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lecture_26 - MA 265 LECTURE NOTES MONDAY MARCH 17...

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MA 265 LECTURE NOTES: MONDAY, MARCH 17 Homogeneous Systems Example. We recall ideas from the previous lecture. We considered the following homogeneous system of linear equations x 1 + x 2 + 4 x 3 + x 4 + 2 x 5 = 0 x 2 + 2 x 3 + x 4 + x 5 = 0 x 4 + 2 x 5 = 0 x 1 - x 2 + 2 x 5 = 0 2 x 1 + x 2 + 6 x 3 + x 5 = 0 Let W denote the set of solutions. We found that the coefficient matrix has the reduced row echelon form: 1 0 2 0 1 0 1 2 0 - 1 0 0 0 1 2 0 0 0 0 0 0 0 0 0 0 = x 1 + 2 x 3 + x 5 = 0 x 2 + 2 x 3 - x 5 = 0 x 4 + 2 x 5 = 0 We see that x 3 = s and x 5 = t are arbitrary variables. Hence the general solution is x = - 2 s - t - 2 s + t s - 2 t t = s v 1 + t v 2 in terms of v 1 = - 2 - 2 1 0 0 , v 2 = - 1 1 0 - 2 1 . Hence S = { v 1 , v 2 } is a basis for the null space W . In particular, the nullity of A is dim W = 2. Note that we have a total of n = 5 variables, where r = 3 are pivot variables and hence p = 2 are free variables. General Remarks. Consider the following homogeneous system of equations: a 11 x 1 + a 12 x 2 + · · · + a 1 n x n = 0 a 21 x 1 + a 22 x 2 + · · · + a 2 n x n = 0 . . . . . . . . . . . . a m 1 x 1 + a m 2 x 2 + · · · + a mn x n = 0 We can express this system in the form A x = 0 in terms of the matrices A = a 11 a 12 . . . a 1 n a 21 a 22 . . . a 2 n . . . . . . . . . . . . a m 1 a m 2 . . . a mn and x = x 1 x 2 .
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