withx5arbitrary. ThenX=x50-x5-3x5x5=x510-1-31,so [1,0,-1,-3,1]tis a basis forN(A).6. In Section 1.6, problem 12, we found that the matrixA=10101010111111000110has reduced row–echelon formB=10011010110011000000.FromBwe read off the following:(a) The three non–zero rows ofBform a basis forR(A).(b) The first three columns ofAform a basis forC(A).(c) To find a basis forN(A), we solveAX= 0 and equivalentlyBX= 0.FromBwe see that the solution isx1=-x4-x5=x4+x5x2=-x4-
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