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Sagot :
To determine if each linear system has a unique solution, we can use the determinant of the coefficient matrix for each system. If the determinant is non-zero, the system has a unique solution. If the determinant is zero, the system does not have a unique solution.
Let's go through each system step by step.
### First System
[tex]\[ \begin{array}{l} x + 3y = 4 \\ 3x - y = 5 \end{array} \][/tex]
The coefficient matrix for this system is:
[tex]\[ \begin{pmatrix} 1 & 3 \\ 3 & -1 \end{pmatrix} \][/tex]
The determinant of this 2x2 matrix is calculated as:
[tex]\[ \text{det} = (1 \cdot -1) - (3 \cdot 3) = -1 - 9 = -10 \][/tex]
Since the determinant [tex]\(\text{det} = -10\)[/tex] is non-zero, the system has a unique solution.
### Second System
[tex]\[ \begin{array}{l} x + 2y - z = 8 \\ x - y + 2z = 0 \\ 2x - 3y + z = 1 \end{array} \][/tex]
The coefficient matrix for this system is:
[tex]\[ \begin{pmatrix} 1 & 2 & -1 \\ 1 & -1 & 2 \\ 2 & -3 & 1 \end{pmatrix} \][/tex]
The determinant of this 3x3 matrix is calculated as:
[tex]\[ \text{det} = \begin{vmatrix} 1 & 2 & -1 \\ 1 & -1 & 2 \\ 2 & -3 & 1 \end{vmatrix} = 12 \][/tex]
Since the determinant [tex]\(\text{det} = 12\)[/tex] is non-zero, the system has a unique solution.
### Third System
[tex]\[ \begin{array}{l} x + 3y + z = 4 \\ 2x + 6y + 2z = 5 \\ x + y + z = 0 \end{array} \][/tex]
The coefficient matrix for this system is:
[tex]\[ \begin{pmatrix} 1 & 3 & 1 \\ 2 & 6 & 2 \\ 1 & 1 & 1 \end{pmatrix} \][/tex]
The determinant of this 3x3 matrix is calculated as:
[tex]\[ \text{det} = \begin{vmatrix} 1 & 3 & 1 \\ 2 & 6 & 2 \\ 1 & 1 & 1 \end{vmatrix} = 0 \][/tex]
Since the determinant [tex]\(\text{det} = 0\)[/tex] is zero, the system does not have a unique solution.
### Summary
- The first system has a unique solution (det = -10).
- The second system has a unique solution (det = 12).
- The third system does not have a unique solution (det = 0).
Let's go through each system step by step.
### First System
[tex]\[ \begin{array}{l} x + 3y = 4 \\ 3x - y = 5 \end{array} \][/tex]
The coefficient matrix for this system is:
[tex]\[ \begin{pmatrix} 1 & 3 \\ 3 & -1 \end{pmatrix} \][/tex]
The determinant of this 2x2 matrix is calculated as:
[tex]\[ \text{det} = (1 \cdot -1) - (3 \cdot 3) = -1 - 9 = -10 \][/tex]
Since the determinant [tex]\(\text{det} = -10\)[/tex] is non-zero, the system has a unique solution.
### Second System
[tex]\[ \begin{array}{l} x + 2y - z = 8 \\ x - y + 2z = 0 \\ 2x - 3y + z = 1 \end{array} \][/tex]
The coefficient matrix for this system is:
[tex]\[ \begin{pmatrix} 1 & 2 & -1 \\ 1 & -1 & 2 \\ 2 & -3 & 1 \end{pmatrix} \][/tex]
The determinant of this 3x3 matrix is calculated as:
[tex]\[ \text{det} = \begin{vmatrix} 1 & 2 & -1 \\ 1 & -1 & 2 \\ 2 & -3 & 1 \end{vmatrix} = 12 \][/tex]
Since the determinant [tex]\(\text{det} = 12\)[/tex] is non-zero, the system has a unique solution.
### Third System
[tex]\[ \begin{array}{l} x + 3y + z = 4 \\ 2x + 6y + 2z = 5 \\ x + y + z = 0 \end{array} \][/tex]
The coefficient matrix for this system is:
[tex]\[ \begin{pmatrix} 1 & 3 & 1 \\ 2 & 6 & 2 \\ 1 & 1 & 1 \end{pmatrix} \][/tex]
The determinant of this 3x3 matrix is calculated as:
[tex]\[ \text{det} = \begin{vmatrix} 1 & 3 & 1 \\ 2 & 6 & 2 \\ 1 & 1 & 1 \end{vmatrix} = 0 \][/tex]
Since the determinant [tex]\(\text{det} = 0\)[/tex] is zero, the system does not have a unique solution.
### Summary
- The first system has a unique solution (det = -10).
- The second system has a unique solution (det = 12).
- The third system does not have a unique solution (det = 0).
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