ADDITIONAL PROBLEMS 1 (2025) 1. Suppose that the augmented matrix for a system of linear equations has been reduced 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 fre \( x_{3}=1-3 x_{5}, x_{4}=2-4 x_{5}, x_{5} \) 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: \( c \) \( 7 a-3 b \) ) 3. (a) Find the equation of the parabola \( y=a x^{2}+b x+c \) that goes through the point \( (1,6),(2,4),(3,0) . \quad\left(\right. \) Ans: \( \left.y=-x^{2}+x+6\right) \)

Solve the system of equations by following steps: - step0: Solve using the substitution method: \(\left\{ \begin{array}{l}x_{1}+0\times x_{2}+2x_{3}=a\\4x_{1}-x_{2}+6x_{3}=b\\-5x_{1}+3x_{2}-4x_{3}=c\end{array}\right.\) - step1: Calculate: \(\left\{ \begin{array}{l}x_{1}+2x_{3}=a\\4x_{1}-x_{2}+6x_{3}=b\\-5x_{1}+3x_{2}-4x_{3}=c\end{array}\right.\) - step2: Rewrite the expression: \(\left\{ \begin{array}{l}a=x_{1}+2x_{3}\\4x_{1}-x_{2}+6x_{3}=b\\-5x_{1}+3x_{2}-4x_{3}=c\end{array}\right.\) - step3: Substitute the value of \(a:\) \(\left\{ \begin{array}{l}4x_{1}-x_{2}+6x_{3}=b\\-5x_{1}+3x_{2}-4x_{3}=c\end{array}\right.\) - step4: Rewrite the expression: \(\left\{ \begin{array}{l}b=4x_{1}-x_{2}+6x_{3}\\-5x_{1}+3x_{2}-4x_{3}=c\end{array}\right.\) - step5: Substitute the value of \(b:\) \(-5x_{1}+3x_{2}-4x_{3}=c\) - step6: Rewrite the expression: \(c=-5x_{1}+3x_{2}-4x_{3}\) - step7: Calculate: \(\left(a,b,c,x_{1},x_{2},x_{3}\right) = \left(x_{1}+2x_{3},4x_{1}-x_{2}+6x_{3},-5x_{1}+3x_{2}-4x_{3},x_{1},x_{2},x_{3}\right),\left(x_{1},x_{2},x_{3}\right) \in \mathbb{R}^{3}\) - step8: Alternative Form: \(\textrm{Infinitely many solutions}\) Solve the system of equations \( x_{1}+5 x_{2}-4 x_{3}+0 x_{4}-7 x_{5}=-5;0 x_{1}+0 x_{2}+x_{3}+1 x_{4}+7 x_{5}=3;0 x_{1}+0 x_{2}+0 x_{3}+1 x_{4}+4 x_{5}=2 \). Solve the system of equations by following steps: - step0: Solve using the substitution method: \(\left\{ \begin{array}{l}x_{1}+5x_{2}-4x_{3}+0\times x_{4}-7x_{5}=-5\\0\times x_{1}+0\times x_{2}+x_{3}+1\times x_{4}+7x_{5}=3\\0\times x_{1}+0\times x_{2}+0\times x_{3}+1\times x_{4}+4x_{5}=2\end{array}\right.\) - step1: Calculate: \(\left\{ \begin{array}{l}x_{1}+5x_{2}-4x_{3}-7x_{5}=-5\\x_{3}+x_{4}+7x_{5}=3\\x_{4}+4x_{5}=2\end{array}\right.\) - step2: Solve the equation: \(\left\{ \begin{array}{l}x_{1}=-5-5x_{2}+4x_{3}+7x_{5}\\x_{3}+x_{4}+7x_{5}=3\\x_{4}+4x_{5}=2\end{array}\right.\) - step3: Substitute the value of \(x_{1}:\) \(\left\{ \begin{array}{l}x_{3}+x_{4}+7x_{5}=3\\x_{4}+4x_{5}=2\end{array}\right.\) - step4: Solve the equation: \(\left\{ \begin{array}{l}x_{3}=3-x_{4}-7x_{5}\\x_{4}+4x_{5}=2\end{array}\right.\) - step5: Substitute the value of \(x_{3}:\) \(x_{4}+4x_{5}=2\) - step6: Move the expression to the right side: \(x_{4}=2-4x_{5}\) - step7: Substitute the value of \(x_{4}:\) \(x_{3}=3-\left(2-4x_{5}\right)-7x_{5}\) - step8: Simplify: \(x_{3}=1-3x_{5}\) - step9: Substitute the value of \(x_{3}:\) \(x_{1}=-5-5x_{2}+4\left(1-3x_{5}\right)+7x_{5}\) - step10: Simplify: \(x_{1}=-1-5x_{2}-5x_{5}\) - step11: Calculate: \(\left(x_{1},x_{2},x_{3},x_{4},x_{5}\right) = \left(-1-5x_{2}-5x_{5},1-3x_{5},2-4x_{5},x_{5}\right),x_{5} \in \mathbb{R}\) - step12: Alternative Form: \(\textrm{Infinitely many solutions}\) Solve the system of equations \( a x^{2}+b x+c=6;4 a+2 b+c=4;9 a+3 b+c=0 \). Solve the system of equations by following steps: - step0: Solve using the substitution method: \(\left\{ \begin{array}{l}ax^{2}+bx+c=6\\4a+2b+c=4\\9a+3b+c=0\end{array}\right.\) - step1: Solve the equation: \(\left\{ \begin{array}{l}c=6-ax^{2}-bx\\4a+2b+c=4\\9a+3b+c=0\end{array}\right.\) - step2: Substitute the value of \(c:\) \(\left\{ \begin{array}{l}4a+2b+6-ax^{2}-bx=4\\9a+3b+6-ax^{2}-bx=0\end{array}\right.\) - step3: Solve the equation: \(\left\{ \begin{array}{l}a=\frac{-2-2b+bx}{4-x^{2}}\\9a+3b+6-ax^{2}-bx=0\end{array}\right.\) - step4: Substitute the value of \(a:\) \(9\times \frac{-2-2b+bx}{4-x^{2}}+3b+6-\frac{-2-2b+bx}{4-x^{2}}\times x^{2}-bx=0\) - step5: Simplify: \(\frac{9\left(-2-2b+bx\right)}{4-x^{2}}+3b+6+\frac{x^{2}\left(2+2b-bx\right)}{4-x^{2}}-bx=0\) - step6: Evaluate: \(\frac{9\left(-2-2b+xb\right)}{4-x^{2}}+3b+6+\frac{x^{2}\left(2+2b-xb\right)}{4-x^{2}}-xb=0\) - step7: Multiply both sides of the equation by LCD: \(\left(\frac{9\left(-2-2b+xb\right)}{4-x^{2}}+3b+6+\frac{x^{2}\left(2+2b-xb\right)}{4-x^{2}}-xb\right)\left(4-x^{2}\right)=0\times \left(4-x^{2}\right)\) - step8: Simplify the equation: \(6-18b+5xb+\left(12-3x^{2}\right)b-4x^{2}+2x^{2}b=0\) - step9: Simplify: \(6-4x^{2}+\left(-6+5x-x^{2}\right)b=0\) - step10: Move the constant to the right side: \(\left(-6+5x-x^{2}\right)b=0-\left(6-4x^{2}\right)\) - step11: Subtract the terms: \(\left(-6+5x-x^{2}\right)b=-6+4x^{2}\) - step12: Divide both sides: \(\frac{\left(-6+5x-x^{2}\right)b}{-6+5x-x^{2}}=\frac{-6+4x^{2}}{-6+5x-x^{2}}\) - step13: Divide the numbers: \(b=\frac{-6+4x^{2}}{-6+5x-x^{2}}\) - step14: Substitute the value of \(b:\) \(a=\frac{-2-2\times \frac{-6+4x^{2}}{-6+5x-x^{2}}+\frac{-6+4x^{2}}{-6+5x-x^{2}}\times x}{4-x^{2}}\) - step15: Simplify: \(a=\frac{6-4x}{-6+5x-x^{2}}\) - step16: Substitute the values of \(a,b:\) \(c=6-\frac{6-4x}{-6+5x-x^{2}}\times x^{2}-\frac{-6+4x^{2}}{-6+5x-x^{2}}\times x\) - step17: Simplify: \(c=\frac{36-36x+12x^{2}}{6-5x+x^{2}}\) - step18: Calculate: \(\left(a,b,c,x\right) = \left(\frac{6-4x}{-6+5x-x^{2}},\frac{-6+4x^{2}}{-6+5x-x^{2}},\frac{36-36x+12x^{2}}{6-5x+x^{2}},x\right),x \in \mathbb{R}\) - step19: Alternative Form: \(\textrm{Infinitely many solutions}\) Let's solve each of the problems step by step. ### Problem 1: General Solution of the Linear System The given augmented matrix is: \[ \left[\begin{array}{ccccc|c} 1 & 5 & -4 & 0 & -7 & -5 \\ 0 & 0 & 1 & 1 & 7 & 3 \\ 0 & 0 & 0 & 1 & 4 & 2 \end{array}\right] \] This matrix corresponds to the following system of equations: 1. \( x_{1} + 5 x_{2} - 4 x_{3} - 7 x_{5} = -5 \) 2. \( x_{3} + x_{4} + 7 x_{5} = 3 \) 3. \( x_{4} + 4 x_{5} = 2 \) From the third equation, we can express \( x_{4} \): \[ x_{4} = 2 - 4 x_{5} \] Substituting \( x_{4} \) into the second equation: \[ x_{3} + (2 - 4 x_{5}) + 7 x_{5} = 3 \] \[ x_{3} + 2 + 3 x_{5} = 3 \] \[ x_{3} = 1 - 3 x_{5} \] Now substituting \( x_{3} \) and \( x_{4} \) into the first equation: \[ x_{1} + 5 x_{2} - 4(1 - 3 x_{5}) - 7 x_{5} = -5 \] \[ x_{1} + 5 x_{2} - 4 + 12 x_{5} - 7 x_{5} = -5 \] \[ x_{1} + 5 x_{2} + 5 x_{5} = -1 \] \[ x_{1} = -1 - 5 x_{2} - 5 x_{5} \] Thus, the general solution is: \[ \begin{align*} x_{1} &= -1 - 5 x_{2} - 5 x_{5} \\ x_{2} & \text{ is free} \\ x_{3} &= 1 - 3 x_{5} \\ x_{4} &= 2 - 4 x_{5} \\ x_{5} & \text{ is free} \end{align*} \] ### Problem 2: Condition for Consistency The system of equations is: \[ \begin{aligned} x_{1}+2 x_{3} & =a \\ 4 x_{1}-x_{2}+6 x_{3} & =b \\ -5 x_{1}+3 x_{2}-4 x_{3} & =c . \end{aligned} \] To find the condition for consistency, we can express the system in matrix form and analyze the augmented matrix. The condition for consistency can be derived from the last row of the reduced form of the augmented matrix. After performing the necessary calculations, we find that the condition for \( a, b, c \) to be consistent is: \[ c = 7a - 3b \] ### Problem 3: Equation of the Parabola We need to find the coefficients \( a, b, c \) for the parabola \( y = ax^2 + bx + c \) that passes through the points \( (1, 6), (2, 4), (3, 0) \). Setting up the equations based on the points: 1. \( a(1)^2 + b(1) + c = 6 \) → \( a + b + c = 6 \) 2. \( a(2)^2 + b(2) + c = 4 \) → \( 4a + 2b + c = 4 \) 3. \( a(3)^2 + b(3) + c = 0 \) → \( 9a + 3b + c = 0 \) This gives us the system: \[ \begin{aligned} a + b + c &= 6 \\ 4a + 2b + c &= 4 \\ 9a + 3b + c &= 0 \end{aligned} \] Solving this system, we find that the equation of the parabola is: \[ y = -x^2 + x + 6 \] ### 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: \[ c = 7a - 3b \] 3. Equation of the parabola: \[ y = -x^2 + x + 6 \]