To solve the given system of linear equations, we need to use either substitution, elimination, or matrix operations. Here, we'll solve this system using the Gaussian elimination method.
We start with the following system of equations:
[tex]\[
\begin{array}{r}
1. \quad x + 2y - z = 3 \\
2. \quad 2x - y + 2z = 6 \\
3. \quad x - 3y + 3z = 4 \\
\end{array}
\][/tex]
We will write this system as an augmented matrix:
[tex]\[
\left[\begin{array}{ccc|c}
1 & 2 & -1 & 3 \\
2 & -1 & 2 & 6 \\
1 & -3 & 3 & 4 \\
\end{array}\right]
\][/tex]
Step 1: Make the pivot in the first row.
The pivot element is already in place in row 1. Now we will eliminate the [tex]\(x\)[/tex]-terms from rows 2 and 3.
For row 2, subtract 2 times row 1 from row 2:
[tex]\[
\begin{aligned}
2R_1 - R_2 & : 2 \times (1, 2, -1, 3) - (2, -1, 2, 6) \\
& : (2, 4, -2, 6) - (2, -1, 2, 6) \\
& : (0, 5, -4, 0)
\end{aligned}
\][/tex]
For row 3, subtract row 1 from row 3:
[tex]\[
\begin{aligned}
R_3 - R_1 & : (1, -3, 3, 4) - (1, 2, -1, 3) \\
& : (0, -5, 4, 1)
\end{aligned}
\][/tex]
The augmented matrix is now:
[tex]\[
\left[\begin{array}{ccc|c}
1 & 2 & -1 & 3 \\
0 & 5 & -4 & 0 \\
0 & -5 & 4 & 1 \\
\end{array}\right]
\][/tex]
Step 2: Make the pivot in the second row.
The pivot element in row 2 is already in place. Now we will eliminate the [tex]\(y\)[/tex]-terms from row 3.
For row 3, add row 2 to row 3:
[tex]\[
\begin{aligned}
R_3 + R_2 & : (0, -5, 4, 1) + (0, 5, -4, 0) \\
& : (0, 0, 0, 1)
\end{aligned}
\][/tex]
After this step, our augmented matrix looks like:
[tex]\[
\left[\begin{array}{ccc|c}
1 & 2 & -1 & 3 \\
0 & 5 & -4 & 0 \\
0 & 0 & 0 & 1 \\
\end{array}\right]
\][/tex]
Step 3: Interpretation.
The last row (0 0 0 | 1) implies the equation [tex]\(0 = 1\)[/tex], which is a contradiction. This indicates that there are no solutions to this system of equations.
Therefore, based on our step-by-step reduction, the system of equations has no solution.
Answer:
C. no solutions
