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Sagot :
Let's solve the given system of linear equations step-by-step.
Given equations:
[tex]\[ 15x + 7y = 10 \tag{1} \][/tex]
[tex]\[ 8x + 6y = -6 \tag{2} \][/tex]
Part 1: Write the system of equations in [tex]\( AX = B \)[/tex] form
In matrix notation, we can represent the system as follows:
[tex]\[ A = \begin{pmatrix} 15 & 7 \\ 8 & 6 \end{pmatrix}, \quad X = \begin{pmatrix} x \\ y \end{pmatrix}, \quad B = \begin{pmatrix} 10 \\ -6 \end{pmatrix} \][/tex]
So the equation [tex]\( AX = B \)[/tex] can be written as:
[tex]\[ \begin{pmatrix} 15 & 7 \\ 8 & 6 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 10 \\ -6 \end{pmatrix} \][/tex]
Part 2: Find [tex]\( A^{-1} \)[/tex]
The inverse of a 2x2 matrix [tex]\( A = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \)[/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]\[ A = \begin{pmatrix} 15 & 7 \\ 8 & 6 \end{pmatrix} \][/tex]
We calculate the determinant [tex]\( \text{det}(A) \)[/tex]:
[tex]\[ \text{det}(A) = 15 \cdot 6 - 7 \cdot 8 = 90 - 56 = 34 \][/tex]
Thus, the inverse [tex]\( A^{-1} \)[/tex] is:
[tex]\[ A^{-1} = \frac{1}{34} \begin{pmatrix} 6 & -7 \\ -8 & 15 \end{pmatrix} \][/tex]
Simplifying, we get:
[tex]\[ A^{-1} = \begin{pmatrix} \frac{6}{34} & \frac{-7}{34} \\ \frac{-8}{34} & \frac{15}{34} \end{pmatrix} = \begin{pmatrix} \frac{3}{17} & \frac{-7}{34} \\ \frac{-4}{17} & \frac{15}{34} \end{pmatrix} \][/tex]
Part 3: Solve the system using the matrix equation
To find [tex]\( X \)[/tex] from [tex]\( AX = B \)[/tex], we use the formula [tex]\( X = A^{-1}B \)[/tex].
Multiply [tex]\( A^{-1} \)[/tex] by [tex]\( B \)[/tex]:
[tex]\[ A^{-1} = \begin{pmatrix} \frac{3}{17} & \frac{-7}{34} \\ \frac{-4}{17} & \frac{15}{34} \end{pmatrix} , \quad B = \begin{pmatrix} 10 \\ -6 \end{pmatrix} \][/tex]
[tex]\[ X = \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} \frac{3}{17} & \frac{-7}{34} \\ \frac{-4}{17} & \frac{15}{34} \end{pmatrix} \begin{pmatrix} 10 \\ -6 \end{pmatrix} \][/tex]
Calculating the product:
[tex]\[ x = \frac{3}{17} \cdot 10 + \frac{-7}{34} \cdot (-6) = \frac{30}{17} + \frac{42}{34} = \frac{30}{17} + \frac{21}{17} = \frac{51}{17} = 3 \][/tex]
[tex]\[ y = \frac{-4}{17} \cdot 10 + \frac{15}{34} \cdot (-6) = \frac{-40}{17} + \frac{-90}{34} = \frac{-40}{17} + \frac{-45}{17} = \frac{-85}{17} = -5 \][/tex]
Therefore, the solution to the system of equations is:
[tex]\[ (x, y) = (3, -5) \][/tex]
Given equations:
[tex]\[ 15x + 7y = 10 \tag{1} \][/tex]
[tex]\[ 8x + 6y = -6 \tag{2} \][/tex]
Part 1: Write the system of equations in [tex]\( AX = B \)[/tex] form
In matrix notation, we can represent the system as follows:
[tex]\[ A = \begin{pmatrix} 15 & 7 \\ 8 & 6 \end{pmatrix}, \quad X = \begin{pmatrix} x \\ y \end{pmatrix}, \quad B = \begin{pmatrix} 10 \\ -6 \end{pmatrix} \][/tex]
So the equation [tex]\( AX = B \)[/tex] can be written as:
[tex]\[ \begin{pmatrix} 15 & 7 \\ 8 & 6 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 10 \\ -6 \end{pmatrix} \][/tex]
Part 2: Find [tex]\( A^{-1} \)[/tex]
The inverse of a 2x2 matrix [tex]\( A = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \)[/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]\[ A = \begin{pmatrix} 15 & 7 \\ 8 & 6 \end{pmatrix} \][/tex]
We calculate the determinant [tex]\( \text{det}(A) \)[/tex]:
[tex]\[ \text{det}(A) = 15 \cdot 6 - 7 \cdot 8 = 90 - 56 = 34 \][/tex]
Thus, the inverse [tex]\( A^{-1} \)[/tex] is:
[tex]\[ A^{-1} = \frac{1}{34} \begin{pmatrix} 6 & -7 \\ -8 & 15 \end{pmatrix} \][/tex]
Simplifying, we get:
[tex]\[ A^{-1} = \begin{pmatrix} \frac{6}{34} & \frac{-7}{34} \\ \frac{-8}{34} & \frac{15}{34} \end{pmatrix} = \begin{pmatrix} \frac{3}{17} & \frac{-7}{34} \\ \frac{-4}{17} & \frac{15}{34} \end{pmatrix} \][/tex]
Part 3: Solve the system using the matrix equation
To find [tex]\( X \)[/tex] from [tex]\( AX = B \)[/tex], we use the formula [tex]\( X = A^{-1}B \)[/tex].
Multiply [tex]\( A^{-1} \)[/tex] by [tex]\( B \)[/tex]:
[tex]\[ A^{-1} = \begin{pmatrix} \frac{3}{17} & \frac{-7}{34} \\ \frac{-4}{17} & \frac{15}{34} \end{pmatrix} , \quad B = \begin{pmatrix} 10 \\ -6 \end{pmatrix} \][/tex]
[tex]\[ X = \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} \frac{3}{17} & \frac{-7}{34} \\ \frac{-4}{17} & \frac{15}{34} \end{pmatrix} \begin{pmatrix} 10 \\ -6 \end{pmatrix} \][/tex]
Calculating the product:
[tex]\[ x = \frac{3}{17} \cdot 10 + \frac{-7}{34} \cdot (-6) = \frac{30}{17} + \frac{42}{34} = \frac{30}{17} + \frac{21}{17} = \frac{51}{17} = 3 \][/tex]
[tex]\[ y = \frac{-4}{17} \cdot 10 + \frac{15}{34} \cdot (-6) = \frac{-40}{17} + \frac{-90}{34} = \frac{-40}{17} + \frac{-45}{17} = \frac{-85}{17} = -5 \][/tex]
Therefore, the solution to the system of equations is:
[tex]\[ (x, y) = (3, -5) \][/tex]
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