5(a) (a) Using Cramer's rule to determine \( x+y \), if \( \left[\begin{array}{cc}5 & -3 \\ 4 & 2\end{array}\right]\left[\begin{array}{l}x \\ y\end{array}\right]=\left[\begin{array}{c}-1 \\ -8\end{array}\right] \) (b) Given IS equation \( 0.4 Y+100 I-255=0 \) and the \( L M \) equation \( 0.25 Y-200 \mathrm{I}-177=0 \). Use matrix form to find the equilibrium level of income and rate of interest.

Let's solve each part of the problem step by step. --- ### **Problem 5(a): Using Cramer's Rule to Determine \( x + y \)** **Given:** \[ \begin{bmatrix} 5 & -3 \\ 4 & 2 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} -1 \\ -8 \end{bmatrix} \] **System of Equations:** \[ \begin{cases} 5x - 3y = -1 \quad \text{(1)} \\ 4x + 2y = -8 \quad \text{(2)} \end{cases} \] **Step 1: Compute the Determinant (D) of the Coefficient Matrix** \[ D = \begin{vmatrix} 5 & -3 \\ 4 & 2 \end{vmatrix} = (5)(2) - (-3)(4) = 10 + 12 = 22 \] **Step 2: Compute Determinants for \( x \) and \( y \) (Dx and Dy)** \[ D_x = \begin{vmatrix} -1 & -3 \\ -8 & 2 \end{vmatrix} = (-1)(2) - (-3)(-8) = -2 - 24 = -26 \] \[ D_y = \begin{vmatrix} 5 & -1 \\ 4 & -8 \end{vmatrix} = (5)(-8) - (-1)(4) = -40 + 4 = -36 \] **Step 3: Solve for \( x \) and \( y \) Using Cramer's Rule** \[ x = \frac{D_x}{D} = \frac{-26}{22} = -\frac{13}{11} \] \[ y = \frac{D_y}{D} = \frac{-36}{22} = -\frac{18}{11} \] **Step 4: Determine \( x + y \)** \[ x + y = -\frac{13}{11} + (-\frac{18}{11}) = -\frac{31}{11} \] **Final Answer for Part (a):** \[ x + y = -\frac{31}{11} \] --- ### **Problem 5(b): Finding the Equilibrium Level of Income and Rate of Interest Using Matrix Methods** **Given Equations:** \[ \begin{cases} 0.4Y + 100I - 255 = 0 \quad \text{(IS Equation)} \\ 0.25Y - 200I - 177 = 0 \quad \text{(LM Equation)} \end{cases} \] **Step 1: Rewrite the Equations in Standard Form** \[ \begin{cases} 0.4Y + 100I = 255 \quad \text{(1)} \\ 0.25Y - 200I = 177 \quad \text{(2)} \end{cases} \] **Step 2: Express in Matrix Form \( AX = B \)** \[ A = \begin{bmatrix} 0.4 & 100 \\ 0.25 & -200 \end{bmatrix}, \quad X = \begin{bmatrix} Y \\ I \end{bmatrix}, \quad B = \begin{bmatrix} 255 \\ 177 \end{bmatrix} \] **Step 3: Compute the Determinant of Matrix \( A \)** \[ D = (0.4)(-200) - (100)(0.25) = -80 - 25 = -105 \] **Step 4: Find the Inverse of Matrix \( A \)** \[ A^{-1} = \frac{1}{D} \begin{bmatrix} -200 & -100 \\ -0.25 & 0.4 \end{bmatrix} = \frac{1}{-105} \begin{bmatrix} -200 & -100 \\ -0.25 & 0.4 \end{bmatrix} \] **Step 5: Multiply \( A^{-1} \) with \( B \) to Find \( X \)** \[ X = A^{-1}B = \frac{1}{-105} \begin{bmatrix} -200 & -100 \\ -0.25 & 0.4 \end{bmatrix} \begin{bmatrix} 255 \\ 177 \end{bmatrix} \] **Calculations:** - For \( Y \): \[ Y = \frac{(-200 \times 255) + (-100 \times 177)}{-105} = \frac{-51,000 - 17,700}{-105} = \frac{-68,700}{-105} = \frac{68,700}{105} = 654 \frac{2}{7} \] - For \( I \): \[ I = \frac{(-0.25 \times 255) + (0.4 \times 177)}{-105} = \frac{-63.75 + 70.8}{-105} = \frac{7.05}{-105} = -\frac{47}{700} \approx -0.067 \] **Final Answer for Part (b):** \[ \begin{cases} \text{Equilibrium Level of Income } Y = 654 \frac{2}{7} \\ \text{Equilibrium Rate of Interest } I = -\frac{47}{700} \approx -0.067 \end{cases} \] --- **Summary:** - **5(a):** \( x + y = -\frac{31}{11} \) - **5(b):** - **Equilibrium Income (Y):** \( 654 \frac{2}{7} \) - **Equilibrium Interest Rate (I):** \( -\frac{47}{700} \) or approximately \(-0.067\)