(a) Consider the following economic model. Where, two countries are engaged in an arms race on defense spending. We denote variable \( r \) as the annual defense expenditure for Country 1 and the variable \( s \) as the annual defense expenditure for Country 2 . We assume that each nation is ready to defend itself and has suffcient budget. We are interested in whether the defense expendituress will reach the equilibrium point(s). The interaction between the two countries expenditures is described by the following system: \[ \frac{d r}{d t}=-a r+b s+c \text {, } \] \[ \frac{d s}{d t}=m r-n s+p \text {. } \] (i) Determine the equilibrium point(s) of this system. Hint: both the r-coordinate and s-coordinate (4) should be written as a fraction with the denominator \( (a n-b m) \neq 0 \)

To determine the equilibrium points of the given system, we set the derivatives equal to zero since, at equilibrium, the rates of change of defense expenditures \( r \) and \( s \) are zero. The system of differential equations is: \[ \frac{dr}{dt} = -a r + b s + c = 0, \] \[ \frac{ds}{dt} = m r - n s + p = 0. \] This gives us the following system of linear equations: 1. \(-a r + b s = -c\), 2. \(m r - n s = -p\). To solve for \( r \) and \( s \), we can use algebraic methods such as substitution or linear algebra (Cramer's rule). Here's how you can solve it step-by-step: 1. **Set Up the Equations:** \[ -a r + b s = -c \quad \text{(Equation 1)}, \] \[ m r - n s = -p \quad \text{(Equation 2)}. \] 2. **Use Cramer's Rule:** The coefficient matrix \( A \) and the constant vector \( \mathbf{B} \) are: \[ A = \begin{pmatrix} -a & b \\ m & -n \end{pmatrix}, \quad \mathbf{B} = \begin{pmatrix} -c \\ -p \end{pmatrix}. \] The determinant of matrix \( A \) is: \[ \text{det}(A) = (-a)(-n) - (b)(m) = a n - b m \neq 0. \] 3. **Compute \( r \) and \( s \):** Using Cramer's rule: \[ r = \frac{\begin{vmatrix} -c & b \\ -p & -n \end{vmatrix}}{\text{det}(A)} = \frac{c n + b p}{a n - b m}, \] \[ s = \frac{\begin{vmatrix} -a & -c \\ m & -p \end{vmatrix}}{\text{det}(A)} = \frac{a p + c m}{a n - b m}. \] **Therefore, the equilibrium point \((r, s)\) is:** \[ \left( \frac{c n + b p}{a n - b m},\ \frac{a p + c m}{a n - b m} \right). \] **Final Answer:** After simplifying, an equilibrium is (r, s) = ( (c n + b p)/(a n – b m), (a p + c m)/(a n – b m) ). Thus, \[ \left( \frac{c n + b p}{a n - b m},\ \frac{a p + c m}{a n - b m} \right) \]