lecture_20 - MA 265 LECTURE NOTES: MONDAY, FEBRUARY 25 Span...

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Unformatted text preview: MA 265 LECTURE NOTES: MONDAY, FEBRUARY 25 Span (cont’d) Example. We finish the example we began in the previous lecture. Consider the homogeneous linear system A x = in terms of the 4 × 4 matrix A = 1 1 2- 2- 2 1- 5 1 1- 1 3 4 4- 1 9 . Recall that the null space for A is the solution set to this system: W = x ∈ R 4 A x = . We showed in the previous lecture that this is a subspace of V = R 4 . We seek a spanning set for W . First we compute all of the solutions to this system of equations. We represent the system A x = by the augmented matrix A , then compute its reduced row echelon form by Gauss-Jordan reduction: 1 1 0 2 0 0 1- 1 0 0 0 0 0 0 = ⇒ x 1 + x 2 + 2 x 4 = 0 x 3- x 4 = 0 We see that x 2 and x 4 are arbitrary variables, so set x 2 = r and x 4 = s . Then we have x 1 x 2 x 3 x 4 = - r- 2 s r s s = r - 1 1 + s - 2 1 1 . This shows that W = span S , where a spanning set for W is S = - 1 1 , - 2 1 1 ....
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This note was uploaded on 03/11/2010 for the course MA 261A 0026100 taught by Professor ... during the Spring '10 term at Purdue University Calumet.

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lecture_20 - MA 265 LECTURE NOTES: MONDAY, FEBRUARY 25 Span...

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