Questions: Use the Gaussian elimination method with back substitution to solve the given system of linear equations. x-4 y+3 z=-3 x-4 y-15 z=15 -x-3 y-8 z=1 Answer x= y= z=

Solution Steps

To solve the given system of linear equations using Gaussian elimination with back substitution, follow these steps:

1. Form the augmented matrix from the system of equations.
2. Use row operations to convert the matrix into an upper triangular form.
3. Perform back substitution to find the values of \(x\), \(y\), and \(z\).

The given system of linear equations is: \[ \begin{cases} x - 4y + 3z = -3 \\ x - 4y - 15z = 15 \\ -x - 3y - 8z = 1 \end{cases} \] We form the augmented matrix: \[ \begin{bmatrix} 1 & -4 & 3 & -3 \\ 1 & -4 & -15 & 15 \\ -1 & -3 & -8 & 1 \end{bmatrix} \]

We perform row operations to convert the matrix into an upper triangular form.

1. Subtract the first row from the second row: \[ R_2 \leftarrow R_2 - R_1 \] \[ \begin{bmatrix} 1 & -4 & 3 & -3 \\ 0 & 0 & -18 & 18 \\ -1 & -3 & -8 & 1 \end{bmatrix} \]

2. Add the first row to the third row: \[ R_3 \leftarrow R_3 + R_1 \] \[ \begin{bmatrix} 1 & -4 & 3 & -3 \\ 0 & 0 & -18 & 18 \\ 0 & -7 & -5 & -2 \end{bmatrix} \]

3. Swap the second and third rows to avoid division by zero: \[ R_2 \leftrightarrow R_3 \] \[ \begin{bmatrix} 1 & -4 & 3 & -3 \\ 0 & -7 & -5 & -2 \\ 0 & 0 & -18 & 18 \end{bmatrix} \]

4. Divide the second row by \(-7\): \[ R_2 \leftarrow \frac{R_2}{-7} \] \[ \begin{bmatrix} 1 & -4 & 3 & -3 \\ 0 & 1 & \frac{5}{7} & \frac{2}{7} \\ 0 & 0 & -18 & 18 \end{bmatrix} \]

5. Eliminate the second element in the third row: \[ R_3 \leftarrow R_3 - \left(\frac{5}{7}\right)R_2 \] \[ \begin{bmatrix} 1 & -4 & 3 & -3 \\ 0 & 1 & \frac{5}{7} & \frac{2}{7} \\ 0 & 0 & -\frac{99}{7} & \frac{120}{7} \end{bmatrix} \]

6. Divide the third row by \(-\frac{99}{7}\): \[ R_3 \leftarrow \frac{R_3}{-\frac{99}{7}} \] \[ \begin{bmatrix} 1 & -4 & 3 & -3 \\ 0 & 1 & \frac{5}{7} & \frac{2}{7} \\ 0 & 0 & 1 & -\frac{120}{99} \end{bmatrix} \]

We perform back substitution to find the values of \(x\), \(y\), and \(z\).

1. From the third row: \[ z = -\frac{120}{99} = -\frac{40}{33} \]

2. Substitute \(z\) into the second row: \[ y + \frac{5}{7}z = \frac{2}{7} \] \[ y + \frac{5}{7}\left(-\frac{40}{33}\right) = \frac{2}{7} \] \[ y - \frac{200}{231} = \frac{2}{7} \] \[ y = \frac{2}{7} + \frac{200}{231} = \frac{66}{231} + \frac{200}{231} = \frac{266}{231} = \frac{38}{33} \]

3. Substitute \(y\) and \(z\) into the first row: \[ x - 4y + 3z = -3 \] \[ x - 4\left(\frac{38}{33}\right) + 3\left(-\frac{40}{33}\right) = -3 \] \[ x - \frac{152}{33} - \frac{120}{33} = -3 \] \[ x - \frac{272}{33} = -3 \] \[ x = -3 + \frac{272}{33} = -\frac{99}{33} + \frac{272}{33} = \frac{173}{33} \]

Final Answer