Question two: The following Matrix gives the number of units of three product ( \( \mathrm{M}, \mathrm{N} \) and O ), that can be processed per hour on three machines ( \( \mathrm{I}, \mathrm{II} \) and III) \[ \begin{array}{lll}\text { I } & \text { II } & \text { III }\end{array} \] \[ N\left[\begin{array}{lll}10 & 12 & 15 \\ 20 & 11 & 20 \\ 16 & 18 & 14\end{array}\right] \] Determine by using matrix algebra, how many units of each product can be produced, if the hours available on machine I, II and III are 50,40 and 42

To solve this problem, we need to use matrix algebra to determine how many units of each product (M, N, and O) can be produced given the constraints of machine hours. ### Step 1: Define the matrices Let: - \( A \) be the matrix representing the number of units of each product that can be processed per hour on each machine. - \( x \) be the column matrix representing the number of units of each product produced. - \( b \) be the column matrix representing the available hours on each machine. The matrices are defined as follows: \[ A = \begin{bmatrix} 10 & 12 & 15 \\ 20 & 11 & 20 \\ 16 & 18 & 14 \end{bmatrix}, \quad b = \begin{bmatrix} 50 \\ 40 \\ 42 \end{bmatrix} \] ### Step 2: Set up the equation The relationship can be expressed as: \[ A \cdot x = b \] ### Step 3: Solve the system of equations We need to solve for \( x \) using the equation \( A \cdot x = b \). This can be done using matrix algebra. Let's proceed to solve the system of equations. Solve the system of equations by following steps: - step0: Solve using the substitution method: \(\left\{ \begin{array}{l}10x_{1}+12x_{2}+15x_{3}=50\\20x_{1}+11x_{2}+20x_{3}=40\\16x_{1}+18x_{2}+14x_{3}=42\end{array}\right.\) - step1: Solve the equation: \(\left\{ \begin{array}{l}x_{1}=\frac{50-12x_{2}-15x_{3}}{10}\\20x_{1}+11x_{2}+20x_{3}=40\\16x_{1}+18x_{2}+14x_{3}=42\end{array}\right.\) - step2: Substitute the value of \(x_{1}:\) \(\left\{ \begin{array}{l}20\times \frac{50-12x_{2}-15x_{3}}{10}+11x_{2}+20x_{3}=40\\16\times \frac{50-12x_{2}-15x_{3}}{10}+18x_{2}+14x_{3}=42\end{array}\right.\) - step3: Simplify: \(\left\{ \begin{array}{l}100-13x_{2}-10x_{3}=40\\\frac{8\left(50-12x_{2}-15x_{3}\right)}{5}+18x_{2}+14x_{3}=42\end{array}\right.\) - step4: Solve the equation: \(\left\{ \begin{array}{l}x_{2}=\frac{60-10x_{3}}{13}\\\frac{8\left(50-12x_{2}-15x_{3}\right)}{5}+18x_{2}+14x_{3}=42\end{array}\right.\) - step5: Substitute the value of \(x_{2}:\) \(\frac{8\left(50-12\times \frac{60-10x_{3}}{13}-15x_{3}\right)}{5}+18\times \frac{60-10x_{3}}{13}+14x_{3}=42\) - step6: Simplify: \(-\frac{8\left(14+15x_{3}\right)}{13}+\frac{18\left(60-10x_{3}\right)}{13}+14x_{3}=42\) - step7: Multiply both sides of the equation by LCD: \(\left(-\frac{8\left(14+15x_{3}\right)}{13}+\frac{18\left(60-10x_{3}\right)}{13}+14x_{3}\right)\times 13=42\times 13\) - step8: Simplify the equation: \(968-118x_{3}=546\) - step9: Move the constant to the right side: \(-118x_{3}=546-968\) - step10: Subtract the numbers: \(-118x_{3}=-422\) - step11: Change the signs: \(118x_{3}=422\) - step12: Divide both sides: \(\frac{118x_{3}}{118}=\frac{422}{118}\) - step13: Divide the numbers: \(x_{3}=\frac{211}{59}\) - step14: Substitute the value of \(x_{3}:\) \(x_{2}=\frac{60-10\times \frac{211}{59}}{13}\) - step15: Simplify: \(x_{2}=\frac{110}{59}\) - step16: Substitute the values of \(x_{2},x_{3}:\) \(x_{1}=\frac{50-12\times \frac{110}{59}-15\times \frac{211}{59}}{10}\) - step17: Simplify: \(x_{1}=-\frac{307}{118}\) - step18: Calculate: \(\left\{ \begin{array}{l}x_{1}=-\frac{307}{118}\\x_{2}=\frac{110}{59}\\x_{3}=\frac{211}{59}\end{array}\right.\) - step19: Check the solution: \(\left\{ \begin{array}{l}x_{1}=-\frac{307}{118}\\x_{2}=\frac{110}{59}\\x_{3}=\frac{211}{59}\end{array}\right.\) - step20: Rewrite: \(\left(x_{1},x_{2},x_{3}\right) = \left(-\frac{307}{118},\frac{110}{59},\frac{211}{59}\right)\) The solution to the system of equations is: \[ (x_1, x_2, x_3) = \left(-\frac{307}{118}, \frac{110}{59}, \frac{211}{59}\right) \] ### Step 4: Interpret the results - \( x_1 \) represents the number of units of product M. - \( x_2 \) represents the number of units of product N. - \( x_3 \) represents the number of units of product O. Calculating the numerical values: 1. \( x_1 = -\frac{307}{118} \approx -2.60 \) (not feasible, as production cannot be negative) 2. \( x_2 = \frac{110}{59} \approx 1.86 \) 3. \( x_3 = \frac{211}{59} \approx 3.58 \) ### Conclusion The results indicate that the production of product M is not feasible under the given constraints, while products N and O can be produced in approximately 1.86 and 3.58 units, respectively. This suggests that the available hours on the machines are insufficient to produce a non-negative quantity of product M.