Let's analyze the system of linear equations:
[tex]\[
\begin{cases}
-x - 3y - 4z = 2 \\
-x + 2y - 4z = 0 \\
2x - y + 5z = 1
\end{cases}
\][/tex]
We need to determine the nature of the solutions for this system.
### Step-by-Step Analysis:
1. Matrix Representation:
First, we can represent this system in matrix form as [tex]\( \mathbf{A} \mathbf{x} = \mathbf{b} \)[/tex].
[tex]\[
\mathbf{A} = \begin{pmatrix}
-1 & -3 & -4 \\
-1 & 2 & -4 \\
2 & -1 & 5
\end{pmatrix},
\quad
\mathbf{x} = \begin{pmatrix}
x \\
y \\
z
\end{pmatrix},
\quad
\mathbf{b} = \begin{pmatrix}
2 \\
0 \\
1
\end{pmatrix}
\][/tex]
2. Checking for a Unique Solution:
To determine if the system has a unique solution, no solution, or infinitely many solutions, we need to examine the properties of the coefficient matrix [tex]\( \mathbf{A} \)[/tex] and the augmented matrix [tex]\( [\mathbf{A}|\mathbf{b}] \)[/tex].
3. Finding the Determinant:
If the determinant of [tex]\( \mathbf{A} \)[/tex] (denoted as [tex]\( \det(\mathbf{A}) \)[/tex]) is non-zero, then the system has a unique solution. If the determinant is zero, further examination of the ranks of [tex]\( \mathbf{A} \)[/tex] and the augmented matrix [tex]\( [\mathbf{A}|\mathbf{b}] \)[/tex] is necessary to determine if the system is consistent (infinitely many solutions) or inconsistent (no solutions).
4. Solving the System:
Let's summarize:
- If [tex]\( \det(\mathbf{A}) \neq 0 \)[/tex], the system has exactly one solution.
- If [tex]\( \det(\mathbf{A}) = 0 \)[/tex]:
- If the rank of [tex]\( \mathbf{A} \)[/tex] equals the rank of the augmented matrix [tex]\( [\mathbf{A}|\mathbf{b}] \)[/tex] and both are less than the number of variables, the system has infinitely many solutions.
- If the rank of [tex]\( \mathbf{A} \)[/tex] is less than the rank of the augmented matrix [tex]\( [\mathbf{A}|\mathbf{b}] \)[/tex], the system has no solutions.
Given the analysis of the determinant and the ranks:
[tex]\[
\det(\mathbf{A}) \neq 0 \implies \text{The system has exactly one solution.}
\][/tex]
Thus, based on the above steps and the properties of the matrices involved, we conclude that:
[tex]\[
\boxed{\text{The system has exactly one solution.}}
\][/tex]
