Represent the following linear transformations in the standa

Represent the following linear transformations in the standard base: (a) T(1, 2, 3) = (2, 1) T(2, 3, 1) = (1, 2) T(1, 1, 1) = (3, 1) (b) R(3, 4, 4) = (2, 1, 2, 1) R(1, 0, 1) = (3, 5, 1, 2) R(0, 7, 2) = (0, 9, -1, 0) (c) Q(1, 1) = (3, 2, 1) Q(6, 7) = (0, 1, 4)

Solution

(a) Let A =

1

2

1

1

0

0

2

3

1

0

1

0

3

1

1

0

0

1

The RREF of A is

1

0

0

-2/3

1/3

1/3

0

1

0

-1/3

2/3

-1/3

0

0

1

7/3

-5/3

1/3

Hence e₁ = (1,0,0)^T = -2/3(1,2,3)^T -1/3(2,3,1)^T+7/3(1,1,1)^T ,

e₂ = (0,1,0)^T = 1/3(1,2,3)^T +2/3(2,3,1)^T-5/3(1,1,1)^T and

e₃ = (0,0,1)^T = 1/3(1,2,3)^T -1/3(2,3,1)^T+1/3(1,1,1)^T

Therefore, T(e₁) = -2/3(2,1)^T-1/3(1,2)^T+7/3(3,1)^T = (16/3,1)^T.

Also, T(e₂) = 1/3(2,1)^T+ 2/3(1,2)^T-5/3(3,1)^T = (-11/3, 0)^T.

(b) Let A =

3

1

0

1

0

0

4

0

7

0

1

0

4

1

2

0

0

1

The RREF of A is

1

0

0

7

2

-7

0

1

0

-20

-6

21

0

0

1

-4

-1

4

Hence e₁ = (1,0,0)^T =7(3,4,4)^T-20(1,0,1)^T -4(0,7,2)^T

Also, e₂ = (0,1,0)^T =2(3,4,4)^T-6(1,0,1)^T -1(0,7,2)^T and

e₃ = (0,0,1)^T =-7(3,4,4)^T+21(1,0,1)^T +4(0,7,2)^T

Therefore, R(e₁) = 7( 2,1,2,1)^T -20( 3,5,1,2)^T -4(0,9,-1,0)^T = (-46,-129,-2,-33)^T

Also, R(e₂) = 2( 2,1,2,1)^T -6( 3,5,1,2)^T -1(0,9,-1,0)^T = (-14,-37,-1,-10)^T and

R(e₃) = -7( 2,1,2,1)^T +21( 3,5,1,2)^T +4(0,9,-1,0)^T = (49,134,3,35)^T

(c) Let A =

1

6

1

0

1

7

0

1

The RREF of A is

1

0

7

-6

0

1

-1

1

Hence e₁ = (1,0)^T =7(1,1)^T-1(6,7)^T and e₂ = (0,1)^T =-6(1,1)^T + 1(6,7)^T

Therefore, Q(e₁) =7(3,2,1)^T-1(0,1,4)^T=(21,13,3)^T and Q(e₂) = -6(3,2,1)^T +1(0,1,4)^T = (-18,11,-2)^T

Note: We know that , for a linear transformation L, we have L( ax+by+c) = aL(x) +bL(y) +cL(z)

1	2	1	1	0	0
2	3	1	0	1	0
3	1	1	0	0	1