Absolutely! We'll use Gaussian elimination to solve the given system of linear equations step-by-step.
The system is:
[tex]\[
\begin{aligned}
x - y + z &= 4 \\
5x - y + z &= 6 \\
3x - y + 5z &= 5
\end{aligned}
\][/tex]
First, we'll write this system in matrix form [tex]\(AX = B\)[/tex] where [tex]\(A\)[/tex] is the coefficient matrix, [tex]\(X\)[/tex] is the vector of variables, and [tex]\(B\)[/tex] is the right-hand side vector:
[tex]\[
A =
\begin{pmatrix}
1 & -1 & 1 \\
5 & -1 & 1 \\
3 & -1 & 5
\end{pmatrix}, \quad
X =
\begin{pmatrix}
x \\
y \\
z
\end{pmatrix}, \quad
B =
\begin{pmatrix}
4 \\
6 \\
5
\end{pmatrix}
\][/tex]
### Step 1: Convert [tex]\(A\)[/tex] to an Upper Triangular Matrix
#### First Iteration (i = 1):
Using the first row to eliminate [tex]\(x\)[/tex] from the 2nd and 3rd rows.
1. Subtract 5 times the 1st row from the 2nd row:
[tex]\[
\begin{pmatrix}
5 & -1 & 1
\end{pmatrix}
- 5 \times
\begin{pmatrix}
1 & -1 & 1
\end{pmatrix}
=
\begin{pmatrix}
5 - 5 & -1 + 5 & 1 - 5
\end{pmatrix}
=
\begin{pmatrix}
0 & 4 & -4
\end{pmatrix}
\][/tex]
Update the 2nd row of [tex]\(B\)[/tex]:
[tex]\[
6 - 5 \times 4 = 6 - 20 = -14
\][/tex]
2. Subtract 3 times the 1st row from the 3rd row:
[tex]\[
\begin{pmatrix}
3 & -1 & 5
\end{pmatrix}
- 3 \times
\begin{pmatrix}
1 & -1 & 1
\end{pmatrix}
=
\begin{pmatrix}
3 - 3 & -1 + 3 & 5 - 3
\end{pmatrix}
=
\begin{pmatrix}
0 & 2 & 2
\end{pmatrix}
\][/tex]
Update the 3rd row of [tex]\(B\)[/tex]:
[tex]\[
5 - 3 \times 4 = 5 - 12 = -7
\][/tex]
The updated system is:
[tex]\[
\begin{aligned}
1x - 1y + 1z &= 4 \\
0x + 4y - 4z &= -14 \\
0x + 2y + 2z &= -7
\end{aligned}
\][/tex]
#### Second Iteration (i = 2):
Using the second row to eliminate [tex]\(y\)[/tex] from the 3rd row.
1. Subtract 0.5 times the 2nd row from the 3rd row:
[tex]\[
\begin{pmatrix}
0 & 2 & 2
\end{pmatrix}
- 0.5 \times
\begin{pmatrix}
0 & 4 & -4
\end{pmatrix}
=
\begin{pmatrix}
0 & 2 - 2 & 2 + 2
\end{pmatrix}
=
\begin{pmatrix}
0 & 0 & 4
\end{pmatrix}
\][/tex]
Update the 3rd row of [tex]\(B\)[/tex]:
[tex]\[
-7 - 0.5 \times (-14) = -7 + 7 = 0
\][/tex]
The updated system is:
[tex]\[
\begin{aligned}
1x - 1y + 1z &= 4 \\
0x + 4y - 4z &= -14 \\
0x + 0y + 4z &= 0
\end{aligned}
\][/tex]
### Step 2: Back Substitution
Starting from the last equation:
[tex]\[
4z = 0 \implies z = 0
\][/tex]
Substituting [tex]\(z = 0\)[/tex] into the second equation:
[tex]\[
4y - 4(0) = -14 \implies 4y = -14 \implies y = -3.5
\][/tex]
Finally, substituting [tex]\(y = -3.5\)[/tex] and [tex]\(z = 0\)[/tex] into the first equation:
[tex]\[
x - (-3.5) + 0 = 4 \implies x + 3.5 = 4 \implies x = 4 - 3.5 = 0.5
\][/tex]
Thus, the solution to the system of equations is:
[tex]\[
\begin{aligned}
x &= 0.5 \\
y &= -3.5 \\
z &= 0
\end{aligned}
\][/tex]
In matrix form, the solution vector [tex]\(X\)[/tex] is:
[tex]\[
X =
\begin{pmatrix}
0.5 \\
-3.5 \\
0
\end{pmatrix}
\][/tex]
