Homework SourseFind the matrix relative to B and B of the linear transforma
Solution
(a). We have T(x,y,z) = (2x,x+y,y+z, x+z) so that T(2,0,1)= (4,2,1,3), T(0,2,1)= (0,2,3,1)and T(1,2,1) = (2,3,3,2). Therefore, the matrix of T relative to B is
4
0
2
2
2
3
1
3
3
3
1
2
Further, since T(x,y,z,w) is not defined, the matrix of T relative to B’ does not exist.
(b). We have [v]STD = (1,-5,2). To determine [T(v)]B, we have to determine [v]STD as a linear combination of the vectors in B. Let A =
2
0
1
1
0
2
2
-5
1
1
1
2
We will reduce A to its RREF as under:
Multiply the 1st row by ½
Add -1 times the 1st row to the 3rd row
Multiply the 2nd row by ½
Add -1 times the 2nd row to the 3rd row
Multiply the 3rd row by -2
Add -1 times the 3rd row to the 2nd row
Add -1/2 times the 3rd row to the 1st row
Then the RREf of A is
1
0
0
9/2
0
1
0
11/2
0
0
1
-8
Hence [v]STD = (1,-5,2) = (9/2)(2,0,1) +(11/2)(0,2,1)-8(1,2,1). Further, since T is a linear transformation, it preserves vector addition and scalar multiplication. Hence [T(v)]B = (9/2) T(2,0,1) +(11/2) T(0,2,1)- 8 T(1,2,1) = (9/2) (4,2,1,3) + (11/2) (0,2,3,1) -8(2,3,3,2)= (18,9,9/2,27/2)+(0,11,33/2,11/2)-(16,24,24,16) = ( 2,-4,-3,3).
However, [T(v)]B’ cannot be determined as [v]STD is not a linear combination of the vectors in B’ and also, the images under T of the vectors in B’ are not defined.
| 4 | 0 | 2 |
| 2 | 2 | 3 |
| 1 | 3 | 3 |
| 3 | 1 | 2 |
