rowreduce - 1 2-5 3 6 14-2 7 12 5-17-29 R 2 -1 / 2 R 2 1...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
MATH 115 STEP BY STEP ROW REDUCTION 1. Write out the augmented matrix for the given system of equations. - 2 x 3 + 7 x 5 = 12 2 x 1 + 4 x 2 - 10 x 3 + 6 x 4 + 12 x 5 = 28 2 x 1 + 4 x 2 - 5 x 3 + 6 x 4 - 5 x 5 = - 1 0 0 - 2 0 7 12 2 4 - 10 6 12 28 2 4 - 5 6 - 5 - 1 2. Locate the leftmost column of the augmented matrix that is not all zeros. 0 0 - 2 0 7 12 2 4 - 10 6 12 28 2 4 - 5 6 - 5 - 1 3. Interchange the top row with another row (if necessary) to put a nonzero entry at the top of this column. R 1 R 3 2 4 - 10 6 12 28 0 0 - 2 0 7 12 2 4 - 5 6 - 5 - 1 4. If the entry that is now at the top of this column is a 6 = 1, multiply the first row by 1 a to get a pivot. R 1 1 2 R 1 1 2 - 5 3 6 14 0 0 - 2 0 7 12 2 4 - 5 6 - 5 - 1 5. Add suitable multiples of the top row to the entries below so that all entries below the pivot become 0’s. R 3 R 3 - 2 R 1 1 - 2 - 5 3 6 14 0 0 - 2 0 7 12 0 0 5 0 - 17 - 29 6. Cover the top row in the matrix and start again with Step 1 on the matrix that remains. Continue until entire matrix is in REF.
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 1 2-5 3 6 14-2 7 12 5-17-29 R 2 -1 / 2 R 2 1 2-5 3 6 14 1-7 2-6 5-17-29 R 3 R 3-5 R 2 1 2-5 3 6 14 1-7 2-6 1 2 1 R 3 2 R 3 1 2-5 3 6 14 1-7 2-6 1 2 7. Beginning with the last non-zero row and working up, add multiples of each row to the rows above to introduce 0s above the leading 1s. R 2 R 2 + 7 2 R 3 1 2-5 3 6 14 1 1 1 2 R 1 R 1-6 R 3 1 2-5 3 2 1 1 1 2 R 1 R 1 + 5 R 2 1 2 3 7 1 1 1 2...
View Full Document

Ask a homework question - tutors are online