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Sagot :
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]
[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]
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