1. Diketahui matriks: $$A = \begin{bmatrix} 2 & -3 \\ -1 & 5 \end{bmatrix}, \quad B = \begin{bmatrix} -1 & 2 \\ 2 & 3 \end{bmatrix}$$ Ditanyakan: Invers dari hasil perkalian matriks $AB$, yaitu $(AB)^{-1}$. 2. Rumus penting: $$(AB)^{-1} = B^{-1} A^{-1}$$ Artinya, invers dari hasil perkalian dua matriks adalah perkalian invers dari matriks tersebut dengan urutan terbalik. 3. Hitung matriks $AB$: $$AB = \begin{bmatrix} 2 & -3 \\ -1 & 5 \end{bmatrix} \begin{bmatrix} -1 & 2 \\ 2 & 3 \end{bmatrix} = \begin{bmatrix} 2 \times (-1) + (-3) \times 2 & 2 \times 2 + (-3) \times 3 \\ -1 \times (-1) + 5 \times 2 & -1 \times 2 + 5 \times 3 \end{bmatrix} = \begin{bmatrix} -2 - 6 & 4 - 9 \\ 1 + 10 & -2 + 15 \end{bmatrix} = \begin{bmatrix} -8 & -5 \\ 11 & 13 \end{bmatrix}$$ 4. Hitung invers matriks $A$ dan $B$ terlebih dahulu. Invers matriks 2x2 $M = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$ adalah: $$M^{-1} = \frac{1}{ad - bc} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$$ 5. Hitung determinan dan invers $A$: $$\det(A) = 2 \times 5 - (-3) \times (-1) = 10 - 3 = 7$$ $$A^{-1} = \frac{1}{7} \begin{bmatrix} 5 & 3 \\ 1 & 2 \end{bmatrix}$$ 6. Hitung determinan dan invers $B$: $$\det(B) = (-1) \times 3 - 2 \times 2 = -3 - 4 = -7$$ $$B^{-1} = \frac{1}{-7} \begin{bmatrix} 3 & -2 \\ -2 & -1 \end{bmatrix} = \begin{bmatrix} -\frac{3}{7} & \frac{2}{7} \\ \frac{2}{7} & \frac{1}{7} \end{bmatrix}$$ 7. Hitung $(AB)^{-1} = B^{-1} A^{-1}$: $$B^{-1} A^{-1} = \begin{bmatrix} -\frac{3}{7} & \frac{2}{7} \\ \frac{2}{7} & \frac{1}{7} \end{bmatrix} \begin{bmatrix} \frac{5}{7} & \frac{3}{7} \\ \frac{1}{7} & \frac{2}{7} \end{bmatrix} = \begin{bmatrix} -\frac{3}{7} \times \frac{5}{7} + \frac{2}{7} \times \frac{1}{7} & -\frac{3}{7} \times \frac{3}{7} + \frac{2}{7} \times \frac{2}{7} \\ \frac{2}{7} \times \frac{5}{7} + \frac{1}{7} \times \frac{1}{7} & \frac{2}{7} \times \frac{3}{7} + \frac{1}{7} \times \frac{2}{7} \end{bmatrix}$$ Sederhanakan: $$= \begin{bmatrix} -\frac{15}{49} + \frac{2}{49} & -\frac{9}{49} + \frac{4}{49} \\ \frac{10}{49} + \frac{1}{49} & \frac{6}{49} + \frac{2}{49} \end{bmatrix} = \begin{bmatrix} -\frac{13}{49} & -\frac{5}{49} \\ \frac{11}{49} & \frac{8}{49} \end{bmatrix}$$ Jadi, invers dari $AB$ adalah: $$\boxed{(AB)^{-1} = \begin{bmatrix} -\frac{13}{49} & -\frac{5}{49} \\ \frac{11}{49} & \frac{8}{49} \end{bmatrix}}$$