To solve the system of equations:
[tex]\[
\begin{cases}
x - 3y + 4z = 17 \\
2x + y + z = 6 \\
-2x + 3y - 3z = -18
\end{cases}
\][/tex]
we will follow these steps:
### Step 1: Express the system as a matrix equation
We can represent the system of equations as:
[tex]\[
\begin{pmatrix}
1 & -3 & 4 \\
2 & 1 & 1 \\
-2 & 3 & -3
\end{pmatrix}
\begin{pmatrix}
x \\
y \\
z
\end{pmatrix}
=
\begin{pmatrix}
17 \\
6 \\
-18
\end{pmatrix}
\][/tex]
### Step 2: Use row reduction to solve the system
Let's denote the matrix of coefficients as [tex]\(A\)[/tex] and the vector of constants as [tex]\(B\)[/tex]:
[tex]\[
A = \begin{pmatrix}
1 & -3 & 4 \\
2 & 1 & 1 \\
-2 & 3 & -3
\end{pmatrix}, \quad B = \begin{pmatrix}
17 \\
6 \\
-18
\end{pmatrix}
\][/tex]
To solve the system using Gaussian elimination, we perform row operations to achieve row-echelon form.
#### Step 2a: Make upper triangular matrix
1. Start with the augmented matrix [tex]\([A|B]\)[/tex]:
[tex]\[
\begin{pmatrix}
1 & -3 & 4 & | & 17 \\
2 & 1 & 1 & | & 6 \\
-2 & 3 & -3 & | & -18
\end{pmatrix}
\][/tex]
2. Eliminate [tex]\(x\)[/tex] from the second and third rows.
[tex]\[
R_2 \rightarrow R_2 - 2R_1
\][/tex]
[tex]\[
R_3 \rightarrow R_3 + 2R_1
\][/tex]
[tex]\[
\begin{pmatrix}
1 & -3 & 4 & | & 17 \\
0 & 7 & -7 & | & -28 \\
0 & -3 & 5 & | & 16
\end{pmatrix}
\][/tex]
3. Next, to eliminate [tex]\(y\)[/tex] from the third row.
[tex]\[
R_3 \rightarrow R_3 + \frac{3}{7}R_2
\][/tex]
[tex]\[
\begin{pmatrix}
1 & -3 & 4 & | & 17 \\
0 & 7 & -7 & | & -28 \\
0 & 0 & \frac{8}{7} & | & 4
\end{pmatrix}
\][/tex]
4. Simplify row 3:
[tex]\[
R_3 \rightarrow R_3 \cdot \frac{7}{8}
\][/tex]
[tex]\[
\begin{pmatrix}
1 & -3 & 4 & | & 17 \\
0 & 7 & -7 & | & -28 \\
0 & 0 & 1 & | & \frac{7}{2}
\end{pmatrix}
\][/tex]
#### Step 2b: Back-substitution
Now that the matrix is in row echelon form, we can back-substitute to find the solutions for [tex]\(x\)[/tex], [tex]\(y\)[/tex], and [tex]\(z\)[/tex].
1. From the third row:
[tex]\[
z = \frac{7}{2}
\][/tex]
2. Substitute [tex]\(z\)[/tex] into the second row:
[tex]\[
7y - 7 \cdot \frac{7}{2} = -28 \\
7y - \frac{49}{2} = -28 \\
7y = -28 + \frac{49}{2} \\
7y = \frac{-56 + 49}{2} \\
7y = \frac{-7}{2} \\
y = \frac{-1}{2}
\][/tex]
3. Substitute [tex]\(y\)[/tex] and [tex]\(z\)[/tex] into the first row:
[tex]\[
x - 3 \left(-\frac{1}{2}\right) + 4 \cdot \frac{7}{2} = 17 \\
x + \frac{3}{2} + 14 = 17 \\
x + 17 \frac{3}{2} = 17 \\
x + 17 = 17 \\
x = 3
\][/tex]
Thus, the solution for the system of equations is:
[tex]\[
x = 3, \; y = -2, \; z = 2
\][/tex]
Hence, the solution is:
[tex]\[
(x, y, z) = (3, -2, 2)
\][/tex]
