Let T : R³→ R³ be defined by T (x1, x2, x3) = (x1 −x3, x2 −x3, x1). Is T invertible? If yes, find a rule for T–¹ like the one which defines T ?

1 Answer
Cesareo R.
Jan 26, 2018

See below.

Explanation:

With the rules

#T(x_1,x_2,x_3) = (x_1-x_3,x_2-x_3,x_1)# we can define

#T = ((1,0,-1),(0,1,-1),(1,0,0))#

Assuming #S# as a linear operator such that

#S(x_1-x_3,x_2-x_3,x_1) = (x_1,x_2,x_3)# due to the linearity of #S# we have

#{(alpha_1(x_1-x_3)+alpha_2(x_2-x_3)+alpha_3 x_1 = x_1),(beta_1(x_1-x_3)+beta_2(x_2-x_3)+beta_3 x_1 = x_2),(gamma_1(x_1-x_3)+gamma_2(x_2-x_3)+gamma_3 x_1 = x_3):}#

or

#{((alpha_1+alpha_3)x_1+alpha_2 x_2 +(alpha_3-alpha_1-alpha_2)x_3 = x_1), ((beta_1+alpha_3)x_1+beta_2 x_2 +(beta_3-beta_1-beta_2)x_3 = x_2),((gamma_1+gamma_3)x_1+gamma_2 x_2 +(gamma_3-gamma_1-gamma_2)x_3 = x_3):}#

then solving the systems

#{(alpha_1+alpha_3=1),(alpha_2=0),(alpha_3-alpha_1-alpha_2=0):}#

#{(beta_1+beta_3=0),(beta_2=1),(alpha_3-alpha_1-alpha_2=0):}#

#{(gamma_1+gamma_3=0),(gamma_2=0),(gamma_3-gamma_1-gamma_2=1):}#

we obtain

#S =((alpha_1,alpha_2,alpha_3),(beta_1,beta_2,beta_3),(gamma_1,gamma_2,gamma_3))= ((0,0,1),(-1,1,1),(-1,0,1)) = T^-1#

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