Linear algebra Show me all steps A 1 1 3 3 0 1 0 2 1 1 2 3

A = [1 1 3 3 0 -1 0 -2 -1 1 2 3 7 8 1 -2 4 0 6 7] has RREF [1 0 2 1 0 0 1 1 2 0 0 0 0 0 1 0 0 0 0 0]. Rank(A) = ____ (A) = _____ Basis of rowspace of A is {_____} basis of colspace of A is {_____} basis of Nullspace of A is {_____}

Solution

From the RREF of A, it is apparent that Rank(A) = number of non-zero rows in the RREF of A = 3. From the RREF of A, it is apparent that a basis for Row(A) is { (1,0,2,1,0),(0,1,1,2,0)} ( we have not selected the corresponding original rows as there might have been some row interchanges in computing the RREF of A). A basis for Col(A) is { (1,-1,2,-2)T,(1,0,3,4)T,(0,1,1,7)T} or{ (1,0,0,0)T,(0,1,0,0)T,(0,0,1,0)T}. Null (A) is the set of solutions to the equation AX = 0. If X = (x1,x2,x3,x4,x5)T, then this equation is equivalent to x1+2x3+x4=0, x2+x3+2x4=0 and x5=0.Let x3=r and x4=t. Then x1=-2r-t, and x2= r-2t so that X = (-2r-t, -r-2t, r,t,0)T= r(-2,-1,1,0,0)T+t(-1,-2,0,1,0)T. Therefore, a basis for Null(A) is {(-2,-1,1,0,0)T,(-1,-2,0,1,0)T }.