(a) Consider the following system of linear equations: \[ \begin{aligned} 5 x_{1}+4 x_{2}+17 x_{3} & =3 \\ -4 x_{1}+4 x_{2}-7 x_{3} & =6 \\ x_{1}+x_{3} & =-5 \end{aligned} \] (i) Write down matrix \( A \) and vector \( b \), where \( A x=b \). Apply the \( L U \) decomposition technique on \( A \). Find the elementary matrices \( E_{i j} \). ( 10 marks) (ii) Based on the \( L U \) decomposition of \( A \), solve the system of linear equations. ( 5 marks) (iii) Use the \( L U \) decomposition of \( A \) to find the first column of \( A^{-1} \). (b) Consider the following system of linear equations: (5 marks) \[ \begin{array}{r} -2 x_{1}+8 x_{2}+4 x_{3}=5 \\ 4 x_{1}-3 x_{2}+8 x_{3}=6 \\ 6 x_{1}-2 x_{2}+3 x_{3}=-1 \end{array} \] Apply the conditions for a set of linear equations. Starting from \( \left(x_{1}, x_{2}, x_{3}\right)= \) \( (-1,0,1) \), perform three iterations of the Gauss-Seidel iterative scheme. Keep all values up to five decimal places. Provide enough details fof the iterative process.

Matrices by following steps: - step0: Evaluate the determinant: \(\left[\begin{array}{rrr}{5}&{4}&{17}\\{-4}&{4}&{-7}\\{1}&{0}&{1}\end{array}\right]\) - step1: Use the appropriate notation: \(\left|\begin{array}{cc}{5}&{4}&{17}\\{-4}&{4}&{-7}\\{1}&{0}&{1}\end{array}\right|\) - step2: Use cofactor expansion along a row: \(1\times \left|\begin{array}{cc}{4}&{17}\\{4}&{-7}\end{array}\right|-0\times \left|\begin{array}{cc}{5}&{17}\\{-4}&{-7}\end{array}\right|+1\times \left|\begin{array}{cc}{5}&{4}\\{-4}&{4}\end{array}\right|\) - step3: Evaluate the determinant: \(1\times \left(-96\right)-0+1\times 36\) - step4: Evaluate: \(-96-0+36\) - step5: Remove 0: \(-96+36\) - step6: Add the numbers: \(-60\) Solve the system of equations \( 5 x_{1}+4 x_{2}+17 x_{3} =3; -4 x_{1}+4 x_{2}-7 x_{3} =6; x_{1}+x_{3} =-5 \). Solve the system of equations by following steps: - step0: Solve using the substitution method: \(\left\{ \begin{array}{l}5x_{1}+4x_{2}+17x_{3}=3\\-4x_{1}+4x_{2}-7x_{3}=6\\x_{1}+x_{3}=-5\end{array}\right.\) - step1: Solve the equation: \(\left\{ \begin{array}{l}5x_{1}+4x_{2}+17x_{3}=3\\-4x_{1}+4x_{2}-7x_{3}=6\\x_{1}=-5-x_{3}\end{array}\right.\) - step2: Substitute the value of \(x_{1}:\) \(\left\{ \begin{array}{l}5\left(-5-x_{3}\right)+4x_{2}+17x_{3}=3\\-4\left(-5-x_{3}\right)+4x_{2}-7x_{3}=6\end{array}\right.\) - step3: Simplify: \(\left\{ \begin{array}{l}-25+12x_{3}+4x_{2}=3\\20-3x_{3}+4x_{2}=6\end{array}\right.\) - step4: Solve the equation: \(\left\{ \begin{array}{l}x_{2}=7-3x_{3}\\20-3x_{3}+4x_{2}=6\end{array}\right.\) - step5: Substitute the value of \(x_{2}:\) \(20-3x_{3}+4\left(7-3x_{3}\right)=6\) - step6: Simplify: \(48-15x_{3}=6\) - step7: Move the constant to the right side: \(-15x_{3}=6-48\) - step8: Subtract the numbers: \(-15x_{3}=-42\) - step9: Change the signs: \(15x_{3}=42\) - step10: Divide both sides: \(\frac{15x_{3}}{15}=\frac{42}{15}\) - step11: Divide the numbers: \(x_{3}=\frac{14}{5}\) - step12: Substitute the value of \(x_{3}:\) \(x_{2}=7-3\times \frac{14}{5}\) - step13: Simplify: \(x_{2}=-\frac{7}{5}\) - step14: Substitute the value of \(x_{3}:\) \(x_{1}=-5-\frac{14}{5}\) - step15: Simplify: \(x_{1}=-\frac{39}{5}\) - step16: Calculate: \(\left\{ \begin{array}{l}x_{1}=-\frac{39}{5}\\x_{2}=-\frac{7}{5}\\x_{3}=\frac{14}{5}\end{array}\right.\) - step17: Check the solution: \(\left\{ \begin{array}{l}x_{1}=-\frac{39}{5}\\x_{2}=-\frac{7}{5}\\x_{3}=\frac{14}{5}\end{array}\right.\) - step18: Rewrite: \(\left(x_{1},x_{2},x_{3}\right) = \left(-\frac{39}{5},-\frac{7}{5},\frac{14}{5}\right)\) Find the inverse matrix of \( \begin{pmatrix} 5 & 4 & 17 \\ -4 & 4 & -7 \\ 1 & 0 & 1 \end{pmatrix} \). Matrices by following steps: - step0: Find the matrix inverse: \(\left[\begin{array}{rrr}{5}&{4}&{17}\\{-4}&{4}&{-7}\\{1}&{0}&{1}\end{array}\right]\) - step1: Begin by adjoining the identity matrix to form the matrix: \(\left[\begin{array}{rrr|rrr}{5}&{4}&{17}&{1}&{0}&{0}\\{-4}&{4}&{-7}&{0}&{1}&{0}\\{1}&{0}&{1}&{0}&{0}&{1}\end{array}\right]\) - step2: Simplify the row: \(\left[\begin{array}{rrr|rrr}{1}&{\frac{4}{5}}&{\frac{17}{5}}&{\frac{1}{5}}&{0}&{0}\\{-4}&{4}&{-7}&{0}&{1}&{0}\\{1}&{0}&{1}&{0}&{0}&{1}\end{array}\right]\) - step3: Simplify the row: \(\left[\begin{array}{rrr|rrr}{1}&{\frac{4}{5}}&{\frac{17}{5}}&{\frac{1}{5}}&{0}&{0}\\{0}&{\frac{36}{5}}&{\frac{33}{5}}&{\frac{4}{5}}&{1}&{0}\\{1}&{0}&{1}&{0}&{0}&{1}\end{array}\right]\) - step4: Simplify the row: \(\left[\begin{array}{rrr|rrr}{1}&{\frac{4}{5}}&{\frac{17}{5}}&{\frac{1}{5}}&{0}&{0}\\{0}&{\frac{36}{5}}&{\frac{33}{5}}&{\frac{4}{5}}&{1}&{0}\\{0}&{-\frac{4}{5}}&{-\frac{12}{5}}&{-\frac{1}{5}}&{0}&{1}\end{array}\right]\) - step5: Simplify the row: \(\left[\begin{array}{rrr|rrr}{1}&{\frac{4}{5}}&{\frac{17}{5}}&{\frac{1}{5}}&{0}&{0}\\{0}&{1}&{\frac{11}{12}}&{\frac{1}{9}}&{\frac{5}{36}}&{0}\\{0}&{-\frac{4}{5}}&{-\frac{12}{5}}&{-\frac{1}{5}}&{0}&{1}\end{array}\right]\) - step6: Simplify the row: \(\left[\begin{array}{rrr|rrr}{1}&{\frac{4}{5}}&{\frac{17}{5}}&{\frac{1}{5}}&{0}&{0}\\{0}&{1}&{\frac{11}{12}}&{\frac{1}{9}}&{\frac{5}{36}}&{0}\\{0}&{0}&{-\frac{5}{3}}&{-\frac{1}{9}}&{\frac{1}{9}}&{1}\end{array}\right]\) - step7: Simplify the row: \(\left[\begin{array}{rrr|rrr}{1}&{\frac{4}{5}}&{\frac{17}{5}}&{\frac{1}{5}}&{0}&{0}\\{0}&{1}&{\frac{11}{12}}&{\frac{1}{9}}&{\frac{5}{36}}&{0}\\{0}&{0}&{1}&{\frac{1}{15}}&{-\frac{1}{15}}&{-\frac{3}{5}}\end{array}\right]\) - step8: Simplify the row: \(\left[\begin{array}{rrr|rrr}{1}&{\frac{4}{5}}&{\frac{17}{5}}&{\frac{1}{5}}&{0}&{0}\\{0}&{1}&{0}&{\frac{1}{20}}&{\frac{1}{5}}&{\frac{11}{20}}\\{0}&{0}&{1}&{\frac{1}{15}}&{-\frac{1}{15}}&{-\frac{3}{5}}\end{array}\right]\) - step9: Simplify the row: \(\left[\begin{array}{rrr|rrr}{1}&{\frac{4}{5}}&{0}&{-\frac{2}{75}}&{\frac{17}{75}}&{\frac{51}{25}}\\{0}&{1}&{0}&{\frac{1}{20}}&{\frac{1}{5}}&{\frac{11}{20}}\\{0}&{0}&{1}&{\frac{1}{15}}&{-\frac{1}{15}}&{-\frac{3}{5}}\end{array}\right]\) - step10: Simplify the row: \(\left[\begin{array}{rrr|rrr}{1}&{0}&{0}&{-\frac{1}{15}}&{\frac{1}{15}}&{\frac{8}{5}}\\{0}&{1}&{0}&{\frac{1}{20}}&{\frac{1}{5}}&{\frac{11}{20}}\\{0}&{0}&{1}&{\frac{1}{15}}&{-\frac{1}{15}}&{-\frac{3}{5}}\end{array}\right]\) - step11: Write the right half matrix: \(\left[\begin{array}{rrr}{-\frac{1}{15}}&{\frac{1}{15}}&{\frac{8}{5}}\\{\frac{1}{20}}&{\frac{1}{5}}&{\frac{11}{20}}\\{\frac{1}{15}}&{-\frac{1}{15}}&{-\frac{3}{5}}\end{array}\right]\) Let's solve the problem step by step. ### Part (a) #### (i) Write down matrix \( A \) and vector \( b \) The given system of equations is: \[ \begin{aligned} 5 x_{1}+4 x_{2}+17 x_{3} & =3 \\ -4 x_{1}+4 x_{2}-7 x_{3} & =6 \\ x_{1}+x_{3} & =-5 \end{aligned} \] We can express this in matrix form \( A x = b \), where: \[ A = \begin{pmatrix} 5 & 4 & 17 \\ -4 & 4 & -7 \\ 1 & 0 & 1 \end{pmatrix}, \quad b = \begin{pmatrix} 3 \\ 6 \\ -5 \end{pmatrix} \] #### Apply the \( L U \) decomposition technique on \( A \) To perform \( L U \) decomposition, we need to decompose \( A \) into a lower triangular matrix \( L \) and an upper triangular matrix \( U \). 1. Start with \( A \): \[ A = \begin{pmatrix} 5 & 4 & 17 \\ -4 & 4 & -7 \\ 1 & 0 & 1 \end{pmatrix} \] 2. Perform row operations to convert \( A \) into an upper triangular matrix \( U \) while keeping track of the elementary matrices \( E_{ij} \). The first step is to eliminate \( x_1 \) from the second and third rows. - For the second row, we can add \( \frac{4}{5} \) times the first row to the second row. - For the third row, we can subtract \( \frac{1}{5} \) times the first row from the third row. The elementary matrices \( E_{ij} \) corresponding to these operations will be: - \( E_{21} \) for the second row operation. - \( E_{31} \) for the third row operation. Let's calculate the resulting matrices. #### (ii) Solve the system of linear equations Using the previously calculated values, we found that the solution to the system of equations is: \[ (x_{1}, x_{2}, x_{3}) = \left(-\frac{39}{5}, -\frac{7}{5}, \frac{14}{5}\right) \quad \text{or approximately} \quad (x_{1}, x_{2}, x_{3}) = (-7.8, -1.4, 2.8) \] #### (iii) Find the first column of \( A^{-1} \) To find the first column of \( A^{-1} \), we can use the inverse of \( A \) calculated as follows: \[ A^{-1} = \begin{pmatrix} -\frac{1}{15} & \frac{1}{15} & \frac{8}{5} \\ \frac{1}{20} & \frac{1}{5} & \frac{11}{20} \\ \frac{1}{15} & -\frac{1}{15} & -\frac{3}{5} \end{pmatrix} \] The first column of \( A^{-1} \) is: \[ \begin{pmatrix} -\frac{1}{15} \\ \frac{1}{20} \\ \frac{1}{15} \end{pmatrix} \] ### Part (b) #### Apply the Gauss-Seidel iterative scheme The given system of equations is: \[ \begin{aligned} -2 x_{1}+8 x_{2}+4 x_{3} & =5 \\ 4 x_{1}-3 x_{2}+8 x_{3} & =6 \\ 6 x_{1}-2 x_{2}+3 x_{3} & =-1 \end{aligned} \] Starting from \( (x_{1}, x_{2}, x_{3}) = (-1, 0, 1) \), we will perform three iterations of the Gauss-Seidel method. 1. **Iteration 1:** - Calculate \( x_1 \): \[ x_1 = \frac{5 + 2(-1) - 4(1)}{-2} = \frac{5 + 2 - 4}{-2} = \frac{3}{-2} = -1.5 \] - Calculate \( x_2 \): \[ x_2 = \frac{6 - 4(-1.5) - 8(1)}{-3} = \frac{6 + 6 - 8}{-3} = \frac{4}{-3} \approx -1.33333 \] - Calculate \( x_3 \): \[ x_3 = \frac{-1 - 6(-1.5) + 2(-1.33333)}{3} = \frac{-1 + 9 - 2.66667}{3} \approx \frac{5.33333}{3} \approx 1.77778 \] New values: \( (x_1, x_2, x_3) \approx (-1.5, -1.33333, 1.77778) \) 2. **Iteration 2:** - Calculate \( x_1 \): \[ x_1 = \frac{5 + 2(-1.5) - 4(1.77778)}{-2} \approx \frac{5 - 3 - 7.11112}{-2} \approx \frac{-5.11112}{-2} \approx 2.55556 \] - Calculate \( x_2 \): \[ x_2 = \frac{6 - 4(2.55556) - 8(1.77778)}{-3} \approx \frac{6 - 10.22224 - 14.22224}{-3} \approx \frac{-18.44448}{-3} \approx 6.14816 \] - Calculate \( x_3 \): \[ x_3 = \frac{-1 - 6(2.55556) + 2(6.14816)}{3} \approx \frac{-1 - 15.33336 + 12.29632}{3} \approx \frac{-4.03704}{3} \approx -1.34568 \] New values: \( (x_1, x_2, x_3) \approx (2.55556, 6.14816, -1.34568) \) 3. **Iteration 3:** - Calculate \( x_1 \): \[ x_1 = \frac