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Sagot :
To solve the given system of linear equations using matrices, let's follow these necessary steps.
1. Write the system of linear equations in matrix form:
The given system of equations can be written as:
[tex]\[ \begin{pmatrix} 2 & 5 \\ 1 & 3 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 7 \\ 5 \end{pmatrix} \][/tex]
2. Identify the coefficient matrix [tex]\(A\)[/tex], the variable matrix [tex]\( \mathbf{x} \)[/tex], and the constants matrix [tex]\( \mathbf{B} \)[/tex]:
[tex]\[ A = \begin{pmatrix} 2 & 5 \\ 1 & 3 \end{pmatrix} , \quad \mathbf{x} = \begin{pmatrix} x \\ y \end{pmatrix} , \quad \mathbf{B} = \begin{pmatrix} 7 \\ 5 \end{pmatrix} \][/tex]
3. Find the inverse of the coefficient matrix [tex]\(A\)[/tex]:
The solution to the system of equations involves finding the inverse of matrix [tex]\(A\)[/tex]. If [tex]\(A^{-1}\)[/tex] is the inverse of [tex]\(A\)[/tex], we multiply both sides of the equation [tex]\(A \mathbf{x} = \mathbf{B}\)[/tex] by [tex]\(A^{-1}\)[/tex] to obtain:
[tex]\[ A^{-1} A \mathbf{x} = A^{-1} \mathbf{B} \][/tex]
Since [tex]\(A^{-1} A\)[/tex] is the identity matrix [tex]\(I\)[/tex], we have:
[tex]\[ \mathbf{x} = A^{-1} \mathbf{B} \][/tex]
After calculating the inverse of matrix [tex]\(A\)[/tex], we get:
[tex]\[ A^{-1} = \begin{pmatrix} 3 & -5 \\ -1 & 2 \end{pmatrix} \][/tex]
4. Multiply the inverse of [tex]\(A\)[/tex] by [tex]\( \mathbf{B} \)[/tex] to find [tex]\( \mathbf{x} \)[/tex]:
[tex]\[ \mathbf{x} = A^{-1} \mathbf{B} = \begin{pmatrix} 3 & -5 \\ -1 & 2 \end{pmatrix} \begin{pmatrix} 7 \\ 5 \end{pmatrix} \][/tex]
Perform the matrix multiplication:
[tex]\[ \mathbf{x} = \begin{pmatrix} 3 \cdot 7 + (-5) \cdot 5 \\ -1 \cdot 7 + 2 \cdot 5 \end{pmatrix} = \begin{pmatrix} 21 - 25 \\ -7 + 10 \end{pmatrix} = \begin{pmatrix} -4 \\ 3 \end{pmatrix} \][/tex]
5. Extract the values of [tex]\(x\)[/tex] and [tex]\(y\)[/tex] from the solution matrix [tex]\( \mathbf{x} \)[/tex]:
[tex]\[ x = -4 \quad \text{and} \quad y = 3 \][/tex]
Therefore, the solution to the system of equations is:
[tex]\[ x = -4 \quad \text{and} \quad y = 3 \][/tex]
1. Write the system of linear equations in matrix form:
The given system of equations can be written as:
[tex]\[ \begin{pmatrix} 2 & 5 \\ 1 & 3 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 7 \\ 5 \end{pmatrix} \][/tex]
2. Identify the coefficient matrix [tex]\(A\)[/tex], the variable matrix [tex]\( \mathbf{x} \)[/tex], and the constants matrix [tex]\( \mathbf{B} \)[/tex]:
[tex]\[ A = \begin{pmatrix} 2 & 5 \\ 1 & 3 \end{pmatrix} , \quad \mathbf{x} = \begin{pmatrix} x \\ y \end{pmatrix} , \quad \mathbf{B} = \begin{pmatrix} 7 \\ 5 \end{pmatrix} \][/tex]
3. Find the inverse of the coefficient matrix [tex]\(A\)[/tex]:
The solution to the system of equations involves finding the inverse of matrix [tex]\(A\)[/tex]. If [tex]\(A^{-1}\)[/tex] is the inverse of [tex]\(A\)[/tex], we multiply both sides of the equation [tex]\(A \mathbf{x} = \mathbf{B}\)[/tex] by [tex]\(A^{-1}\)[/tex] to obtain:
[tex]\[ A^{-1} A \mathbf{x} = A^{-1} \mathbf{B} \][/tex]
Since [tex]\(A^{-1} A\)[/tex] is the identity matrix [tex]\(I\)[/tex], we have:
[tex]\[ \mathbf{x} = A^{-1} \mathbf{B} \][/tex]
After calculating the inverse of matrix [tex]\(A\)[/tex], we get:
[tex]\[ A^{-1} = \begin{pmatrix} 3 & -5 \\ -1 & 2 \end{pmatrix} \][/tex]
4. Multiply the inverse of [tex]\(A\)[/tex] by [tex]\( \mathbf{B} \)[/tex] to find [tex]\( \mathbf{x} \)[/tex]:
[tex]\[ \mathbf{x} = A^{-1} \mathbf{B} = \begin{pmatrix} 3 & -5 \\ -1 & 2 \end{pmatrix} \begin{pmatrix} 7 \\ 5 \end{pmatrix} \][/tex]
Perform the matrix multiplication:
[tex]\[ \mathbf{x} = \begin{pmatrix} 3 \cdot 7 + (-5) \cdot 5 \\ -1 \cdot 7 + 2 \cdot 5 \end{pmatrix} = \begin{pmatrix} 21 - 25 \\ -7 + 10 \end{pmatrix} = \begin{pmatrix} -4 \\ 3 \end{pmatrix} \][/tex]
5. Extract the values of [tex]\(x\)[/tex] and [tex]\(y\)[/tex] from the solution matrix [tex]\( \mathbf{x} \)[/tex]:
[tex]\[ x = -4 \quad \text{and} \quad y = 3 \][/tex]
Therefore, the solution to the system of equations is:
[tex]\[ x = -4 \quad \text{and} \quad y = 3 \][/tex]
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