Here T(x,y,z)=(x−y+z,x+y,y+z)
v1=(1,1,1) , v2=(0,1,1) , v3=(0,0,1)
So T(1,1,1)=(1−1+1,1+1,1+1)=(1,2,2)
T(0,1,1)=(0−1+1,0+1,1+1)=(0,1,2)
T(0,0,1)=(0−0+1,0+0,0+1)=(1,0,1)
So Matrix of T with respect to basis {v1,v2,v3} is :
A=⎣⎡101210221⎦⎤
Here |A|=1(1−0)−2(0−2)+2(0−1)=1+4−2=5−2=3 which is not equal to .So matrix is invertible.
Comments