Linear Algebra Solutions 31

Linear Algebra Solutions 31 - 3 1 2 1 x 6 1 1 1 so[3 1 2 1...

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7. Let A be the following matrix over Z 5 : A = 1 1 2 0 1 3 2 1 4 0 3 2 0 0 0 1 3 0 3 0 2 4 3 2 . We Fnd that A has reduced row–echelon form B : B = 1 0 0 0 2 4 0 1 0 0 4 4 0 0 1 0 0 0 0 0 0 1 3 0 . ±rom B we read o² the following: (a) The four rows of B form a basis for R ( A ). (Consequently the rows of A also form a basis for R ( A ). (b) The Frst four columns of A form a basis for C ( A ). (c) To Fnd a basis for N ( A ), we solve AX = 0 and equivalently BX = 0. ±rom B we see that the solution is x 1 = - 2 x 5 - 4 x 6 = 3 x 5 + x 6 x 2 = - 4 x 5 - 4 x 6 = x 5 + x 6 x 3 = 0 x 4 = - 3 x 5 = 2 x 5 , where x 5 and x 6 are arbitrary elements of Z 5 . Hence X = x 5
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Unformatted text preview: 3 1 2 1 + x 6 1 1 1 , so [3 , 1 , , 2 , 1 , 0] t and [1 , 1 , , , , 1] t form a basis for R ( A ). 8. Let F = { , 1 , a, b } be a Feld and let A be the following matrix over F : A = 1 a b a a b b 1 1 1 1 a . 34...
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