Define: R^3→R^3 by
T(x,y,z)=(x-y+z,x+y,y+z)
Let v1= (1,1,1), v2= (0,1,1), v3= (0,0,1). Find a matrix of T with respect to the basis {v1,v2,v3}. Futher check T is invertible or not.
Here T(x,y,z)=(x−y+z,x+y,y+z)x,y,z)=(x-y+z,x+y,y+z)x,y,z)=(x−y+z,x+y,y+z)
v1=(1,1,1)v_1=(1,1,1)v1=(1,1,1) , v2=(0,1,1)v_2=(0,1,1)v2=(0,1,1) , v3=(0,0,1)v_3=(0,0,1)v3=(0,0,1)
So T(1,1,1)=(1−1+1,1+1,1+1)=(1,2,2)T(1,1,1)=(1-1+1,1+1,1+1)=(1,2,2)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(0,1,1)=(0-1+1,0+1,1+1)=(0,1,2)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)T(0,0,1)=(0-0+1,0+0,0+1)=(1,0,1)T(0,0,1)=(0−0+1,0+0,0+1)=(1,0,1)
So Matrix of T with respect to basis {v1,v2,v3v_1,v_2,v_3v1,v2,v3} is :
A=[122012101]A=\begin{bmatrix} 1& 2 &2 \\ 0 & 1 & 2\\ 1 &0&1 \end{bmatrix}A=⎣⎡101210221⎦⎤
Here |A|=1(1−0)−2(0−2)+2(0−1)=1+4−2=5−2=31(1-0)-2(0-2)+2(0-1)=1+4-2=5-2=31(1−0)−2(0−2)+2(0−1)=1+4−2=5−2=3 which is not equal to .So matrix is invertible.
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