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Sagot :
To solve the given system of linear equations using Cramer's rule, we need to follow these steps:
The system of equations is:
1. [tex]\( x + y - z = 6 \)[/tex]
2. [tex]\( 3x - 2y + z = -5 \)[/tex]
3. [tex]\( x + 3y - 2z = 14 \)[/tex]
### Step 1: Write the system in matrix form
[tex]\[ A \mathbf{x} = \mathbf{B} \][/tex]
where
[tex]\[ A = \begin{pmatrix} 1 & 1 & -1 \\ 3 & -2 & 1 \\ 1 & 3 & -2 \end{pmatrix} \][/tex]
[tex]\[ \mathbf{x} = \begin{pmatrix} x \\ y \\ z \end{pmatrix} \][/tex]
[tex]\[ \mathbf{B} = \begin{pmatrix} 6 \\ -5 \\ 14 \end{pmatrix} \][/tex]
### Step 2: Calculate the determinant of matrix [tex]\( A \)[/tex]
[tex]\[ \text{det}(A) = -3 \][/tex]
Since the determinant of [tex]\( A \)[/tex] is non-zero, we know the system has a unique solution.
### Step 3: Replace each column of [tex]\( A \)[/tex] with [tex]\( \mathbf{B} \)[/tex] and calculate the determinants
1. Replace the first column of [tex]\(A\)[/tex] with [tex]\(\mathbf{B}\)[/tex]:
[tex]\[ A_1 = \begin{pmatrix} 6 & 1 & -1 \\ -5 & -2 & 1 \\ 14 & 3 & -2 \end{pmatrix} \][/tex]
[tex]\[ \text{det}(A_1) = -3 \][/tex]
2. Replace the second column of [tex]\(A\)[/tex] with [tex]\(\mathbf{B}\)[/tex]:
[tex]\[ A_2 = \begin{pmatrix} 1 & 6 & -1 \\ 3 & -5 & 1 \\ 1 & 14 & -2 \end{pmatrix} \][/tex]
[tex]\[ \text{det}(A_2) = -9 \][/tex]
3. Replace the third column of [tex]\(A\)[/tex] with [tex]\(\mathbf{B}\)[/tex]:
[tex]\[ A_3 = \begin{pmatrix} 1 & 1 & 6 \\ 3 & -2 & -5 \\ 1 & 3 & 14 \end{pmatrix} \][/tex]
[tex]\[ \text{det}(A_3) = 6 \][/tex]
### Step 4: Apply Cramer's Rule to find [tex]\( x \)[/tex], [tex]\( y \)[/tex], and [tex]\( z \)[/tex]
Using Cramer's rule, the solutions for [tex]\(x\)[/tex], [tex]\(y\)[/tex], and [tex]\(z\)[/tex] are:
[tex]\[ x = \frac{\text{det}(A_1)}{\text{det}(A)} \][/tex]
[tex]\[ y = \frac{\text{det}(A_2)}{\text{det}(A)} \][/tex]
[tex]\[ z = \frac{\text{det}(A_3)}{\text{det}(A)} \][/tex]
Substituting the values we calculated:
[tex]\[ x = \frac{-3}{-3} = 1 \][/tex]
[tex]\[ y = \frac{-9}{-3} = 3 \][/tex]
[tex]\[ z = \frac{6}{-3} = -2 \][/tex]
Therefore, the solution to the system of equations is:
[tex]\[ x = 1 \][/tex]
[tex]\[ y = 3 \][/tex]
[tex]\[ z = -2 \][/tex]
The system of equations is:
1. [tex]\( x + y - z = 6 \)[/tex]
2. [tex]\( 3x - 2y + z = -5 \)[/tex]
3. [tex]\( x + 3y - 2z = 14 \)[/tex]
### Step 1: Write the system in matrix form
[tex]\[ A \mathbf{x} = \mathbf{B} \][/tex]
where
[tex]\[ A = \begin{pmatrix} 1 & 1 & -1 \\ 3 & -2 & 1 \\ 1 & 3 & -2 \end{pmatrix} \][/tex]
[tex]\[ \mathbf{x} = \begin{pmatrix} x \\ y \\ z \end{pmatrix} \][/tex]
[tex]\[ \mathbf{B} = \begin{pmatrix} 6 \\ -5 \\ 14 \end{pmatrix} \][/tex]
### Step 2: Calculate the determinant of matrix [tex]\( A \)[/tex]
[tex]\[ \text{det}(A) = -3 \][/tex]
Since the determinant of [tex]\( A \)[/tex] is non-zero, we know the system has a unique solution.
### Step 3: Replace each column of [tex]\( A \)[/tex] with [tex]\( \mathbf{B} \)[/tex] and calculate the determinants
1. Replace the first column of [tex]\(A\)[/tex] with [tex]\(\mathbf{B}\)[/tex]:
[tex]\[ A_1 = \begin{pmatrix} 6 & 1 & -1 \\ -5 & -2 & 1 \\ 14 & 3 & -2 \end{pmatrix} \][/tex]
[tex]\[ \text{det}(A_1) = -3 \][/tex]
2. Replace the second column of [tex]\(A\)[/tex] with [tex]\(\mathbf{B}\)[/tex]:
[tex]\[ A_2 = \begin{pmatrix} 1 & 6 & -1 \\ 3 & -5 & 1 \\ 1 & 14 & -2 \end{pmatrix} \][/tex]
[tex]\[ \text{det}(A_2) = -9 \][/tex]
3. Replace the third column of [tex]\(A\)[/tex] with [tex]\(\mathbf{B}\)[/tex]:
[tex]\[ A_3 = \begin{pmatrix} 1 & 1 & 6 \\ 3 & -2 & -5 \\ 1 & 3 & 14 \end{pmatrix} \][/tex]
[tex]\[ \text{det}(A_3) = 6 \][/tex]
### Step 4: Apply Cramer's Rule to find [tex]\( x \)[/tex], [tex]\( y \)[/tex], and [tex]\( z \)[/tex]
Using Cramer's rule, the solutions for [tex]\(x\)[/tex], [tex]\(y\)[/tex], and [tex]\(z\)[/tex] are:
[tex]\[ x = \frac{\text{det}(A_1)}{\text{det}(A)} \][/tex]
[tex]\[ y = \frac{\text{det}(A_2)}{\text{det}(A)} \][/tex]
[tex]\[ z = \frac{\text{det}(A_3)}{\text{det}(A)} \][/tex]
Substituting the values we calculated:
[tex]\[ x = \frac{-3}{-3} = 1 \][/tex]
[tex]\[ y = \frac{-9}{-3} = 3 \][/tex]
[tex]\[ z = \frac{6}{-3} = -2 \][/tex]
Therefore, the solution to the system of equations is:
[tex]\[ x = 1 \][/tex]
[tex]\[ y = 3 \][/tex]
[tex]\[ z = -2 \][/tex]
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