1. **Problem Statement:** We have three chemical species $A_1$, $A_2$, and $A_3$ undergoing the following reactions with first-order kinetics: $$ A_1 \xrightarrow{k_1} A_2, \quad A_2 \xrightleftharpoons[k_{-2}]{k_2} A_3, \quad A_3 \xrightarrow{k_3} A_1 $$ We want to write the differential equations for the concentrations $[A_i]$ in matrix form. 2. **Writing the rate equations:** - For $A_1$: - It is consumed by the reaction $A_1 \to A_2$ at rate $k_1 [A_1]$. - It is produced by the reaction $A_3 \to A_1$ at rate $k_3 [A_3]$. So, $$\frac{d[A_1]}{dt} = -k_1 [A_1] + k_3 [A_3]$$ - For $A_2$: - It is produced by $A_1 \to A_2$ at rate $k_1 [A_1]$. - It is consumed by the forward reaction $A_2 \to A_3$ at rate $k_2 [A_2]$. - It is produced by the backward reaction $A_3 \to A_2$ at rate $k_{-2} [A_3]$. So, $$\frac{d[A_2]}{dt} = k_1 [A_1] - k_2 [A_2] + k_{-2} [A_3]$$ - For $A_3$: - It is produced by $A_2 \to A_3$ at rate $k_2 [A_2]$. - It is consumed by the backward reaction $A_3 \to A_2$ at rate $k_{-2} [A_3]$. - It is consumed by $A_3 \to A_1$ at rate $k_3 [A_3]$. So, $$\frac{d[A_3]}{dt} = k_2 [A_2] - (k_{-2} + k_3) [A_3]$$ 3. **Matrix form:** Define the concentration vector: $$\mathbf{C} = \begin{bmatrix} [A_1] \\ [A_2] \\ [A_3] \end{bmatrix}$$ The system can be written as: $$\frac{d\mathbf{C}}{dt} = \mathbf{M} \mathbf{C}$$ where the matrix $\mathbf{M}$ is: $$\mathbf{M} = \begin{bmatrix} -k_1 & 0 & k_3 \\ k_1 & -k_2 & k_{-2} \\ 0 & k_2 & -(k_{-2} + k_3) \end{bmatrix}$$ 4. **Explanation:** - Each diagonal element of $\mathbf{M}$ is negative and represents the total rate of consumption of that species. - Off-diagonal elements represent production rates from other species. - This matrix form compactly represents the coupled differential equations for the system. **Final answer:** $$\frac{d}{dt} \begin{bmatrix} [A_1] \\ [A_2] \\ [A_3] \end{bmatrix} = \begin{bmatrix} -k_1 & 0 & k_3 \\ k_1 & -k_2 & k_{-2} \\ 0 & k_2 & -(k_{-2} + k_3) \end{bmatrix} \begin{bmatrix} [A_1] \\ [A_2] \\ [A_3] \end{bmatrix}$$