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 \[ B=\{(1,3),(-2,4)\}, C=\{(1,1,1),(2,2,0),(3,0,0)\} \]

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: \[ \boxed{\begin{pmatrix} 0 & 0 \\ -\frac{1}{2} & 1 \\ \frac{8}{3} & \frac{4}{3} \end{pmatrix}} \]