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 \[ B=\{(1,3),(-2,4)\}, C=\{(1,1,1),(2,2,0),(3,0,0)\} \] For the linear transformation \( L: \mathbf{R}^{4} \rightarrow \mathbf{R}^{3} \) \( L\left(x_{1}, x_{2}, x_{3}, x_{4}\right)=\left(4 x_{1}+x_{2}-2 x_{3}-3 x_{4}, 2 x_{1}+x_{2}+x_{3}-4 x_{4}, 6 x_{1}-9 x_{3}+9 x_{4}\right) \) find a basis for \( \operatorname{ker}(L) \) and a basis for \( \operatorname{range}(L) \)

To find the matrix \( A_{B C} \) for the linear transformation \( L((x, y))=(x+2y,-x, 0) \) with respect to the bases \( B=\{(1,3),(-2,4)\} \) and \( C=\{(1,1,1),(2,2,0),(3,0,0)\} \), we first evaluate \( L \) on the basis vectors in \( B \). 1. For \( B_1=(1,3) \): \[ L((1,3)) = (1 + 2 \cdot 3, -1, 0) = (7, -1, 0) \] 2. For \( B_2=(-2,4) \): \[ L((-2,4)) = (-2 + 2 \cdot 4, 2, 0) = (6, 2, 0) \] Next, we need to express these results in terms of the basis \( C \). 1. To express \( (7, -1, 0) \) in terms of \( C \): \[ (7, -1, 0) = a(1,1,1) + b(2,2,0) + c(3,0,0) \] This yields the system: - \( a + 2b + 3c = 7 \) - \( a + 2b = -1 \) - \( a = 0 \) (from the third equation) Solving gives \( c = 2 \) and \( b = -\frac{1}{2} \). Thus, \( (7, -1, 0) = 0(1,1,1) - \frac{1}{2}(2,2,0) + 2(3,0,0) \). 2. To express \( (6, 2, 0) \) in terms of \( C \): \[ (6, 2, 0) = a(1,1,1) + b(2,2,0) + c(3,0,0) \] Solving gives \( a + 2b + 3c = 6 \), \( a + 2b = 2 \) leading again to \( a = 0\), \( b = 1 \), \( c = 1 \). Thus, \( (6, 2, 0) = 0(1,1,1) + 1(2,2,0) + 1(3,0,0) \). Putting it all together, the matrix \( A_{B C} \) will have the columns as the coefficients we calculated: \[ A_{B C} = \begin{bmatrix} 0 & 0 \\ -\frac{1}{2} & 1 \\ 2 & 1 \end{bmatrix}. \] Now, for the transformation \( L: \mathbb{R}^{4} \to \mathbb{R}^{3} \) defined by \[ L\left(x_{1}, x_{2}, x_{3}, x_{4}\right)=\left(4 x_{1}+x_{2}-2 x_{3}-3 x_{4}, 2 x_{1}+x_{2}+x_{3}-4 x_{4}, 6 x_{1}-9 x_{3}+9 x_{4}\right), \] To find a basis for \( \ker(L) \), we need to solve \( L(x_{1}, x_{2}, x_{3}, x_{4}) = (0, 0, 0) \). This results in the equations: 1. \( 4x_1 + x_2 - 2x_3 - 3x_4 = 0 \) 2. \( 2x_1 + x_2 + x_3 - 4x_4 = 0 \) 3. \( 6x_1 - 9x_3 + 9x_4 = 0 \) By solving these equations, you will find that the kernel has a structure allowing you to write the basis as: \[ \text{Basis for } \ker(L) = \{ (1, -2, 1, 0), (0, -\frac{1}{2}, 0, 1) \}. \] Next, to find a basis for the range of \( L \), we can determine the rank by computing the row echelon form of the matrix