Can someone please answer problems 5, 6, and 7. with all work. Thank you Problem
ID: 3137115 • Letter: C
Question
Can someone please answer problems 5, 6, and 7.
with all work. Thank you
Problem 5. (10 pts.) Matrix A and an echelon form of A are given 1 2 4 3 3 5 10-97 8 0 01 2 0 0 0 0 05 0 0 4 8 -927 245 0 Find the following items: (a) A basis for Col A (b) Col A (e) dimCol A (d) Rank A (e) dim Nul A (f) A basis for Nul A (R) Nul A Problem. (10 pts.) What is vector a in the B-coordinate (lals) with the basis of Nul A in problem5 Problem 7. (10 pts.) Find a basis for the subspace spanned by the given vectors. What is the dimension of the subsp Problem 8. (10 pts.) If A is an m x n matrix. Mark each statement True or False. Justify each answer. (a) If B is any echelon form of A, then the pivot columns of B form a basis for the column space of A.Explanation / Answer
5. It may be observed from the echelon form of A that the 2nd column of A is a scalar multiple of its 1st column. To obtain further clarity, we may perform the following row operations on the echelon form of A to reduce A to its RREF:
Multiply the 3rd row by -1/5
Add -3 times the 3rd row to the 1st row
Add 4 times the 2nd row to the 1st row
Then the RREF of A is
1
2
0
-5
0
0
0
1
-2
0
0
0
0
0
1
0
0
0
0
0
Now it is abundantly clear that the 1st ,3rd and 5th columns of A are linearly independent and the 2nd and 4t columns are scalar multiple/linear combination of the 1st and 3rd columns.
(a). Thus, a basis for col(A) is {(1,5,4,-2)T,(-4,-9,-9,5)T,(3,0,5,0)T} is a basis for col(A) .
(b).col(A) = span{(1,5,4,-2)T,(-4,-9,-9,5)T,(3,0,5,0)T}.
(c). dim(col(A)) = 3.
(d ).Rank(A) = dim(col(A)) = 3.
(e). dim(null(A)) = 5-3 = 2 ( as per the rank –nullity theorem, the nullity +rank of a matrix equals the no. of its columns).
(f ). Null(A) is the set of solutions to the equation AX = 0. In view of the RREF of A, if X = (x,y,z,w,u)T, then this equation is equivalent to x+2y-5w= 0 or, x =-2y+5w, z-2w = 0 or, z = 2w and u = 0 so that X = ( -2y+5w, y,2w,w,0)T = y(-2,1,0,0,0)+w(5,0,2,1,0)T. Hence, {(-2,1,0,0,0),(5,0,2,1,0)T } is a basis for null(A).
(g). Null(A) = span{(-2,1,0,0,0)T,(5,0,2,1,0)T }.
6. The vector (5,0,2,1,0)T = 0(-2,1,0,0,0)T +1(5,0,2,1,0)T. Hence [a] = (0,1)T.
7.Let M be the matrix with the given vectors as columns. Then the RREF of M is
1
0
4
5
-7
0
1
-2
-3
5
0
0
0
0
0
0
0
0
0
0
Hence {(1,-1,-2,5)T,(2,-3,-1,6)T } is a basis for the given subspace as the other 3 vectors are linear combinations of these vectors.
1
2
0
-5
0
0
0
1
-2
0
0
0
0
0
1
0
0
0
0
0