5. (a). Let M = [A|b] =
1 |
-2 |
3 |
11 |
2 |
-1 |
3 |
10 |
4 |
1 |
-1 |
4 |
The RREF of M is
1 |
0 |
0 |
2 |
0 |
1 |
0 |
-3 |
0 |
0 |
1 |
1 |
It implies that b = (11,10,4)T = 2(1,2,4)T-3(-2,-1,1)T+(3,3,-1)T. Hence b ? im(A) = col(A).
(b). Let M = [A|b] =
0 |
5 |
15 |
9 |
2 |
-2 |
-4 |
-2 |
3 |
-3 |
-6 |
-4 |
The RREF of M is
1 |
0 |
1 |
0 |
0 |
1 |
3 |
0 |
0 |
0 |
0 |
1 |
It implies that b = (9,-2,-4)T ? im(A) = col(A).
6. (a). T(e1) = T(1,0)T = (2,0)T and T(e2) = T(0,1)T = (0,3)T. Hence the standard matrix of T is A =
2 |
0 |
0 |
3 |
The RREF of A is I2 which implies that the columns of A are linearly independent. Hence T is one-to-one.
(b). T(e1) = T(1,0,0)T =(1,0)T,T(e2) = T(0,1,0)T =(1,0)T and T(e3) = T(0,0,1)T =(0,1)T. Hence the standard matrix of T is A =
1 |
1 |
0 |
0 |
0 |
1 |
The columns of A are not linearly independent. Hence T is not one-to-one.
(c ). T(e1) = T(1,0)T =(1,2,0)T, T(e2) = T(0,1)T =(1,1,1)T. Hence the standard matrix of T is A =
1 |
1 |
2 |
1 |
0 |
1 |
The RREF of A is
1 |
0 |
0 |
1 |
0 |
0 |
It implies that the columns of A are linearly independent. Hence T is one-to-one.
(d). T(e1) = T(1,0,0,0)T =(1,0,0)T, T(e2) = T(0,1,0,0)T =(1,1,0)T T(e3) = T(0,0,1,0)T =(0,1,1)T T(e4) = T(0,0,0,1)T =(0,0,1)T. Hence the standard matrix of T is A =
1 |
1 |
0 |
0 |
0 |
1 |
1 |
0 |
0 |
0 |
1 |
1 |
A has 4 column vectors. Since dim(R3)= 3, hence the columns of are not linearly independent. Hence T is not one-to-one.
Please post the remaining questions again.
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