1. **Problem Statement:** Verify that $$(ABC)^{-1} = C^{-1} B^{-1} A^{-1}$$ for the given matrices: $$A = \begin{bmatrix}3 & 1 \\ 5 & 2\end{bmatrix}, B = \begin{bmatrix}2 & -3 \\ 4 & 4\end{bmatrix}, C = \begin{bmatrix}6 & 4 \\ -2 & -1\end{bmatrix}$$ 2. **Recall the property of inverses of matrix products:** For invertible matrices, the inverse of a product is the product of the inverses in reverse order: $$ (ABC)^{-1} = C^{-1} B^{-1} A^{-1} $$ 3. **Calculate $A^{-1}$:** The inverse of a 2x2 matrix $\begin{bmatrix}a & b \\ c & d\end{bmatrix}$ is $$ \frac{1}{ad - bc} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix} $$ For $A$: $$ \det(A) = 3 \times 2 - 1 \times 5 = 6 - 5 = 1 $$ So, $$ A^{-1} = \begin{bmatrix} 2 & -1 \\ -5 & 3 \end{bmatrix} $$ 4. **Calculate $B^{-1}$:** $$ \det(B) = 2 \times 4 - (-3) \times 4 = 8 + 12 = 20 $$ $$ B^{-1} = \frac{1}{20} \begin{bmatrix} 4 & 3 \\ -4 & 2 \end{bmatrix} = \begin{bmatrix} 0.2 & 0.15 \\ -0.2 & 0.1 \end{bmatrix} $$ 5. **Calculate $C^{-1}$:** $$ \det(C) = 6 \times (-1) - 4 \times (-2) = -6 + 8 = 2 $$ $$ C^{-1} = \frac{1}{2} \begin{bmatrix} -1 & -4 \\ 2 & 6 \end{bmatrix} = \begin{bmatrix} -0.5 & -2 \\ 1 & 3 \end{bmatrix} $$ 6. **Calculate $ABC$:** First, compute $AB$: $$ AB = \begin{bmatrix}3 & 1 \\ 5 & 2\end{bmatrix} \begin{bmatrix}2 & -3 \\ 4 & 4\end{bmatrix} = \begin{bmatrix}3\times2 + 1\times4 & 3\times(-3) + 1\times4 \\ 5\times2 + 2\times4 & 5\times(-3) + 2\times4\end{bmatrix} = \begin{bmatrix}10 & -5 \\ 18 & -7\end{bmatrix} $$ Then, compute $ABC$: $$ ABC = \begin{bmatrix}10 & -5 \\ 18 & -7\end{bmatrix} \begin{bmatrix}6 & 4 \\ -2 & -1\end{bmatrix} = \begin{bmatrix}10\times6 + (-5)\times(-2) & 10\times4 + (-5)\times(-1) \\ 18\times6 + (-7)\times(-2) & 18\times4 + (-7)\times(-1)\end{bmatrix} = \begin{bmatrix}60 + 10 & 40 + 5 \\ 108 + 14 & 72 + 7\end{bmatrix} = \begin{bmatrix}70 & 45 \\ 122 & 79\end{bmatrix} $$ 7. **Calculate $(ABC)^{-1}$:** $$ \det(ABC) = 70 \times 79 - 45 \times 122 = 5530 - 5490 = 40 $$ $$ (ABC)^{-1} = \frac{1}{40} \begin{bmatrix}79 & -45 \\ -122 & 70 \end{bmatrix} = \begin{bmatrix}1.975 & -1.125 \\ -3.05 & 1.75 \end{bmatrix} $$ 8. **Calculate $C^{-1} B^{-1} A^{-1}$:** First, compute $B^{-1} A^{-1}$: $$ B^{-1} A^{-1} = \begin{bmatrix}0.2 & 0.15 \\ -0.2 & 0.1\end{bmatrix} \begin{bmatrix}2 & -1 \\ -5 & 3\end{bmatrix} = \begin{bmatrix}0.2\times2 + 0.15\times(-5) & 0.2\times(-1) + 0.15\times3 \\ -0.2\times2 + 0.1\times(-5) & -0.2\times(-1) + 0.1\times3\end{bmatrix} = \begin{bmatrix}0.4 - 0.75 & -0.2 + 0.45 \\ -0.4 - 0.5 & 0.2 + 0.3\end{bmatrix} = \begin{bmatrix}-0.35 & 0.25 \\ -0.9 & 0.5\end{bmatrix} $$ Then, compute $C^{-1} (B^{-1} A^{-1})$: $$ C^{-1} B^{-1} A^{-1} = \begin{bmatrix}-0.5 & -2 \\ 1 & 3\end{bmatrix} \begin{bmatrix}-0.35 & 0.25 \\ -0.9 & 0.5\end{bmatrix} = \begin{bmatrix}-0.5\times(-0.35) + (-2)\times(-0.9) & -0.5\times0.25 + (-2)\times0.5 \\ 1\times(-0.35) + 3\times(-0.9) & 1\times0.25 + 3\times0.5\end{bmatrix} = \begin{bmatrix}0.175 + 1.8 & -0.125 - 1 \\ -0.35 - 2.7 & 0.25 + 1.5\end{bmatrix} = \begin{bmatrix}1.975 & -1.125 \\ -3.05 & 1.75\end{bmatrix} $$ 9. **Conclusion:** We have $$ (ABC)^{-1} = \begin{bmatrix}1.975 & -1.125 \\ -3.05 & 1.75\end{bmatrix} = C^{-1} B^{-1} A^{-1} $$ Thus, the property is verified for the given matrices.