1. **State the problem:** We need to find the 2x2 matrix \( T = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \) that transforms the point \((2,1)\) to \((1,4)\) and the point \((1,-3)\) to \((4,9)\). 2. **Write the transformation equations:** \[ \begin{pmatrix} x' \\ y' \end{pmatrix} = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} \] For the first point \((2,1)\) to \((1,4)\): \[ \begin{cases} 2a + b = 1 \\ 2c + d = 4 \end{cases} \] For the second point \((1,-3)\) to \((4,9)\): \[ \begin{cases} a - 3b = 4 \\ c - 3d = 9 \end{cases} \] 3. **Solve for \(a\) and \(b\):** Multiply the second equation by \(-2\) to eliminate \(a\): \[ \begin{cases} 2a + b = 1 \\ -2a + 6b = -8 \end{cases} \] Add the two equations: \[ (2a - 2a) + (b + 6b) = 1 - 8 \Rightarrow 7b = -7 \] So, \[ b = -1 \] Substitute \(b = -1\) into \(2a + b = 1\): \[ 2a - 1 = 1 \Rightarrow 2a = 2 \Rightarrow a = 1 \] 4. **Solve for \(c\) and \(d\):** Multiply the second equation by \(-2\) to eliminate \(c\): \[ \begin{cases} 2c + d = 4 \\ -2c + 6d = -18 \end{cases} \] Add the two equations: \[ (2c - 2c) + (d + 6d) = 4 - 18 \Rightarrow 7d = -14 \] So, \[ d = -2 \] Substitute \(d = -2\) into \(2c + d = 4\): \[ 2c - 2 = 4 \Rightarrow 2c = 6 \Rightarrow c = 3 \] 5. **Write the transformation matrix:** \[ T = \begin{pmatrix} 1 & -1 \\ 3 & -2 \end{pmatrix} \] **Final answer:** The matrix that performs the transformation is \( \boxed{\begin{pmatrix} 1 & -1 \\ 3 & -2 \end{pmatrix}} \).