Homework SourseLet A 1 3 0 0 0 0 0 1 2 1 0 0 3 3 0 1 3 2 4 2 3 9 7 5 2 1 3
Let A = [1 3 0 0 0 0 0 -1 -2 1 0 0 -3 -3 0 -1 -3 2 4 -2 -3 -9 -7 -5 -2 -1 -3 -1 -2 1]. Find a basis for the row space of A, a basis for the column space of A, a basis for the null space of A, the rank of A, and the nullity of A. (Note that the reduced row echelon form of A is [1 0 0 0 0 0 1 0 0 0 0 0 1 0 0 -1 -2 0 0 0 -3 -2 3 0 0 -1 1 0 0 0].) Row Space basis: Column Space basis: Null Space basis: Rank: Nullity: ![Let A = [1 3 0 0 0 0 0 -1 -2 1 0 0 -3 -3 0 -1 -3 2 4 -2 -3 -9 -7 -5 -2 -1 -3 -1 -2 1]. Find a basis for the row space of A, a basis for the column space of A,](/WebImages/34/let-a-1-3-0-0-0-0-0-1-2-1-0-0-3-3-0-1-3-2-4-2-3-9-7-5-2-1-3-1100090-1761581053-0.webp "Let A = [1 3 0 0 0 0 0 -1 -2 1 0 0 -3 -3 0 -1 -3 2 4 -2 -3 -9 -7 -5 -2 -1 -3 -1 -2 1]. Find a basis for the row space of A, a basis for the column space of A,")
Solution
The rank of A ,being the number of non-zero rows in its RREF is 3. Since A has 6 columns, by the Rank-Nullity(Dimension) theorem, the nullity of A is 6-3=3. A basis for Row(A) is {(1,0,0,-1,-3,-1) ,(0,1,0,-2,-2,1), (0,0,1,0,3,0)}. Also, a basis for Col(A) is {(1,3,0,0,0)T, (0,0,-1,-2,1)T,(0,0,-3,,-3,0)T}. Null (A) is the set of solutions to the equation AX = 0. Now,letX=(x1,x2,x3,x4,x5,x6)T . Then this equation is equivalent to x1 –x4-3x5-x6 = 0, x2 -2x4-2x5+x6 = 0 and x3+3x5 = 0. Let x4 =r, x5 =s and x6 = t. Then x1 = r+3s+t, x2 = 2r+2s-t and x3 = -3s so that X = (r+3s+t ,2r+2s-t, -3s,r,s,t)T = r(1,2,0,1,0,0)T +s(3,2,-3,0,1,0)T+t(1,-1,0,0,0,1)T}. Thus, a basis for Null(A) is {(1,2,0,1,0,0)T, (3,2,-3,0,1,0)T, (1,-1,0,0,0,1)T}.
