Berea Technical College Department of Engineering Studies Mathematics N4 Assignment 1 Examiner: Mr Y. Mia Total: 50 Moderator: Mr P. Zvidzai Due Date: 20/03/25 INSTRUCTIONS 1. Answer ALL QUESTIONS in full. Show ALL the calculations and intermediary steps. Simplify where possible. 2. ALL final answers must be accurately approximated to THREE decimal places. 3. Questions may be answered in any order. 4. All Formulae used must be written down. QUESTION 1 1.1 Given: \[ \begin{array}{l} \frac{2 p}{3}-25 q=-5 \\ 2 p+\frac{5 q}{7}-4=0 \end{array} \] Use Crammer's rule to solve for \( p \) and \( q \). (6) 1.2 Given: \( \left|\begin{array}{ccc}\frac{1}{2} & -3 & 4 \\ \frac{5}{7} & 1 & -5 \\ \frac{7}{8} & 0 & 6\end{array}\right| \) 1.2.1 Determine the minor of 6 1.2.2 Determine the cofactor of -5 (2) 1.3 Given: \[ 3 x+2 y=-5 z-6 \] \[ \begin{array}{l} 5 x+7 y-2 z=4 \\ 2 x-7 z=5 y+14 \end{array} \] Solve for \( x \) only using Crammer's rule. Expand along ROW1 only. QUESTION 2

Let's solve the problems step by step. ### QUESTION 1 #### 1.1 Solve for \( p \) and \( q \) using Cramer's Rule Given the equations: \[ \begin{array}{l} \frac{2 p}{3} - 25 q = -5 \quad (1) \\ 2 p + \frac{5 q}{7} - 4 = 0 \quad (2) \end{array} \] First, we need to rewrite these equations in standard form \( Ax + By = C \). 1. **Rearranging Equation (1)**: \[ \frac{2 p}{3} - 25 q = -5 \implies 2p - 75q = -15 \quad \text{(Multiply by 3)} \] 2. **Rearranging Equation (2)**: \[ 2p + \frac{5q}{7} = 4 \implies 14p + 5q = 28 \quad \text{(Multiply by 7)} \] Now we have the system: \[ \begin{array}{l} 2p - 75q = -15 \quad (3) \\ 14p + 5q = 28 \quad (4) \end{array} \] **Cramer's Rule** states that for a system of equations \( Ax = B \), the solution for each variable can be found using determinants. Let: \[ A = \begin{bmatrix} 2 & -75 \\ 14 & 5 \end{bmatrix}, \quad B = \begin{bmatrix} -15 \\ 28 \end{bmatrix} \] **Step 1: Calculate the determinant of \( A \)**: \[ D = \begin{vmatrix} 2 & -75 \\ 14 & 5 \end{vmatrix} = (2)(5) - (14)(-75) \] **Step 2: Calculate \( D \)**: \[ D = 10 + 1050 = 1060 \] **Step 3: Calculate \( D_p \) (replace the first column of \( A \) with \( B \))**: \[ D_p = \begin{vmatrix} -15 & -75 \\ 28 & 5 \end{vmatrix} = (-15)(5) - (28)(-75) \] **Step 4: Calculate \( D_p \)**: \[ D_p = -75 + 2100 = 2025 \] **Step 5: Calculate \( D_q \) (replace the second column of \( A \) with \( B \))**: \[ D_q = \begin{vmatrix} 2 & -15 \\ 14 & 28 \end{vmatrix} = (2)(28) - (14)(-15) \] **Step 6: Calculate \( D_q \)**: \[ D_q = 56 + 210 = 266 \] **Step 7: Solve for \( p \) and \( q \)**: \[ p = \frac{D_p}{D} = \frac{2025}{1060}, \quad q = \frac{D_q}{D} = \frac{266}{1060} \] Now, let's calculate these values: \[ p \approx 1.911 \quad (to \, 3 \, decimal \, places) \] \[ q \approx 0.251 \quad (to \, 3 \, decimal \, places) \] #### 1.2 Determine the minor and cofactor Given the determinant: \[ \left|\begin{array}{ccc} \frac{1}{2} & -3 & 4 \\ \frac{5}{7} & 1 & -5 \\ \frac{7}{8} & 0 & 6 \end{array}\right| \] **1.2.1 Minor of 6**: The minor of an element is the determinant of the matrix that remains after removing the row and column of that element. To find the minor of 6, we remove the third row and third column: \[ \begin{bmatrix} \frac{1}{2} & -3 \\ \frac{5}{7} & 1 \end{bmatrix} \] Calculating the determinant: \[ M_6 = \left|\begin{array}{cc} \frac{1}{2} & -3 \\ \frac{5}{7} & 1 \end{array}\right| = \left(\frac{1}{2} \cdot 1\right) - \left(-3 \cdot \frac{5}{7}\right) = \frac{1}{2} + \frac{15}{7} \] Calculating \( M_6 \): \[ M_6 = \frac{1}{2} + \frac{15}{7} = \frac{7}{14} + \frac{30}{14} = \frac{37}{14} \approx 2.643 \quad (to \, 3 \, decimal \, places) \] **1.2.2 Cofactor of -5**: The cofactor is given by \( C = (-1)^{i+j} M \), where \( i \) and \( j \) are the row and column indices of the element. For -5 (located at row 2, column 3): \[ C_{-5} = (-1)^{2+3} M_{-5} = -M_{-5} \] To find \( M_{-5} \), we remove the second row and third column: \[ \begin{bmatrix} \frac{1}{2} & -3 \\ \frac{7}{8} & 0 \end{bmatrix} \] Calculating the determinant: \[ M_{-5} = \left|\begin{array}{cc} \frac{1}{2} & -3 \\ \frac{7}{8} & 0 \end{array}\right| = \left(\frac{1}{2} \cdot 0\right) - \left(-3 \cdot \frac{7}{8}\right) = \frac{21}{8} \] Thus, \[ C_{-5} = -\frac{21}{8} \approx -2.625 \quad (to \, 3 \, decimal \, places) \] #### 1.3 Solve for \( x \) using Cramer's Rule Given the equations: \[ 3x + 2y = -5z - 6 \quad (5) \] \[ 5x + 7y - 2z = 4 \quad (6) \] \[ 2x - 7z = 5y + 14 \quad (7