T: R²->R² the linear operator given by T(x,y)=(x-y,x+2y). Determine the T^-1 matrix, related to the canonical basis of R².using this matrix, calculate T^-1(-1,8).?

1 Answer
Steve M
Apr 14, 2018

The matrix associated with # bb(T)^(-1)# is:

# \ \ \ \ \ \ \ \ ((2/3,1/3),(-1/3,1/3)) #

This leads to the inverse transformation:

# bb(T)^(-1):RR^2 rarr RR^2# where #bb(T)^(-1)(x,y)=(2/3x+1/3y,1/3y-1/3x)#

Using this result, we have:

# bb(T)^(-1)(-1,8)=(2,3)#

Explanation:

We have a linear operator:

# bb(T):RR^2 rarr RR^2# defined by #bb(T)(x,y)=(x-y,x+2y)#

Which, if we use a standard basis, we can represent in matrix form by:

# bb(T) ((x),(y)) = ((x-y),(x+2y)) #

# \ \ \ \ \ \ \ \ \ \ \ \= ((1,-1),(1,2)) ((x),(y)) #

# \ \ \ \ \ \ \ \ \ \ \ \= bb(M) ((x),(y))#, say with #bb(M) ((1,-1),(1,2)) #

Thus we can form the inverse operator #bb(T)^(-1):RR^2 rarr RR^2# by calculating the inverse #bb(M)^(-1)#, of the above matrix, thus:

# bb(M)^(-1) = 1/ || bb(M) || adj(bb(M)) #

# \ \ \ \ \ \ \ \ = 1/ (2-(-1)) ((2,1),(-1,1)) #

# \ \ \ \ \ \ \ \ = 1/3 ((2,1),(-1,1)) #

# \ \ \ \ \ \ \ \ = ((2/3,1/3),(-1/3,1/3)) #

And so the inverse operator in matrix form is

# bb(T)^(-1) ((x),(y)) = ((2/3,1/3),(-1/3,1/3)) ((x),(y)) #

# \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = ((2/3x+1/3y),(1/3y-1/3x)) #

So we can write:

# bb(T)^(-1):RR^2 rarr RR^2# where #bb(T)^(-1)(x,y)=(2/3x+1/3y,1/3y-1/3x)#

Using this result, we have:

# bb(T)^(-1)(-1,8)=(-2/3+8/3,8/3+1/3)#
# \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = (2,3)#

As a validation check we would expect:

# bb(T)(2,3) = (-1,8)#

And:

# bb(T)(2,3) = (2-3,2+6) = (-1,8)#

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