Let
$\mathcal{E}=\left\{e_1,e_2,e_3\right\}$
be our canonical base. With this base, transormation T has representation
$T=\left(
\begin{array}{ccc}
3 & 1 & 0 \\
1 & 0 & 1 \\
1 & 0 & -1 \\
\end{array}
\right)$.
Now we have got a new base:
$\mathcal{F}=\left\{e_1,e_1+e_2,e_1+e_2+e_3\right\}$.
Let
$M_{\mathcal{F}}=\left(
\begin{array}{ccc}
1 & 1 & 1 \\
0 & 1 & 1 \\
0 & 0 & 1 \\
\end{array}
\right)$
be the transition between the two bases.
Then canonical coordinates are transormed in new coordinates
(with respect to base $\mathcal{F}$ ) by inverse matrix, which is
$N_{\mathcal{F}}=\left(
\begin{array}{ccc}
1 & -1 & 0 \\
0 & 1 & -1 \\
0 & 0 & 1 \\
\end{array}
\right)$.
Take
$A=\left\{a_1,a_2,a_3\right\}$
and get new coordinates
$B=N_{\mathcal{F}}.A$.
Then, with $S=T.M_{\mathcal{F}}$
we see:
$T.A=T.M_{\mathcal{F}}.N_{\mathcal{F}}.A=S.B$.
It's not a miracle, only lin. Algebra.
Key is transformation of basis, which implies
transformation of coordinates. That's all.
By the way: Calculating without inverses is not
possible. Your transformation with bases must be
regular. They must be invertible, otherwise it didn't
work.
Let's see. Other basis
$\mathcal{B}=\left\{2 e_1+5 e_3,e_1+e_2+6 e_3,3 e_1+9 e_3\right\}$,
another transition:
$M_{\mathcal{B}}=\left(
\begin{array}{ccc}
2 & 1 & 3 \\
0 & 1 & 0 \\
5 & 6 & 9 \\
\end{array}
\right)$.
The inverse:
$N_{\mathcal{B}}=\left(
\begin{array}{ccc}
3 & 3 & -1 \\
0 & 1 & 0 \\
-\frac{5}{3} & -\frac{7}{3} & \frac{2}{3} \\
\end{array}
\right)$.
Old transformation T
$T=\left(
\begin{array}{ccc}
3 & 1 & 0 \\
1 & 0 & 1 \\
1 & 0 & -1 \\
\end{array}
\right)$.
Transformed T:
$S=T.M_{\mathcal{B}}=\left(
\begin{array}{ccc}
6 & 4 & 9 \\
7 & 7 & 12 \\
-3 & -5 & -6 \\
\end{array}
\right)$
Transformed A:
$B=N_{\mathcal{B}}.A$.
$T.A=T.M_{\mathcal{B}}.N_{\mathcal{B}}.A=S.B$
Like before.
Given two bases $B,B'\in \mathbb{R}^3$ and a linear transformation $$\varphi:\mathbb{R}^3_B \to \mathbb{R}^3_{B'},$$ what you want is to somehow express the images of your basis vectors $B$ with respect to $B'$.
Let $e_1,e_2,e_3 \in B$ denote the basis vectors of $B$ and $b_1,b_2,b_3\in B'$ the basis vectors of $B'$. Your goal is to find the coordinates of $\varphi(e_1)$ in $\mathbb{R}^3_{B'}$ with respect to the basis $B'$, that is
$$\varphi(e_i) = \lambda_1b_1+ \lambda_2b_2+ \lambda_3b_3, \ \text{for}\ i=1,2,3$$
That leads to solving three linear equations $$\varphi(e_i) = \begin{pmatrix} 1 & 1 & 1\\ 0 & 1 & 1\\ 0 & 0 & 1 \end{pmatrix}\cdot\begin{pmatrix} \lambda_1 \\ \lambda_2 \\ \lambda_3\end{pmatrix}$$
Note that the columns of the matrix $A_\varphi$ are precisely the basis vectors $b_1,b_2,b_2$ of $B'$ and $\varphi(e_i)$ are simply the images of the basis vectors of $B$.
Best Answer
Since $\ker T\neq\{0\}$, $\dim\operatorname{Im}Y\leqslant2$. On the other hand, you know that $(1,1,1)+3(1,1,0)+2(1,0,1)\bigl(=(6,4,3)\bigr)$ and $2(1,1,0)+(1,0,1)\bigl(=(3,2,1)\bigr)$ belong to $\operatorname{Im}Y$. Since they are linearly independent, they form a basis of $\operatorname{Im}Y$.