Sure, let's solve the system of equations using the Gaussian elimination method step-by-step. The system of equations is:
[tex]\[
\begin{array}{l}
x - y + z = 4 \\
5x - y + z = 6 \\
3x - y + 5z = 5
\end{array}
\][/tex]
1. Form the augmented matrix:
[tex]\[
\left[\begin{array}{ccc|c}
1 & -1 & 1 & 4 \\
5 & -1 & 1 & 6 \\
3 & -1 & 5 & 5
\end{array}\right]
\][/tex]
2. Eliminate the first element of the second and third rows:
- For the second row, we subtract [tex]\( 5 \)[/tex] times the first row from the second row:
[tex]\[
\left[\begin{array}{ccc|c}
1 & -1 & 1 & 4 \\
5 & -1 & 1 & 6 \\
3 & -1 & 5 & 5
\end{array}\right]
\][/tex]
[tex]\[
\left[\begin{array}{ccc|c}
1 & -1 & 1 & 4 \\
5 - 5 \cdot 1 & -1 - 5 \cdot (-1) & 1 - 5 \cdot 1 & 6 - 5 \cdot 4 \\
3 & -1 & 5 & 5
\end{array}\right]
\][/tex]
Which simplifies to:
[tex]\[
\left[\begin{array}{ccc|c}
1 & -1 & 1 & 4 \\
0 & 4 & -4 & -14 \\
3 & -1 & 5 & 5
\end{array}\right]
\][/tex]
- For the third row, we subtract [tex]\( 3 \)[/tex] times the first row from the third row:
[tex]\[
\left[\begin{array}{ccc|c}
1 & -1 & 1 & 4 \\
0 & 4 & -4 & -14 \\
3 - 3 \cdot 1 & -1 - 3 \cdot (-1) & 5 - 3 \cdot 1 & 5 - 3 \cdot 4
\end{array}\right]
\][/tex]
Which simplifies to:
[tex]\[
\left[\begin{array}{ccc|c}
1 & -1 & 1 & 4 \\
0 & 4 & -4 & -14 \\
0 & 2 & 2 & -7
\end{array}\right]
\][/tex]
3. Eliminate the second element of the third row:
- For the third row, we subtract [tex]\( \frac{2}{4} \)[/tex] (or 0.5) times the second row from the third row:
[tex]\[
\left[\begin{array}{ccc|c}
1 & -1 & 1 & 4 \\
0 & 4 & -4 & -14 \\
0 & 2 & 2 & -7
\end{array}\right]
\][/tex]
[tex]\[
\left[\begin{array}{ccc|c}
1 & -1 & 1 & 4 \\
0 & 4 & -4 & -14 \\
0 & 2 - 0.5 \cdot 4 & 2 - 0.5 \cdot (-4) & -7 - 0.5 \cdot (-14)
\end{array}\right]
\][/tex]
Which simplifies to:
[tex]\[
\left[\begin{array}{ccc|c}
1 & -1 & 1 & 4 \\
0 & 4 & -4 & -14 \\
0 & 0 & 4 & 0
\end{array}\right]
\][/tex]
4. Back-substitution to find the solutions:
- From row 3, solve for [tex]\( z \)[/tex]:
[tex]\[
4z = 0 \implies z = 0
\][/tex]
- Substitute [tex]\( z = 0 \)[/tex] into row 2 to solve for [tex]\( y \)[/tex]:
[tex]\[
4y - 4(0) = -14 \implies 4y = -14 \implies y = -3.5
\][/tex]
- Substitute [tex]\( y = -3.5 \)[/tex] and [tex]\( z = 0 \)[/tex] into row 1 to solve for [tex]\( x \)[/tex]:
[tex]\[
x - (-3.5) + 0 = 4 \implies x + 3.5 = 4 \implies x = 0.5
\][/tex]
Therefore, the solution to the system of equations is:
[tex]\[
\boxed{(x, y, z) = \left(0.5, -3.5, 0\right)}
\][/tex]
