To determine the nature of the solutions for the given system of linear equations:
[tex]\[
\left\{
\begin{array}{l}
4x - y + 2z = -1 \\
-x + 2y + 5z = 2 \\
-x + y - 3z = 1
\end{array}
\right.
\][/tex]
We can represent this system in matrix form [tex]\( AX = B \)[/tex], where:
[tex]\[
A = \begin{pmatrix}
4 & -1 & 2 \\
-1 & 2 & 5 \\
-1 & 1 & -3
\end{pmatrix}
\][/tex]
[tex]\( X = \begin{pmatrix} x \\ y \\ z \end{pmatrix} \)[/tex]
[tex]\[
B = \begin{pmatrix} -1 \\ 2 \\ 1 \end{pmatrix}
\][/tex]
To find the solution to the system, we need to solve the matrix equation [tex]\( AX = B \)[/tex].
### Steps to Solve the System
1. Check for Invertibility of Matrix [tex]\( A \)[/tex]:
- A system of linear equations has a unique solution if and only if the coefficient matrix [tex]\( A \)[/tex] is invertible. Matrix [tex]\( A \)[/tex] is invertible if its determinant is non-zero.
2. Find the Determinant of [tex]\( A \)[/tex]:
- Calculate [tex]\(\text{det}(A)\)[/tex]:
[tex]\[
A = \begin{pmatrix}
4 & -1 & 2 \\
-1 & 2 & 5 \\
-1 & 1 & -3
\end{pmatrix}
\][/tex]
If the determinant [tex]\(\text{det}(A) \neq 0\)[/tex], then matrix [tex]\( A \)[/tex] is invertible and the system has exactly one solution.
3. Solve the System [tex]\( AX = B \)[/tex]:
- If [tex]\( A \)[/tex] is invertible, use matrix methods such as Gaussian elimination or matrix inversion to solve for [tex]\( X \)[/tex].
- Alternatively, use numerical methods to directly solve for [tex]\( X \)[/tex].
### Conclusion:
Given that the system:
[tex]\[
\left\{
\begin{array}{l}
4x - y + 2z = -1 \\
-x + 2y + 5z = 2 \\
-x + y - 3z = 1
\end{array}
\right.
\][/tex]
has been examined and it is determined that the coefficient matrix [tex]\( A \)[/tex] is invertible, the system therefore has exactly one solution.
Thus, the correct answer is:
[tex]\[
\boxed{\text{The system has exactly one solution.}}
\][/tex]
