1. **Problem Statement:** Find the standard matrix $A$ of the linear transformation $T: \mathbb{R}^2 \to \mathbb{R}^2$ such that $$ T\begin{pmatrix}-1 \\ 2\end{pmatrix} = \begin{pmatrix}-1 \\ 4\end{pmatrix}, \quad T\begin{pmatrix}-2 \\ 3\end{pmatrix} = \begin{pmatrix}-1 \\ 5\end{pmatrix}. $$ 2. **Recall:** The standard matrix $A$ satisfies $T(\mathbf{x}) = A\mathbf{x}$ for any vector $\mathbf{x}$. If $A = \begin{pmatrix}a & b \\ c & d\end{pmatrix}$, then $$ A \begin{pmatrix}x \\ y\end{pmatrix} = \begin{pmatrix}a x + b y \\ c x + d y\end{pmatrix}. $$ 3. **Set up equations using given vectors:** From $T(u) = A u$: $$ A \begin{pmatrix}-1 \\ 2\end{pmatrix} = \begin{pmatrix}a(-1) + b(2) \\ c(-1) + d(2)\end{pmatrix} = \begin{pmatrix}-1 \\ 4\end{pmatrix}. $$ This gives two equations: $$ - a + 2 b = -1 \quad (1) $$ $$ - c + 2 d = 4 \quad (2) $$ From $T(v) = A v$: $$ A \begin{pmatrix}-2 \\ 3\end{pmatrix} = \begin{pmatrix}a(-2) + b(3) \\ c(-2) + d(3)\end{pmatrix} = \begin{pmatrix}-1 \\ 5\end{pmatrix}. $$ This gives two more equations: $$ - 2 a + 3 b = -1 \quad (3) $$ $$ - 2 c + 3 d = 5 \quad (4) $$ 4. **Solve for $a$ and $b$ using equations (1) and (3):** From (1): $$ - a + 2 b = -1 \implies a = 2 b + 1. $$ Substitute into (3): $$ - 2 (2 b + 1) + 3 b = -1 $$ Simplify: $$ -4 b - 2 + 3 b = -1 $$ $$ - b - 2 = -1 $$ $$ - b = 1 \implies b = -1. $$ Then, $$ a = 2(-1) + 1 = -2 + 1 = -1. $$ 5. **Solve for $c$ and $d$ using equations (2) and (4):** From (2): $$ - c + 2 d = 4 \implies c = 2 d - 4. $$ Substitute into (4): $$ - 2 (2 d - 4) + 3 d = 5 $$ Simplify: $$ -4 d + 8 + 3 d = 5 $$ $$ - d + 8 = 5 $$ $$ - d = -3 \implies d = 3. $$ Then, $$ c = 2(3) - 4 = 6 - 4 = 2. $$ 6. **Write the standard matrix $A$:** $$ A = \begin{pmatrix}a & b \\ c & d\end{pmatrix} = \begin{pmatrix}-1 & -1 \\ 2 & 3\end{pmatrix}. $$ **Final answer:** $$ \boxed{\begin{pmatrix}-1 & -1 \\ 2 & 3\end{pmatrix}}. $$