Linear Algebra Solutions 30

Linear Algebra Solutions 30 - with x5 arbitrary. Then X= x5...

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with x 5 arbitrary. Then X = x 5 0 - x 5 - 3 x 5 x 5 = x 5 1 0 - 1 - 3 1 , so [1 , 0 , - 1 , - 3 , 1] t is a basis for N ( A ). 6. In Section 1.6, problem 12, we found that the matrix A = 1 0 1 0 1 0 1 0 1 1 1 1 1 1 0 0 0 1 1 0 has reduced row–echelon form B = 1 0 0 1 1 0 1 0 1 1 0 0 1 1 0 0 0 0 0 0 . From B we read o± the following: (a) The three non–zero rows of B form a basis for R ( A ). (b) The ²rst three columns of A form a basis for C ( A ). (c) To ²nd a basis for N ( A ), we solve AX = 0 and equivalently BX = 0. From B we see that the solution is x 1 = - x 4 - x 5 = x 4 + x 5 x 2 = - x 4 - x 5 = x 4 +
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