Basis for an invertible linear transformation I'm working through Hoffman's linear algebra book right now for a proof-based class on Linear Algebra and I'm a little confused about some of the calculations regarding linear transformations.
For this question in particular, I am not sure if I have done it correctly or how to do part (c),

Let $T$ be the linear operation on $\mathbb{R}^3$ defined by, $$T(x_{1},x_{2},x_{3})=(3x_{1} + x_{3},-2x_{1} + x_{2}, -x_{1} + 2x_{2} + 4x_{3})$$
  $(a)$ What is the matrix of $T$ in the standard ordered basis for $\mathbb{R}^3$?
$(b)$ What is the matrix of $T$ in the ordered basis $\{\alpha_{1},\alpha_{2},\alpha_{3}\}$, where $\alpha_{1}=(1,0,1), \alpha_{2}=(-1,2,1),$ and $\alpha_{3}=(2,1,1)$?
$(c)$ Prove that $T$ is invertible and give a rule for $T^{-1}$ like the one which defines $T$.

For $(a)$, I think you simply compute $T(1,0,0), T(0,1,0)$, and $T(0,0,1)$ and then whatever you get will be the entries in $[T]_{\mathbb{B}}$, where $\mathbb{B}$ denotes the standard basis for $\mathbb{R}^3$. That is,
$$[T]_{\mathbb{B}} =\left[
\begin{array}{ccc}
3 & -2 & -1 \\
0 & 1 & 2 \\
1 & 0 & 4 \\
\end{array}
\right]$$
Now, for $(b)$ I am pretty sure that you just compute $T(1,0,1), T(-1,2,1),$ and $T(2,1,1)$ to find the matrix of $T$ in the ordered basis $\{\alpha_{1},\alpha_{2},\alpha_{3}\}$. Then, we get for $1 \leq i \leq 3$,
$$[T]_{\mathbb{\alpha_{i}}} =\left[
\begin{array}{ccc}
4 & -2 & 3 \\
-2 & 4 & 9 \\
7 & -3 & 4 \\
\end{array}
\right]$$
Let me know if I have gone wrong so far. My idea for $(c)$ was to invert this matrix and then somehow deduce a rule from there, but I am not sure.
My question is: How can I answer (c)? (and also, if I have done anything wrong thus far let me know).
This is not homework.
 A: If your coordinate vectors are row vectors, then $[T]_{\mathbb{B}}$ is correct. Otherwise, it should be transposed. I will continue to work in the world of row vectors (assuming your first matrix is following the conventions of your text)...
Let's call $\mathbb{A}=\{\alpha_1,\alpha_2,\alpha_3\}$. Then $T(\alpha_1)=T(1,0,1)=(4,-2,3)$ etc. and so your second matrix is $[T]_\mathbb{A}^\mathbb{B}$ (this matrix takes vectors in $\mathbb{A}$ coordinates and spits out answers in $\mathbb{B}$ coordinates. So that's not the matrix you want.
Instead, consider $P = [I]_\mathbb{A}^\mathbb{B} = \begin{bmatrix} 1 & 0 & 1 \\ -1 & 2 & 1 \\ 2 & 1 & 1 \end{bmatrix}$ (we get this by applying the identity map to each element of $\mathbb{A}$ and then writing them in $\mathbb{B}$ coordinates, so for example: $I(1,0,1)=(1,0,1)$ and $[(1,0,1)]_\mathbb{B}=[1\;0\;1]$). This matrix transforms $\mathbb{A}$ coordinate vectors to $\mathbb{B}$-coordinates. For example: $(0,2,2)=(1,0,1)+(-1,2,1)$ so $[(0,2,2)]_\mathbb{A}=[1\;1\;0]$. Notice that $[1\;1\;0]P = [0\;2\;2] = [(0,2,2)]_\mathbb{B}$.
Thus to get $T$ written in $\mathbb{A}$ coordinates, we need to shift from $\mathbb{A}$ to $\mathbb{B}$ coordinates, apply $T$ in $\mathbb{B}$ coordinates, and then shift back.
Therefore, $[T]_\mathbb{A} = P[T]_\mathbb{B}P^{-1}$ (since $P^{-1}=[I]^\mathbb{A}_\mathbb{B}$).
Now to get $T^{-1}$, just compute $([T]_\mathbb{B})^{-1}$ this will be the coordinate matrix of $T^{-1}$ in standard coordinates. To get a formula for $T^{-1}$ reverse the process you went through to get $[T]_\mathbb{B}$ in the first place.
Once you have a formula for $T^{-1}$ you can show that $T^{-1}(T(x_1,x_2,x_3))=(x_1,x_2,x_3)$ and $T(T^{-1}(y_1,y_2,y_3))=(y_1,y_2,y_3)$ thus showing that $T^{-1}$ is in fact the inverse of $T$ (which the notation $T^{-1}$ kind of assumes). Alternatively, one can show that a linear transformation (between finite dimensional vector spaces) is invertible if and only if (any of) its coordinate matrix is invertible which in turn is true iff the determinant of $[T]_\mathbb{B}$ is not zero.
