Let $L : \mathbb{R}^3\rightarrow\mathbb{R}^3$ be a linear transformation such that its matrix with respect to the standard basis is: $$[L] = \begin{bmatrix}1 & 0 & 1 \\ -1 & -1 & 0 \\0&1&-1\end{bmatrix}$$ Consider the two following bases of $\mathbb{R}^3$: $$\beta = \left(\begin{bmatrix}1 \\ 1 \\ 0\end{bmatrix},\begin{bmatrix}0 \\ 1 \\ 1\end{bmatrix},\begin{bmatrix}0 \\ 0 \\ 1\end{bmatrix}\right)$$ $$\gamma = \left(\begin{bmatrix}1 \\ 2 \\ 3\end{bmatrix},\begin{bmatrix}0 \\ 1 \\ 2\end{bmatrix},\begin{bmatrix}0 \\ 0 \\ 1\end{bmatrix}\right)$$ Find the matrix associated with the transformation with respect to $\beta$ in input and to $\gamma$ in output.
I seem to have some troubles with this kind of exercises. I have been explained a way to do it through multiplying $3$ matrices, but I can't seem to quite remember it every time I do an exercise. This is what I have tried:
The first thing is to rewrite the vectors in $\beta$ as linear combinations of those of the standard basis. Let's enumerate the vectors in $\beta$ and $\gamma$ with an index that goes from $1$ to $3$. Also, $e_1, e_2, e_3$ are the vectors of the standard basis of $\mathbb{R}^3$.
We observe that $\beta_1 = e_1 + e_2, \beta_2 = e_2+e_3, \beta_3 = e_3$, therefore the matrix that uses $\beta$ as the input basis and outputs vectors in the standard basis will be: $$[L]^{\beta}= \begin{bmatrix}1 & 1 & 1 \\ -2 & -1 & 0 \\0&0&-1\end{bmatrix}$$
Now, if my understanding is correct, I need to write the vectors that make up the columns of that matrix as linear combinations of vectors in $\gamma$.
Using Gauss' algorithm, we find that: $$\left[\begin{array}{ccc|ccc}1&0&0&1 & 1 & 1 \\ 2&1&0&-2 & -1 & 0 \\3&2&1&0&0&-1\end{array}\right]\rightarrow \left[\begin{array}{ccc|ccc}1 & 0 & 0 & 1 & 1 & 1 \\ 0 & 1 & 0 & -4 & -3 & -2 \\0 & 0 & 1 & 5 & 6 & 2\end{array}\right]$$
Which finally means that: $$[L]^{\beta}_{\gamma} = \left[\begin{array} & 1 & 1 & 1 \\ -4 & -3 & -2 \\5 & 6 & 2\end{array}\right]$$
Is my reasoning correct? Did I make any mistakes?