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Sagot :
Sure, let's solve the given system of linear equations using the given methods:
The system of equations is:
[tex]\[ 3m + 2y = 1 \][/tex]
[tex]\[ 4m - 5y = -14 \][/tex]
### a) Using Inverses
Step 1: Write the system in matrix form:
[tex]\[ A \mathbf{x} = \mathbf{b} \][/tex]
where
[tex]\[ A = \begin{pmatrix} 3 & 2 \\ 4 & -5 \\ \end{pmatrix} \][/tex],
[tex]\[ \mathbf{x} = \begin{pmatrix} m \\ y \\ \end{pmatrix} \][/tex],
and
[tex]\[ \mathbf{b} = \begin{pmatrix} 1 \\ -14 \\ \end{pmatrix}. \][/tex]
Step 2: Find the inverse [tex]\( A^{-1} \)[/tex] of matrix [tex]\( A \)[/tex]:
[tex]\[ A^{-1} = \begin{pmatrix} 0.2173913 & 0.08695652 \\ 0.17391304 & -0.13043478 \\ \end{pmatrix} \][/tex]
Step 3: Multiply the inverse of the matrix [tex]\( A \)[/tex] by the vector [tex]\( \mathbf{b} \)[/tex] to get [tex]\( \mathbf{x} \)[/tex]:
[tex]\[ \mathbf{x} = A^{-1} \mathbf{b} = \begin{pmatrix} -1 \\ 2 \\ \end{pmatrix} \][/tex]
So, the solution using the inverse method is:
[tex]\[ m = -1, \quad y = 2 \][/tex]
### b) Using Determinants (Cramer's Rule)
Step 1: Calculate the determinant of matrix [tex]\( A \)[/tex]:
[tex]\[ \text{det}(A) = \begin{vmatrix} 3 & 2 \\ 4 & -5 \\ \end{vmatrix} = (3 \cdot (-5)) - (2 \cdot 4) = -15 - 8 = -23 \][/tex]
Step 2: Compute the determinant of matrix obtained by replacing the first column of [tex]\( A \)[/tex] with [tex]\( \mathbf{b} \)[/tex] to find [tex]\( \text{det}(A_m) \)[/tex]:
[tex]\[ A_m = \begin{pmatrix} 1 & 2 \\ -14 & -5 \\ \end{pmatrix} \][/tex]
[tex]\[ \text{det}(A_m) = \begin{vmatrix} 1 & 2 \\ -14 & -5 \\ \end{vmatrix} = (1 \cdot (-5)) - (2 \cdot (-14)) = -5 + 28 = 23 \][/tex]
Step 3: Compute the determinant of matrix obtained by replacing the second column of [tex]\( A \)[/tex] with [tex]\( \mathbf{b} \)[/tex] to find [tex]\( \text{det}(A_y) \)[/tex]:
[tex]\[ A_y = \begin{pmatrix} 3 & 1 \\ 4 & -14 \\ \end{pmatrix} \][/tex]
[tex]\[ \text{det}(A_y) = \begin{vmatrix} 3 & 1 \\ 4 & -14 \\ \end{vmatrix} = (3 \cdot (-14)) - (1 \cdot 4) = -42 - 4 = -46 \][/tex]
Step 4: Use Cramer's Rule to find [tex]\( m \)[/tex] and [tex]\( y \)[/tex]:
[tex]\[ m = \frac{\text{det}(A_m)}{\text{det}(A)} = \frac{23}{-23} = -1 \][/tex]
[tex]\[ y = \frac{\text{det}(A_y)}{\text{det}(A)} = \frac{-46}{-23} = 2 \][/tex]
So, the solution using determinants (Cramer's Rule) is:
[tex]\[ m = -1, \quad y = 2 \][/tex]
In both methods, we obtained the same solution:
[tex]\[ m = -1, \quad y = 2 \][/tex]
The system of equations is:
[tex]\[ 3m + 2y = 1 \][/tex]
[tex]\[ 4m - 5y = -14 \][/tex]
### a) Using Inverses
Step 1: Write the system in matrix form:
[tex]\[ A \mathbf{x} = \mathbf{b} \][/tex]
where
[tex]\[ A = \begin{pmatrix} 3 & 2 \\ 4 & -5 \\ \end{pmatrix} \][/tex],
[tex]\[ \mathbf{x} = \begin{pmatrix} m \\ y \\ \end{pmatrix} \][/tex],
and
[tex]\[ \mathbf{b} = \begin{pmatrix} 1 \\ -14 \\ \end{pmatrix}. \][/tex]
Step 2: Find the inverse [tex]\( A^{-1} \)[/tex] of matrix [tex]\( A \)[/tex]:
[tex]\[ A^{-1} = \begin{pmatrix} 0.2173913 & 0.08695652 \\ 0.17391304 & -0.13043478 \\ \end{pmatrix} \][/tex]
Step 3: Multiply the inverse of the matrix [tex]\( A \)[/tex] by the vector [tex]\( \mathbf{b} \)[/tex] to get [tex]\( \mathbf{x} \)[/tex]:
[tex]\[ \mathbf{x} = A^{-1} \mathbf{b} = \begin{pmatrix} -1 \\ 2 \\ \end{pmatrix} \][/tex]
So, the solution using the inverse method is:
[tex]\[ m = -1, \quad y = 2 \][/tex]
### b) Using Determinants (Cramer's Rule)
Step 1: Calculate the determinant of matrix [tex]\( A \)[/tex]:
[tex]\[ \text{det}(A) = \begin{vmatrix} 3 & 2 \\ 4 & -5 \\ \end{vmatrix} = (3 \cdot (-5)) - (2 \cdot 4) = -15 - 8 = -23 \][/tex]
Step 2: Compute the determinant of matrix obtained by replacing the first column of [tex]\( A \)[/tex] with [tex]\( \mathbf{b} \)[/tex] to find [tex]\( \text{det}(A_m) \)[/tex]:
[tex]\[ A_m = \begin{pmatrix} 1 & 2 \\ -14 & -5 \\ \end{pmatrix} \][/tex]
[tex]\[ \text{det}(A_m) = \begin{vmatrix} 1 & 2 \\ -14 & -5 \\ \end{vmatrix} = (1 \cdot (-5)) - (2 \cdot (-14)) = -5 + 28 = 23 \][/tex]
Step 3: Compute the determinant of matrix obtained by replacing the second column of [tex]\( A \)[/tex] with [tex]\( \mathbf{b} \)[/tex] to find [tex]\( \text{det}(A_y) \)[/tex]:
[tex]\[ A_y = \begin{pmatrix} 3 & 1 \\ 4 & -14 \\ \end{pmatrix} \][/tex]
[tex]\[ \text{det}(A_y) = \begin{vmatrix} 3 & 1 \\ 4 & -14 \\ \end{vmatrix} = (3 \cdot (-14)) - (1 \cdot 4) = -42 - 4 = -46 \][/tex]
Step 4: Use Cramer's Rule to find [tex]\( m \)[/tex] and [tex]\( y \)[/tex]:
[tex]\[ m = \frac{\text{det}(A_m)}{\text{det}(A)} = \frac{23}{-23} = -1 \][/tex]
[tex]\[ y = \frac{\text{det}(A_y)}{\text{det}(A)} = \frac{-46}{-23} = 2 \][/tex]
So, the solution using determinants (Cramer's Rule) is:
[tex]\[ m = -1, \quad y = 2 \][/tex]
In both methods, we obtained the same solution:
[tex]\[ m = -1, \quad y = 2 \][/tex]
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