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Sagot :
To solve a system of linear equations using inverse matrices, we can follow these steps:
1. Write the system of equations in matrix form:
[tex]\[ \left\{ \begin{array}{l} x + 5y = -3 \\ -4x - y = -7 \end{array} \right. \][/tex]
This can be represented as [tex]\( A\mathbf{x} = \mathbf{b} \)[/tex], where
[tex]\[ A = \begin{pmatrix} 1 & 5 \\ -4 & -1 \end{pmatrix}, \quad \mathbf{x} = \begin{pmatrix} x \\ y \end{pmatrix}, \quad \mathbf{b} = \begin{pmatrix} -3 \\ -7 \end{pmatrix} \][/tex]
2. Find the inverse of matrix [tex]\( A \)[/tex]:
The inverse of a [tex]\( 2 \times 2 \)[/tex] matrix [tex]\( \begin{pmatrix} a & b \\ c & d \end{pmatrix} \)[/tex] is given by:
[tex]\[ A^{-1} = \frac{1}{ad - bc} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix} \][/tex]
For our matrix [tex]\( A \)[/tex]:
[tex]\[ A = \begin{pmatrix} 1 & 5 \\ -4 & -1 \end{pmatrix} \][/tex]
First, we compute the determinant of [tex]\( A \)[/tex]:
[tex]\[ \det(A) = (1 \cdot -1) - (5 \cdot -4) = -1 + 20 = 19 \][/tex]
Thus, the inverse of matrix [tex]\( A \)[/tex] is:
[tex]\[ A^{-1} = \frac{1}{19} \begin{pmatrix} -1 & -5 \\ 4 & 1 \end{pmatrix} = \begin{pmatrix} -\frac{1}{19} & -\frac{5}{19} \\ \frac{4}{19} & \frac{1}{19} \end{pmatrix} \][/tex]
3. Multiply both sides of [tex]\( A\mathbf{x} = \mathbf{b} \)[/tex] by [tex]\( A^{-1} \)[/tex] to solve for [tex]\( \mathbf{x} \)[/tex]:
[tex]\[ \mathbf{x} = A^{-1}\mathbf{b} \][/tex]
Performing the matrix multiplication:
[tex]\[ \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} -\frac{1}{19} & -\frac{5}{19} \\ \frac{4}{19} & \frac{1}{19} \end{pmatrix} \begin{pmatrix} -3 \\ -7 \end{pmatrix} \][/tex]
This gives us:
[tex]\[ x = \left( -\frac{1}{19} \cdot -3 \right) + \left( -\frac{5}{19} \cdot -7 \right) = \frac{3}{19} + \frac{35}{19} = \frac{38}{19} = 2 \][/tex]
And:
[tex]\[ y = \left( \frac{4}{19} \cdot -3 \right) + \left( \frac{1}{19} \cdot -7 \right) = -\frac{12}{19} - \frac{7}{19} = -\frac{19}{19} = -1 \][/tex]
Thus, the solution to the system of equations is:
[tex]\[ (x, y) = (2, -1) \][/tex]
1. Write the system of equations in matrix form:
[tex]\[ \left\{ \begin{array}{l} x + 5y = -3 \\ -4x - y = -7 \end{array} \right. \][/tex]
This can be represented as [tex]\( A\mathbf{x} = \mathbf{b} \)[/tex], where
[tex]\[ A = \begin{pmatrix} 1 & 5 \\ -4 & -1 \end{pmatrix}, \quad \mathbf{x} = \begin{pmatrix} x \\ y \end{pmatrix}, \quad \mathbf{b} = \begin{pmatrix} -3 \\ -7 \end{pmatrix} \][/tex]
2. Find the inverse of matrix [tex]\( A \)[/tex]:
The inverse of a [tex]\( 2 \times 2 \)[/tex] matrix [tex]\( \begin{pmatrix} a & b \\ c & d \end{pmatrix} \)[/tex] is given by:
[tex]\[ A^{-1} = \frac{1}{ad - bc} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix} \][/tex]
For our matrix [tex]\( A \)[/tex]:
[tex]\[ A = \begin{pmatrix} 1 & 5 \\ -4 & -1 \end{pmatrix} \][/tex]
First, we compute the determinant of [tex]\( A \)[/tex]:
[tex]\[ \det(A) = (1 \cdot -1) - (5 \cdot -4) = -1 + 20 = 19 \][/tex]
Thus, the inverse of matrix [tex]\( A \)[/tex] is:
[tex]\[ A^{-1} = \frac{1}{19} \begin{pmatrix} -1 & -5 \\ 4 & 1 \end{pmatrix} = \begin{pmatrix} -\frac{1}{19} & -\frac{5}{19} \\ \frac{4}{19} & \frac{1}{19} \end{pmatrix} \][/tex]
3. Multiply both sides of [tex]\( A\mathbf{x} = \mathbf{b} \)[/tex] by [tex]\( A^{-1} \)[/tex] to solve for [tex]\( \mathbf{x} \)[/tex]:
[tex]\[ \mathbf{x} = A^{-1}\mathbf{b} \][/tex]
Performing the matrix multiplication:
[tex]\[ \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} -\frac{1}{19} & -\frac{5}{19} \\ \frac{4}{19} & \frac{1}{19} \end{pmatrix} \begin{pmatrix} -3 \\ -7 \end{pmatrix} \][/tex]
This gives us:
[tex]\[ x = \left( -\frac{1}{19} \cdot -3 \right) + \left( -\frac{5}{19} \cdot -7 \right) = \frac{3}{19} + \frac{35}{19} = \frac{38}{19} = 2 \][/tex]
And:
[tex]\[ y = \left( \frac{4}{19} \cdot -3 \right) + \left( \frac{1}{19} \cdot -7 \right) = -\frac{12}{19} - \frac{7}{19} = -\frac{19}{19} = -1 \][/tex]
Thus, the solution to the system of equations is:
[tex]\[ (x, y) = (2, -1) \][/tex]
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