1. **Problem:** Show that the linear operator $T: \mathbb{R}^2 \to \mathbb{R}^2$ defined by $$u_1 = 2x_1 + x_2, \quad u_2 = 3x_1 + 4x_2$$ is one-to-one and find $T^{-1}(u_1, u_2)$. 2. **Formula and rules:** A linear transformation $T$ is one-to-one if $T(\mathbf{x}) = \mathbf{0}$ implies $\mathbf{x} = \mathbf{0}$. 3. **Step 1: Write $T$ as a matrix:** $$T = \begin{bmatrix} 2 & 1 \\ 3 & 4 \end{bmatrix}$$ 4. **Step 2: Check if $T$ is one-to-one by checking invertibility:** Calculate determinant: $$\det(T) = 2 \times 4 - 3 \times 1 = 8 - 3 = 5 \neq 0$$ Since determinant is nonzero, $T$ is invertible and hence one-to-one. 5. **Step 3: Find $T^{-1}$:** Inverse of $2 \times 2$ matrix $\begin{bmatrix} a & b \\ c & d \end{bmatrix}$ is $$\frac{1}{ad - bc} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$$ 6. **Step 4: Apply formula:** $$T^{-1} = \frac{1}{5} \begin{bmatrix} 4 & -1 \\ -3 & 2 \end{bmatrix}$$ 7. **Step 5: Express $T^{-1}(u_1, u_2)$:** $$\begin{bmatrix} x_1 \\ x_2 \end{bmatrix} = T^{-1} \begin{bmatrix} u_1 \\ u_2 \end{bmatrix} = \frac{1}{5} \begin{bmatrix} 4 & -1 \\ -3 & 2 \end{bmatrix} \begin{bmatrix} u_1 \\ u_2 \end{bmatrix} = \begin{bmatrix} \frac{4u_1 - u_2}{5} \\ \frac{-3u_1 + 2u_2}{5} \end{bmatrix}$$ **Final answer:** $T$ is one-to-one and $$T^{-1}(u_1, u_2) = \left( \frac{4u_1 - u_2}{5}, \frac{-3u_1 + 2u_2}{5} \right)$.