1. **Problem statement:** We have a system of linear equations derived from the equilibrium condition $\frac{d[A_i]}{dt} = 0$ for all $i$. We want to solve for the concentrations $[A_i]$ in terms of the total concentration $[A] = [A_1] + [A_2] + [A_3]$, which is constant. 2. **Approach:** The system can be written as a matrix equation $\mathbf{M} \mathbf{x} = \mathbf{0}$ where $\mathbf{x} = \begin{bmatrix} [A_1] \\ [A_2] \\ [A_3] \end{bmatrix}$. We use Gaussian elimination (transforming to echelon form) to find relations between $[A_i]$. 3. **Step 1: Write the system in matrix form.** Suppose the system is: $$ \begin{cases} a_{11}[A_1] + a_{12}[A_2] + a_{13}[A_3] = 0 \\ a_{21}[A_1] + a_{22}[A_2] + a_{23}[A_3] = 0 \\ a_{31}[A_1] + a_{32}[A_2] + a_{33}[A_3] = 0 \end{cases} $$ 4. **Step 2: Apply Gaussian elimination to get echelon form.** This will give two independent equations relating $[A_1], [A_2], [A_3]$. 5. **Step 3: Express two variables in terms of the third, say $[A_3]$.** For example, $$ [A_1] = \alpha [A_3], \quad [A_2] = \beta [A_3] $$ where $\alpha, \beta$ are constants derived from the coefficients. 6. **Step 4: Use the total concentration constraint:** $$ [A] = [A_1] + [A_2] + [A_3] = \alpha [A_3] + \beta [A_3] + [A_3] = (\alpha + \beta + 1)[A_3] $$ 7. **Step 5: Solve for $[A_3]$:** $$ [A_3] = \frac{[A]}{\alpha + \beta + 1} $$ 8. **Step 6: Substitute back to find $[A_1]$ and $[A_2]$:** $$ [A_1] = \alpha \frac{[A]}{\alpha + \beta + 1}, \quad [A_2] = \beta \frac{[A]}{\alpha + \beta + 1} $$ **Summary:** By transforming the system to echelon form, expressing variables in terms of one variable, and using the total concentration constraint, we express all $[A_i]$ in terms of $[A]$. This method ensures the solution satisfies both the equilibrium and total concentration conditions.