matrix A is row equivalent to B. Without calculations, list rank A and dim Nul A
ID: 3186278 • Letter: M
Question
matrix A is row equivalent to B. Without calculations, list rank A and dim Nul A. Then find bases for Col A Row A, and Nul A 13 -4 1-2 3 9-14 5-9 -2-6 4 2 8 2 6 5-1 13-4 1-2 80 0 1-1-1 o o o 0-5 rank A . dim Nul A = A basis for Col A is (Use a comma to separate vectors as needed.) A basis for Row A is (Use a comma to separate vectors as needed.) L A basis for Nul A is (Use a comma to separate vectors as needed.) 01 0-2 -1 0 0-6 0 12 A basis for the null space is (Use a comma to separate answers as needed.) htt 1/8Explanation / Answer
2. Since the matrices A and B are row-equivalent, hence rank(A) = rank(B) = 3 , as B has 3 non-zero rows.
Thus, rank(A) = 3.
A has 5 columns so that , as per the rank-nullity theorem, the nullity of A = 5-rank(A) = 5-3 = 2.
Thus, dim (Null(A)) = 2.
It may be observed from B that its 1st ,4th and 5th columns are linearly independent and that its remaining 2 columns are linear combinations of these 3 columns. Hence, a basis for Col(A) is {(1,3,-2,2)T, (1,5,2,-1)T, (-2,-9,8,0)T}.
It is apparent from B that a basis for Row(A) is {(1,3,-4,1,-2),(0,0,1,-1,-1),(0,0,0,0,-5)}. ( we have not chosen the rows from A as we do not know whether there were any row-interchanges involved in arriving at B from A).
Null(A) is the set of solutions to the equation AX = 0 or, BX = 0. If X = (x,y,z,w,u)T, then this equation is equivalent to -5u = 0 or, u = 0, z-w-u = 0 or, z = w+u or, z = w and x+3y-4z+w -2u = 0 or, x = -3y+4z-w+2u =-3y+4w-w= -3y+3w so that X = (-3y+3w,y,w,w,0)T = y(-3,1,0,0,0)T+w(3,0,1,1,0)T. Thus, a basis for Null (A) is {(-3,1,0,0,0)T,(3,0,1,1,0)T }.
3. Let the given matrix be denoted by A. Then, Null(A) is the set of solutions to the equation AX = 0. If X = (x,y,z,w,u)T, then this equation is equivalent to -6z+12u = 0 or, z = 2u, y-2w-u = 0 or, y = 2w+u and x+y-2z -w +5u = 0 or, x = -y+2z+w-5u = -2w-u+4u+w-5u = -w-2u. Then X =(-w-2u,2w+u,2u,w,u)T= w(-1,2,0,1,0)T + u(-2,1,2,0,1)T. Thus, a basis for the null space of the given matrix is {(-1,2,0,1,0)T,(-2,1,2,0,1)T
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