{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Math416_GaussJordan

Math416_GaussJordan - Remark 2 Let us check that ± 2-1 ²...

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

Math 416 - Abstract Linear Algebra Fall 2011, section E1 Preview of Gauss-Jordan elimination ( § 2.2) Consider the linear system ( x 1 - x 2 = 3 4 x 1 + 3 x 2 = 5 whose augmented matrix is 1 - 1 3 4 3 5 . We solve the system using Gauss-Jordan elimination: 1 - 1 3 4 3 5 R 2 - 4 R 1 1 - 1 3 0 7 - 7 1 7 R 2 1 - 1 3 0 1 - 1 R 1 + R 2 1 0 2 0 1 - 1 . The unique solution is -→ x = x 1 x 2 = 2 - 1 . Remark 1: The reduced row echelon form of the augmented matrix (last step) is saying
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Remark 2: Let us check that ± 2-1 ² is indeed a solution, in all three points of view on linear systems. As a system of linear equations: ( (2)-(-1) = 3 4(2) + 3(-1) = 5 . As a matrix-vector equation: ± 1-1 4 3 ²± 2-1 ² = ± 3 5 ² . As a vector equation: 2 ± 1 4 ²-1 ±-1 3 ² = ± 2 8 ² + ± 1-3 ² = ± 3 5 ² . 1...
View Full Document

{[ snackBarMessage ]}