Question

1. Use Gassian elimination method to solve the linear system
3x+4y+3z = 3
2x+2y+3z = 1
x+3y+2z = −1

Answer 1

To solve the linear system using Gaussian elimination, we will follow these steps:

1. **Set up the augmented matrix:**
The given equations are:
\[
\begin{align*}
3x + 4y + 3z &= 3 \\
2x + 2y + 3z &= 1 \\
x + 3y + 2z &= -1
\end{align*}
\]
The corresponding augmented matrix is:
\[
\begin{pmatrix}
3 & 4 & 3 & | & 3 \\
2 & 2 & 3 & | & 1 \\
1 & 3 & 2 & | & -1
\end{pmatrix}
\]

2. **Perform row operations to obtain an upper triangular matrix:**

- First, we can simplify the first row by dividing by 3:
\[
R_1 \leftarrow \frac{1}{3}R_1 \implies \begin{pmatrix}
1 & \frac{4}{3} & 1 & | & 1 \\
2 & 2 & 3 & | & 1 \\
1 & 3 & 2 & | & -1
\end{pmatrix}
\]

- Next, we eliminate the first column entries below the first row:
- For \( R_2 \):
\[
R_2 \leftarrow R_2 - 2R_1 \implies 2 - 2(1) = 0, \quad 2 - 2\left(\frac{4}{3}\right) = 2 - \frac{8}{3} = -\frac{2}{3}, \quad 3 - 2(1) = 1, \quad 1 - 2(1) = -1
\]
So, \( R_2 \) becomes:
\[
\begin{pmatrix} 0 & -\frac{2}{3} & 1 & | & -1 \end{pmatrix}
\]

- For \( R_3 \):
\[
R_3 \leftarrow R_3 - R_1 \implies 1-1=0, \quad 3 - \frac{4}{3} = \frac{9}{3} - \frac{4}{3} = \frac{5}{3}, \quad 2 - 1 = 1, \quad -1 - 1 = -2
\]
So, \( R_3 \) becomes:
\[
\begin{pmatrix} 0 & \frac{5}{3} & 1 & | & -2 \end{pmatrix}
\]

The augmented matrix now looks like this:
\[
\begin{pmatrix}
1 & \frac{4}{3} & 1 & | & 1 \\
0 & -\frac{2}{3} & 1 & | & -1 \\
0 & \frac{5}{3} & 1 & | & -2
\end{pmatrix}
\]

- Next, we multiply \( R_2 \) by \(-\frac{3}{2}\) for simplification:
\[
R_2 \leftarrow -\frac{3}{2}R_2 \implies \begin{pmatrix} 0 & 1 & -\frac{3}{2} & | & \frac{3}{2} \end{pmatrix}
\]

The augmented matrix now looks like this:
\[
\begin{pmatrix}
1 & \frac{4}{3} & 1 & | & 1 \\
0 & 1 & -\frac{3}{2} & | & \frac{3}{2} \\
0 & \frac{5}{3} & 1 & | & -2
\end{pmatrix}
\]

- Now we will eliminate the second column from \( R_3 \):
\[
R_3 \leftarrow R_3 - \frac{5}{3}R_2 \implies 0 - \frac{5}{3}(0) = 0, \quad \frac{5}{3} - \frac{5}{3}(1) = 0, \quad 1 - \frac{5}{3}\left(-\frac{3}{2}\right) = 1 + \frac{15}{6} = \frac{21}{6} = \frac{7}{2}, \quad -2 - \frac{5}{3}\left(\frac{3}{2}\right)= -2 - \frac{15}{6} = -2 - \frac{5}{2} = -4.5
\]

This results in:
\[
R_3: \begin{pmatrix} 0 & 0 & \frac{7}{2} & | & -4.5 \end{pmatrix}
\]

The augmented matrix in upper triangular form is now:
\[
\begin{pmatrix}
1 & \frac{4}{3} & 1 & | & 1 \\
0 & 1 & -\frac{3}{2} & | & \frac{3}{2} \\
0 & 0 & \frac{7}{2} & | & -4.5
\end{pmatrix}
\]

3. **Back-substitution:**
- From the third row, we find:
\[
\frac{7}{2} z = -4.5 \implies z = -\frac{4.5 \times 2}{7} = -\frac{9}{7}
\]

- Now substituting \( z = -\frac{9}{7} \) back into the second row:
\[
1y - \frac{3}{2}\left(-\frac{9}{7}\right) = \frac{3}{2} \implies y + \frac{27}{14} = \frac{3}{2}
\]
Multiply \( \frac{3}{2} \) by \( \frac{7}{7}\) to get the same denominator:
\[
y + \frac{27}{14} = \frac{21}{14} \implies y = \frac{21}{14} - \frac{27}{14} = -\frac{6}{14} = -\frac{3}{7}
\]

- Now substituting \( y = -\frac{3}{7} \) and \( z = -\frac{9}{7} \) into the first row:
\[
1x + \frac{4}{3}\left(-\frac{3}{7}\right) + 1\left(-\frac{9}{7}\right) = 1
\]
\[
x - \frac{12}{21} - \frac{9}{7} = 1 \implies x - \frac{12}{21} - \frac{27}{21} = 1
\]
\[
x - \frac{39}{21} = 1 \implies x = 1 + \frac{39}{21} \implies x = 1 + \frac{13}{7} = 1 + 1.857 = 2.857 \approx \frac{7}{7} + \frac{13}{7} = \frac{20}{7}
\]

Finally, the solution to the system is:
\[
\begin{align*}
x &= \frac{20}{7} \\
y &= -\frac{3}{7} \\
z &= -\frac{9}{7}
\end{align*}
\]

Thus, the solution to the system is:
\[
(x, y, z) = \left( \frac{20}{7}, -\frac{3}{7}, -\frac{9}{7} \right)
\]