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Find the inverse of the matrix by utilizing Gauss-Jordan elimination. |1 2 - 1 1 1 1

Find the inverse of the matrix by utilizing Gauss-Jordan elimination. |1 2 - 1 1 1 1

Advanced Engineering Mathematics
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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Find the inverse of the matrix by utilizing Gauss-Jordan elimination. |1 2 -2| 1 lo o 1 Follow the guide below for the format: Let us see an example. Suppose that 1 2 -3 A = 5 -8 -3 -5 First we make the super augmented matrix -3 | 10 0 -8 | 0 1 0 8 | 0 0 1, 1 -3 -5 Now apply Gaussian elimination. Multiply the first row by –2 and 3 and add them to the second and third row. (1 2 0 1 0 1 -3 | 1 0 0 -2 | -2 1 0 -1 | 3 0 1 Now multiply the second row by –1 and add it to the third row to get 1 2 -3 | 1 0 1 -2 | -2 1 0 0 1 | 5 -1 1
Transcribed Image Text:Find the inverse of the matrix by utilizing Gauss-Jordan elimination. |1 2 -2| 1 lo o 1 Follow the guide below for the format: Let us see an example. Suppose that 1 2 -3 A = 5 -8 -3 -5 First we make the super augmented matrix -3 | 10 0 -8 | 0 1 0 8 | 0 0 1, 1 -3 -5 Now apply Gaussian elimination. Multiply the first row by –2 and 3 and add them to the second and third row. (1 2 0 1 0 1 -3 | 1 0 0 -2 | -2 1 0 -1 | 3 0 1 Now multiply the second row by –1 and add it to the third row to get 1 2 -3 | 1 0 1 -2 | -2 1 0 0 1 | 5 -1 1
This completes the Gaussian elimination. Now we continue Gauss Jor- dan elimination Multiply the third row by 2 and 3 and add it to the second and first row, to get (1 2 0 | 16 -3 3 0 1 0 | 8 -1 2 0 0 1 | 5 -1 1 The last step is to multiply the second row by –2 and add it to the first row, ´1 0 0 | 0 -1 -1' 0 1 0 |8 -1 0 0 1 | 5 -1 -1 The inverse matrix is 0, B = | 8 -1 5 -1 - -1 One can check that indeed AB = BA= I3.
Transcribed Image Text:This completes the Gaussian elimination. Now we continue Gauss Jor- dan elimination Multiply the third row by 2 and 3 and add it to the second and first row, to get (1 2 0 | 16 -3 3 0 1 0 | 8 -1 2 0 0 1 | 5 -1 1 The last step is to multiply the second row by –2 and add it to the first row, ´1 0 0 | 0 -1 -1' 0 1 0 |8 -1 0 0 1 | 5 -1 -1 The inverse matrix is 0, B = | 8 -1 5 -1 - -1 One can check that indeed AB = BA= I3.
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