Question

Show that MDB(T) is invertible and use the fact that MBD(T^-1)=[MBD(T)]^-1 to determine the action of T^-1. T: P2-->R3, T(a+bx+cx^2)=(a+c,c,b-c); B={1,x,x^2}, D=standard

Answer 1

Given, T(a+bx+cx²) = (a+c,c,b-c) and B = {1,x,x²} and D = {(1,0,0),(0,1,0),(0,0,1)}.

Now, T(1) = T(1+0x+0x²) = (1+0,0,0-0) = (1,0,0)

T(x) = T(0+1x+0x²) = (0+0,0,1-0) = (0,0,1)

T(x²) = T(0+0x+1x²) = (0+1,1,0-1) = (1,1,-1)

Here, (1,0,0) = 1*(1,0,0)+0*(0,1,0)+0*(0,0,1)

(0,0,1) =0*(1,0,0)+0*(0,1,0)+1*(0,0,1)

(1,1,-1) = 1*(1,0,0)+1*(0,1,0)+(-1)*(0,0,1)

Therefore, = .

Since, determinant of = -1 0, therefore, the matrix is invertible.

Now, we have, =

i.e., =

This implies that,

T^-1(1,0,0) = 1*1+0*x+0*x²

T^-1(0,1,0) = (-1)*1+1*x+1*x²

T^-1(0,0,1) = 0*1+1*x+0*x²

i.e., T^-1(1,0,0) = 1

T^-1(0,1,0) = 1+x+x²

T^-1(0,0,1) = x

i.e., T^-1(1,0,0) = (1+0)*1+0*x+(0+0)*x²

T^-1(0,1,0) = (0+1)*1+1*x+(1+0)*x²

T^-1(0,0,1) = (0+0)*1+0*x+(0+1)*x²

This implies that T^-1(a,b,c) = (a+b)*1+b*x+(b+c)*x².