For the linear transformation $$ L: \mathbb{R}^{2} \rightarrow \mathbb{R}^{3} $$ given by $$ L((x, y))=(x+2 y,-x, 0) $$ find the matrix $$ A_{B C} $$ for $$ L $$ with respect to the bases $$ B=\{(1,3),(-2,4)\}, C=\{(1,1,1),(2,2,0),(3,0,0)\} $$

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To find the matrix $$ A_{B C} $$ for the linear transformation $$ L: \mathbb{R}^{2} \rightarrow \mathbb{R}^{3} $$ with respect to the bases $$ B $$ and $$ C $$, we will follow these steps:

1. **Apply the transformation $$ L $$ to the basis vectors of $$ B $$**.
2. **Express the results in terms of the basis $$ C $$**.
3. **Form the matrix $$ A_{B C} $$ using the coefficients from the previous step**.

### Step 1: Apply $$ L $$ to the basis vectors of $$ B $$

The basis $$ B $$ consists of the vectors $$ (1, 3) $$ and $$ (-2, 4) $$.

1. For the first basis vector $$ (1, 3) $$:
   $$
   L((1, 3)) = (1 + 2 \cdot 3, -1, 0) = (1 + 6, -1, 0) = (7, -1, 0)
   $$

2. For the second basis vector $$ (-2, 4) $$:
   $$
   L((-2, 4)) = (-2 + 2 \cdot 4, -(-2), 0) = (-2 + 8, 2, 0) = (6, 2, 0)
   $$

### Step 2: Express the results in terms of the basis $$ C $$

The basis $$ C $$ consists of the vectors $$ (1, 1, 1) $$, $$ (2, 2, 0) $$, and $$ (3, 0, 0) $$.

We need to express $$ (7, -1, 0) $$ and $$ (6, 2, 0) $$ as linear combinations of the vectors in $$ C $$.

#### Expressing $$ (7, -1, 0) $$

We want to find scalars $$ a, b, c $$ such that:
$$
a(1, 1, 1) + b(2, 2, 0) + c(3, 0, 0) = (7, -1, 0)
$$
This gives us the system of equations:
1. $$ a + 2b + 3c = 7 $$
2. $$ a + 2b = -1 $$
3. $$ a = 0 $$

From the third equation, we have $$ a = 0 $$. Substituting $$ a = 0 $$ into the second equation:
$$
0 + 2b = -1 \implies b = -\frac{1}{2}
$$
Now substituting $$ a = 0 $$ and $$ b = -\frac{1}{2} $$ into the first equation:
$$
0 + 2(-\frac{1}{2}) + 3c = 7 \implies -1 + 3c = 7 \implies 3c = 8 \implies c = \frac{8}{3}
$$
Thus, we have:
$$
(7, -1, 0) = 0(1, 1, 1) - \frac{1}{2}(2, 2, 0) + \frac{8}{3}(3, 0, 0)
$$
This corresponds to the coefficients $$ (0, -\frac{1}{2}, \frac{8}{3}) $$.

#### Expressing $$ (6, 2, 0) $$

Now we find $$ a', b', c' $$ such that:
$$
a'(1, 1, 1) + b'(2, 2, 0) + c'(3, 0, 0) = (6, 2, 0)
$$
This gives us the system:
1. $$ a' + 2b' + 3c' = 6 $$
2. $$ a' + 2b' = 2 $$
3. $$ a' = 0 $$

From the third equation, $$ a' = 0 $$. Substituting into the second equation:
$$
0 + 2b' = 2 \implies b' = 1
$$
Now substituting $$ a' = 0 $$ and $$ b' = 1 $$ into the first equation:
$$
0 + 2(1) + 3c' = 6 \implies 2 + 3c' = 6 \implies 3c' = 4 \implies c' = \frac{4}{3}
$$
Thus, we have:
$$
(6, 2, 0) = 0(1, 1, 1) + 1(2, 2, 0) + \frac{4}{3}(3, 0, 0)
$$
This corresponds to the coefficients $$ (0, 1, \frac{4}{3}) $$.

### Step 3: Form the matrix $$ A_{B C} $$

The matrix $$ A_{B C} $$ is formed by taking the coefficients we found:
$$
A_{B C} = \begin{pmatrix}
0 & 0 \
-\frac{1}{2} & 1 \
\frac{8}{3} & \frac{4}{3}
\end{pmatrix}
$$

Thus, the final answer is:
$$
\boxed{\begin{pmatrix}
0 & 0 \
-\frac{1}{2} & 1 \
\frac{8}{3} & \frac{4}{3}
\end{pmatrix}}
$$
The matrix $$ A_{B C} $$ for the linear transformation $$ L $$ is: $$ \boxed{\begin{pmatrix} 0 & 0 \ -\frac{1}{2} & 1 \ \frac{8}{3} & \frac{4}{3} \end{pmatrix}} $$

For the linear transformation \( L: \mathbb{R}^{2} \rightarrow \mathbb{R}^{3} \) given by \[ L((x, y))=(x+2 y,-x, 0) \] find the matrix \( A_{B C} \) for \( L \) with respect to the bases \[ B=\{(1,3),(-2,4)\}, C=\{(1,1,1),(2,2,0),(3,0,0)\} \]

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