Let T1 R2 rightarrow R3 where T1x y 2x y 3x 4y Let T2 R3 r

Let T_1: R^2 rightarrow R^3, where T_1([x y]) = [2x + y 3x -4y]. Let T_2: R^3 rightarrow R^2, where T_2 ([x y z]) = [y - 3z x - 2z]. (a) What are the standard matrices of T_1 and T_2? (b) Let T = T_1 compositefunction T_2. Compute the standard matrix of T. (c) Is T invertible? If yes, find a formula for T^-1 (d) Let L = T_2 compositefunction T_1. Compute the standard matrix of L. (e) Is L invertible? If yes, find a formula for L^-1.

Solution

(a). We know that the standard matrix of a linear transformation T has columns which are images, under T, of the vectors in the standard basis. Here,T₁(e₁)= T₁(1,0)^T=(2,3,0)^T and T₁(e₂)=T₁(0,1)^T=(1,0,-4)^T. Hence the standard matrix of T₁ is A =

2

1

3

0

0

-4

Also, T₁(e₁)= T₂(1,0,0)^T=(0,1)^T ,T₂(e₂)=T₂(0,1,0)^T=(1,0)^T and T₂(e₃)=T₂(0,0,1)^T=(-3,-2)^T. Hence the standard matrix of T₂ is B=

0

1

-3

1

0

-2

(b) (T₁ oT₂)(x,y,z)^T= T₁(T₂(x,y,z)^T) = T₁( y-3z,x-2z)^T = (x+2y-8z, 3y-9z,-4x+8z)^T. Hence (T₁oT₂)(e₁) =                   (T₁oT₂)(1,0,0)^T=(1,0,-4)^T,(T₁oT₂)(e₂) =(T₁oT₂)(0,1,0)^T=(2,3,0)^T,and (T₁oT₂)(e₃)=(T₁oT₂)(0,0,1)^T=(-8,-9,8)^T. Hence the standard matrix of T₁oT₂ is C=

1

2

-8

0

3

-9

-4

0

8

(c ) A linear transformation T is invertible if its standard mastrix is invertible. Since det ( C) = 0, hence T is not invertible.

(d) (T₂ oT₁)(x,y)^T= T₂(T₁(x,y)^T) = T₂( 2x+y, 3x, -4y)^T = (3x+12y,2x+9y)^T. Now, (T₂oT₁)(e₁) =(T₂oT₁)(1,0)^T         = (3,2)^T, and (T₂oT₁)(e₂) =(T₂oT₁)(0,1)^T= (12,9)^T. Hence the standard matrix of T₂oT₁ is D =

3

12

2

9

( e) Since det(D) = 27-24 = 3 0, hence L = T₂ oT₁ is invertible. The standard matrix of L^-1 is D^-1=

3

-4

-2/3

1

The formula for L^-1 is L^-1 (X) = D^-1 X.

2	1
3	0
0	-4