For each of the following choices of A and b, determine if b is in the column sp
ID: 3036654 • Letter: F
Question
For each of the following choices of A and b, determine if b is in the column space of A and state whether the system Ax = b is consistent (a) A = [2 1 3 4], b = [4 6] b in column space of A? yes no Ax = b is consistent? yes no (b) A = [1 1 2 1 1 2 1 1 2], b = [1 2 3] b in column space of A? yes no Ax = b is consistent? yes no (c) A = [1 2 2 4 1 2], b = [5 10 5] b in column space of A? yes no (d) A = [0 1 1 0 0 1], b = [2 5 2] b in column space A? yes no Ax = b is consistent? Yes no (e) For each of the consistent systems in parts (a)-(d), determine which have infinitely many solutions. Enter your answer as a list of letters (lower case) separated by commas, e.g., b, c, d.Explanation / Answer
(a) Let B =
2
1
4
3
4
6
The RREF of B is
1
0
2
0
1
0
Thus, (4,6)T = 2(2,3)T is in Col(A).The equation AX = b is consistent. If X = (x,y)T, then x = 2,y = 0 so that X = (2,0)T.
(b) Let B =
1
1
2
1
1
1
2
2
1
1
2
3
The RREF of B is
1
1
2
0
0
0
0
1
0
0
0
0
Apparently b is not in Col(A) . Also, the equation AX = b is not consistent.
(c ) Let B =
1
2
5
2
4
10
1
2
5
The RREF of B is
1
2
5
0
0
0
0
0
0
Apparently, b = (1,2,1)T + 2(2,4,2)T. Hence b is in Col(A). Further, the equation AX = b is consistent. If X = (x,y)T, then x+2y = 5, which has infinite solutions.
( d) Let B =
0
1
2
1
0
5
0
1
2
The RREF of B is
1
0
5
0
1
2
0
0
0
Then b = 5(0,1,0)T+2(1,0,1)T is in Col(A). Further, if X = (x,y)T, then x =5,y = 2, The equation AX = b is thus consistent.
(e) Only ( c )has infinite solutions.
2
1
4
3
4
6
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