Let's solve the system of linear equations:
[tex]\[
\begin{cases}
x + 4y - z = 5 \\
3x - 2y + 2z = 7 \\
2x - z = -2
\end{cases}
\][/tex]
### Step 1: Set Up the Coefficient Matrix and the Constant Matrix
First, write the augmented matrix for the system of equations.
The coefficient matrix [tex]\(A\)[/tex] and the constant matrix [tex]\(B\)[/tex] are:
[tex]\[
A = \begin{pmatrix}
1 & 4 & -1 \\
3 & -2 & 2 \\
2 & 0 & -1
\end{pmatrix}
\][/tex]
[tex]\[
B = \begin{pmatrix}
5 \\
7 \\
-2
\end{pmatrix}
\][/tex]
### Step 2: Systematic Elimination of Variables
We use methods like Gaussian elimination to transform matrix [tex]\(A\)[/tex] to Row Echelon Form (REF) and solve the equations step-by-step. Alternatively, we can compare potential solutions to verify which one satisfies all the equations.
### Step 3: Check Each Proposed Solution
Let’s check each given option by substituting back into the original equations to see which one satisfies all three equations.
1. For (1, 0, -4):
[tex]\[
\begin{cases}
1 + 4(0) - (-4) = 5 & \quad \text{(True: } 1 + 0 + 4 = 5 \text{)} \\
3(1) - 2(0) + 2(-4) = 7 & \quad \text{(False: } 3 - 8 \neq 7 \text{)} \\
2(1) - (-4) = -2 & \quad \text{(False: } 2 + 4 \neq -2 \text{)}
\end{cases}
\][/tex]
Thus, (1, 0, -4) is not a solution.
2. For (-1, 1.5, 0):
[tex]\[
\begin{cases}
-1 + 4(1.5) - 0 = 5 & \quad \text{(True: } -1 + 6 = 5 \text{)} \\
3(-1) - 2(1.5) + 2(0) = 7 & \quad \text{(False: } -3 - 3 \neq 7 \text{)} \\
2(-1) - 0 = -2 & \quad \text{(True: } -2 = -2 \text{)}
\end{cases}
\][/tex]
Thus, (-1, 1.5, 0) is not a solution even though it satisfies the first and third equations, it fails the second one.
3. For (1, 2, 4):
[tex]\[
\begin{cases}
1 + 4(2) - 4 = 5 & \quad \text{(True: } 1 + 8 - 4 = 5 \text{)} \\
3(1) - 2(2) + 2(4) = 7 & \quad \text{(True: } 3 - 4 + 8 = 7 \text{)} \\
2(1) - 4 = -2 & \quad \text{(True: } 2 - 4 = -2 \text{)}
\end{cases}
\][/tex]
Thus, (1, 2, 4) is the correct solution.
4. For (13, 5, 28):
[tex]\[
\begin{cases}
13 + 4(5) - 28 = 5 & \quad \text{(False: } 13 + 20 - 28 \neq 5 \text{)} \\
3(13) - 2(5) + 2(28) = 7 & \quad \text{(False: } 39 - 10 + 56 \neq 7 \text{)} \\
2(13) - 28 = -2 & \quad \text{(False: } 26 - 28 \neq -2 \text{)}
\end{cases}
\][/tex]
Thus, (13, 5, 28) is not a solution.
### Conclusion
The solution to the system of equations
[tex]\[
\begin{cases}
x + 4y - z = 5 \\
3x - 2y + 2z = 7 \\
2x - z = -2
\end{cases}
\][/tex]
is [tex]\((1, 2, 4)\)[/tex].
