Answer
See below
Work Step by Step
Assume $x=(x_1,x_2,x_3)\\
y=(y_1,y_2,y_3)\\
c\in R$
$T(x+y)=T((x_1,x_2,x_3)+(y_1,y_2,y_3))\\
=T(x_1+y_1,x_2+y_2,x_3+y_3)\\
=(2(x_1+y_1)-3(x_2+y_2),-(x_1+y_1))\\
=(2x_1++2y_1-3x_2-3y_2,-x_1-y_1)\\
=(2x_1-3x_2,-x_1)+(2y_1-3y_2,-y_1)\\
=T(x_1,x_2,x_3)+T(y_1,y_2,y_3)\\
=T(x)+T(y)$
$T(cx)=T(c(x_1,x_2,x_3))\\
=T(cx_1,cx_2,cx_3)\\
=(2cx_1-3cx_2,-cx_1)\\
=c(2x_1-3x_2,-x_1)\\
=cT(x_1x_2,x_3)\\
=cT(x)$
Hence, $T$ is a linear transformation
Obtain: $Ker(T)=\{x \in R^3:T (x)=0\}\\
=\{(x,y,z) \in R^3: T(x,y,z)=0\}\\
=\{(z,y,z) \in R^3: (2x-3y,-x)=(0,0)\\
\rightarrow x=y=0$
then $Ker(T)=\{(0,0,z)\}=\{z(0,0,1)\}=span \{(0,0,1)\}\\
\rightarrow \dim [Ker(T)]=1 \ne 0$
$T$ is not one-to-one (1)
Apply Rank Nullity Theorem:
$\dim [Ker(T)]+\dim [Rng(T)]=\dim R^3\\
1+\dim [Rng(T)]=3\\
\dim [Rng(T)]=2$
$Rng(T) \subset R^2$ and $\dim R^2=2=\dim [Rng(T)]$
$T$ is onto (2)
Basic for $Rng(T)$ is $\{(1,0),(0,1)$
