Let A = [1 -1 -2 7 1 6 -6 2 -4 -18 -3 -26 4 -1 4 10 2 17 3 -1 2 9 1 12], B = [1
ID: 3038136 • Letter: L
Question
Let A = [1 -1 -2 7 1 6 -6 2 -4 -18 -3 -26 4 -1 4 10 2 17 3 -1 2 9 1 12], B = [1 0 2 1 0 3 0 1 4 -6 0 -1 0 0 0 0 1 2 0 0 0 0 0 0] where B = rref(A). (a) Find bases for the column space and row space of B. What is the rank of B? What is the nullity of B? Give an example of a non-zero vector in R^4 which is not in the column space of B. (b) Find bases for the row space and column space of A. What is the rank, and nullity of A. (c) Give an explanation why the column space of A does not span R^4 and the row space of A does not span R^5. Does Ax = b have a solution for any b belongs to R^4? Why or why not? (d) Extra Credit: Prove A and B are row equivalent and find an invertible matrix P such that PA = B.Explanation / Answer
(a). It may be observed from the RREF of A that only its 1st, 2nd and the 5th columns are linearly independent and that the 3rd, 4th and 6th columns are linear combinations of the 1st, 2nd and the 5th columns.Hence,a basis for Col(B) is {(1,0,0,0)T,(0,1,0,0)T, (0,0,1,0)T}. Also, a basis for Row(B) is { (1,0,2,1,0,3), (0,1,4,-6,0,-1),(0,0,0,0,1,2)}.Any vector of the form (0,0,0,x)T in R4, where x 0 is not in Col(B).
(b). A basis for Col(A) is { (1,-6,4,3)T, (-1,2,-1,-1)T, (1,-3,2,1)T } or, {(1,0,0,0)T,(0,1,0,0)T, (0,0,1,0)T}. Since we do not know whetherthere was a row interchange during the process of reduction of A to its RREFB, hence, a basis of Row(A) is also the same as a basis for Row(B) i.e. { (1,0,2,1,0,3), (0,1,4,-6,0,-1), (0,0,0,0,1,2)}. The rank of A, being the number of non-zero rows in its RREF is 3. Further, by the rank-nullity theorem, the nullity of A is 5(the no. of columns of A)- 3 = 2.
(c ). The vector (0,0,0,x )T in R4 where x 0 is not in Col(A).Therefore, Col(A0 does not span R4. The basis elements of Row (A0 are 6-vectors. Therefore, Row(A) canot span R5.
If the vector b is of the form (0,0,0,x )T , where x 0, then the equation Ax = b will not have a solution as b cannot then be expressed as a linear combination of vectors in Col(A.
(d). We know that 2 matrices are row equivalent if one can be changed to the other by a sequence of elementary row operations. Here, since B is the RREF of A, hence A and B are row ewuivalent.
The row operations done to row-reduce A to B are as under:
1.Add 6 times the 1st row to the 2nd row
2.Add -4 times the 1st row to the 3rd row
3.Add -3 times the 1st row to the 4th row
4.Multiply the 2nd row by -1/4
5.Add -3 times the 2nd row to the 3rd row
6.Add -2 times the 2nd row to the 4th row
7.Multiply the 3rd row by 4
8.Add 1/2 times the 3rd row to the 4th row
9.Add 3/4 times the 3rd row to the 2nd row
10.Add -1 times the 3rd row to the 1st row
11.Add 1 times the 2nd row to the 1st row
Now, let ENow, let Ei (1 i 11) be the elementary matrix obtained by the ith row operation stated above. Then P = E11.E10….E2E1.
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