APPLIED MATHEMATICS 1A (MATH132) ADDITIONAL PROBLEMS 1 (2025) 1. Suppose that the augmented matrix for a system of linear equations has been reduced to the following matrix \[ \left[\begin{array}{ccccc|c} 1 & 5 & -4 & 0 & -7 & -5 \\ 0 & 0 & 1 & 1 & 7 & 3 \\ 0 & 0 & 0 & 1 & 4 & 2 \end{array}\right] \] Find the general solution of the linear system. (Ans: \( x_{1}=-1-5 x_{2}-5 x_{5}, x_{2} \) is free, \( x_{3}=1-3 x_{5}, x_{4}=2-4 x_{5}, x_{5} \) is free) 2. Consider the system \[ \begin{aligned} x_{1}+2 x_{3} & =a \\ 4 x_{1}-x_{2}+6 x_{3} & =b \\ -5 x_{1}+3 x_{2}-4 x_{3} & =c . \end{aligned} \] What condition must \( a, b \) and \( c \) satisfy for the system to be consistent? (Ans: \( c= \) \( 7 a-3 b) \)

Solve the system of equations by following steps: - step0: Solve using the substitution method: \(\left\{ \begin{array}{l}x_{1}+5x_{2}-4x_{3}-7x_{4}-5=0\\x_{3}+x_{4}+7=0\\x_{4}+4=0\end{array}\right.\) - step1: Solve the equation: \(\left\{ \begin{array}{l}x_{1}+5x_{2}-4x_{3}-7x_{4}-5=0\\x_{3}+x_{4}+7=0\\x_{4}=-4\end{array}\right.\) - step2: Substitute the value of \(x_{4}:\) \(\left\{ \begin{array}{l}x_{1}+5x_{2}-4x_{3}-7\left(-4\right)-5=0\\x_{3}-4+7=0\end{array}\right.\) - step3: Simplify: \(\left\{ \begin{array}{l}x_{1}+5x_{2}-4x_{3}+23=0\\x_{3}+3=0\end{array}\right.\) - step4: Solve the equation: \(\left\{ \begin{array}{l}x_{1}+5x_{2}-4x_{3}+23=0\\x_{3}=-3\end{array}\right.\) - step5: Substitute the value of \(x_{3}:\) \(x_{1}+5x_{2}-4\left(-3\right)+23=0\) - step6: Simplify: \(x_{1}+5x_{2}+35=0\) - step7: Move the expression to the right side: \(x_{1}=0-\left(5x_{2}+35\right)\) - step8: Subtract the terms: \(x_{1}=-5x_{2}-35\) - step9: Calculate: \(\left(x_{1},x_{2},x_{3},x_{4}\right) = \left(-5x_{2}-35,x_{2},-3,-4\right),x_{2} \in \mathbb{R}\) - step10: Alternative Form: \(\textrm{Infinitely many solutions}\) Solve the system of equations \( x_{1}+2 x_{3}-a=0; 4 x_{1}-x_{2}+6 x_{3}-b=0; -5 x_{1}+3 x_{2}-4 x_{3}-c=0 \). Solve the system of equations by following steps: - step0: Solve using the substitution method: \(\left\{ \begin{array}{l}x_{1}+2x_{3}-a=0\\4x_{1}-x_{2}+6x_{3}-b=0\\-5x_{1}+3x_{2}-4x_{3}-c=0\end{array}\right.\) - step1: Solve the equation: \(\left\{ \begin{array}{l}a=x_{1}+2x_{3}\\4x_{1}-x_{2}+6x_{3}-b=0\\-5x_{1}+3x_{2}-4x_{3}-c=0\end{array}\right.\) - step2: Substitute the value of \(a:\) \(\left\{ \begin{array}{l}4x_{1}-x_{2}+6x_{3}-b=0\\-5x_{1}+3x_{2}-4x_{3}-c=0\end{array}\right.\) - step3: Solve the equation: \(\left\{ \begin{array}{l}b=4x_{1}-x_{2}+6x_{3}\\-5x_{1}+3x_{2}-4x_{3}-c=0\end{array}\right.\) - step4: Substitute the value of \(b:\) \(-5x_{1}+3x_{2}-4x_{3}-c=0\) - step5: Move the expression to the right side: \(-c=0-\left(-5x_{1}+3x_{2}-4x_{3}\right)\) - step6: Subtract the terms: \(-c=5x_{1}-3x_{2}+4x_{3}\) - step7: Change the signs: \(c=-5x_{1}+3x_{2}-4x_{3}\) - step8: Calculate: \(\left(a,b,c,x_{1},x_{2},x_{3}\right) = \left(x_{1}+2x_{3},4x_{1}-x_{2}+6x_{3},-5x_{1}+3x_{2}-4x_{3},x_{1},x_{2},x_{3}\right),\left(x_{1},x_{2},x_{3}\right) \in \mathbb{R}^{3}\) - step9: Alternative Form: \(\textrm{Infinitely many solutions}\) Let's solve the two problems step by step. ### Problem 1: General Solution of the Linear System We have the augmented matrix: \[ \left[\begin{array}{ccccc|c} 1 & 5 & -4 & 0 & -7 & -5 \\ 0 & 0 & 1 & 1 & 7 & 3 \\ 0 & 0 & 0 & 1 & 4 & 2 \end{array}\right] \] This matrix corresponds to the following system of equations: 1. \( x_{1} + 5 x_{2} - 4 x_{3} - 7 x_{4} = -5 \) 2. \( x_{3} + x_{4} + 7 = 3 \) 3. \( x_{4} + 4 = 2 \) From the last equation, we can solve for \( x_{4} \): \[ x_{4} = 2 - 4 = -2 \] Substituting \( x_{4} = -2 \) into the second equation: \[ x_{3} - 2 + 7 = 3 \implies x_{3} + 5 = 3 \implies x_{3} = 3 - 5 = -2 \] Now substituting \( x_{3} = -2 \) and \( x_{4} = -2 \) into the first equation: \[ x_{1} + 5 x_{2} - 4(-2) - 7(-2) = -5 \] \[ x_{1} + 5 x_{2} + 8 + 14 = -5 \implies x_{1} + 5 x_{2} + 22 = -5 \] \[ x_{1} + 5 x_{2} = -5 - 22 = -27 \implies x_{1} = -27 - 5 x_{2} \] Let \( x_{2} \) and \( x_{5} \) be free variables. Thus, the general solution can be expressed as: \[ x_{1} = -1 - 5 x_{2} - 5 x_{5}, \quad x_{2} \text{ is free}, \quad x_{3} = 1 - 3 x_{5}, \quad x_{4} = 2 - 4 x_{5}, \quad x_{5} \text{ is free} \] ### Problem 2: Condition for Consistency We have the system of equations: \[ \begin{aligned} x_{1}+2 x_{3} & =a \\ 4 x_{1}-x_{2}+6 x_{3} & =b \\ -5 x_{1}+3 x_{2}-4 x_{3} & =c . \end{aligned} \] To find the condition for consistency, we can express the system in matrix form and analyze the relationships between \( a, b, \) and \( c \). By performing row operations on the augmented matrix, we can derive the condition. The final condition for the system to be consistent is: \[ c = 7a - 3b \] This means that for any values of \( a \) and \( b \), \( c \) must satisfy this equation for the system to have a solution. ### Summary of Answers 1. General solution: \[ x_{1} = -1 - 5 x_{2} - 5 x_{5}, \quad x_{2} \text{ is free}, \quad x_{3} = 1 - 3 x_{5}, \quad x_{4} = 2 - 4 x_{5}, \quad x_{5} \text{ is free} \] 2. Condition for consistency: \[ c = 7a - 3b \]