Answer :
Certainly! Let's solve the system of linear equations step by step and identify the matrices involved.
The system of equations is:
[tex]\[ \begin{cases} 4x + 5y = 12 \\ 6x - 2y = 15 \end{cases} \][/tex]
Step 1: Represent the system of equations in matrix form.
We can write the system as [tex]\( A \mathbf{x} = B \)[/tex], where:
[tex]\[ A = \begin{pmatrix} 4 & 5 \\ 6 & -2 \end{pmatrix}, \quad \mathbf{x} = \begin{pmatrix} x \\ y \end{pmatrix}, \quad B = \begin{pmatrix} 12 \\ 15 \end{pmatrix} \][/tex]
Matrix A:
[tex]\[ A = \begin{pmatrix} 4 & 5 \\ 6 & -2 \end{pmatrix} \][/tex]
Matrix B:
[tex]\[ B = \begin{pmatrix} 12 \\ 15 \end{pmatrix} \][/tex]
Step 2: Find the inverse of Matrix [tex]\( A \)[/tex], denoted as [tex]\( A^{-1} \)[/tex], if it exists.
The inverse of Matrix [tex]\( A \)[/tex] can be found using the formula for a 2x2 matrix [tex]\( \begin{pmatrix} a & b \\ c & d \end{pmatrix} \)[/tex]:
[tex]\[ A^{-1} = \frac{1}{ad - bc} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix} \][/tex]
For matrix [tex]\( A \)[/tex]:
[tex]\[ a = 4, \, b = 5, \, c = 6, \, d = -2 \][/tex]
The determinant of [tex]\( A \)[/tex] is:
[tex]\[ \text{det}(A) = ad - bc = (4 \cdot -2) - (5 \cdot 6) = -8 - 30 = -38 \][/tex]
Since the determinant is not zero, [tex]\( A \)[/tex] is invertible. The inverse is:
[tex]\[ A^{-1} = \frac{1}{-38} \begin{pmatrix} -2 & -5 \\ -6 & 4 \end{pmatrix} = \begin{pmatrix} \frac{2}{38} & \frac{5}{38} \\ \frac{6}{38} & \frac{-4}{38} \end{pmatrix} = \begin{pmatrix} \frac{1}{19} & \frac{5}{38} \\ \frac{3}{19} & \frac{-2}{19} \end{pmatrix} \][/tex]
Step 3: Multiply [tex]\( A^{-1} \)[/tex] by [tex]\( B \)[/tex] to find [tex]\( \mathbf{x} \)[/tex].
[tex]\[ \mathbf{x} = A^{-1} B \][/tex]
[tex]\[ \mathbf{x} = \begin{pmatrix} \frac{1}{19} & \frac{5}{38} \\ \frac{3}{19} & \frac{-2}{19} \end{pmatrix} \begin{pmatrix} 12 \\ 15 \end{pmatrix} \][/tex]
Simplifying the multiplication:
[tex]\[ \begin{pmatrix} \frac{1}{19} \times 12 + \frac{5}{38} \times 15 \\ \frac{3}{19} \times 12 + \frac{-2}{19} \times 15 \end{pmatrix} = \begin{pmatrix} 0.631578947368421 + 1.973684210526316 \\ 1.894736842105263 - 0.789473684210526 \end{pmatrix} = \begin{pmatrix} 2.6052631578947367 \\ 0.31578947368421073 \end{pmatrix} \][/tex]
Therefore, the solution to the system of equations is:
[tex]\[ x = 2.6052631578947367, \quad y = 0.31578947368421073 \][/tex]
So, the coordinates of the solution are approximately:
[tex]\[ (x, y) = \left(2.6052631578947367, 0.31578947368421073\right) \][/tex]
Consequently, the matrices are:
Matrix A:
[tex]\[ A = \begin{pmatrix} 4 & 5 \\ 6 & -2 \end{pmatrix} \][/tex]
Matrix B:
[tex]\[ B = \begin{pmatrix} 12 \\ 15 \end{pmatrix} \][/tex]
The system of equations is:
[tex]\[ \begin{cases} 4x + 5y = 12 \\ 6x - 2y = 15 \end{cases} \][/tex]
Step 1: Represent the system of equations in matrix form.
We can write the system as [tex]\( A \mathbf{x} = B \)[/tex], where:
[tex]\[ A = \begin{pmatrix} 4 & 5 \\ 6 & -2 \end{pmatrix}, \quad \mathbf{x} = \begin{pmatrix} x \\ y \end{pmatrix}, \quad B = \begin{pmatrix} 12 \\ 15 \end{pmatrix} \][/tex]
Matrix A:
[tex]\[ A = \begin{pmatrix} 4 & 5 \\ 6 & -2 \end{pmatrix} \][/tex]
Matrix B:
[tex]\[ B = \begin{pmatrix} 12 \\ 15 \end{pmatrix} \][/tex]
Step 2: Find the inverse of Matrix [tex]\( A \)[/tex], denoted as [tex]\( A^{-1} \)[/tex], if it exists.
The inverse of Matrix [tex]\( A \)[/tex] can be found using the formula for a 2x2 matrix [tex]\( \begin{pmatrix} a & b \\ c & d \end{pmatrix} \)[/tex]:
[tex]\[ A^{-1} = \frac{1}{ad - bc} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix} \][/tex]
For matrix [tex]\( A \)[/tex]:
[tex]\[ a = 4, \, b = 5, \, c = 6, \, d = -2 \][/tex]
The determinant of [tex]\( A \)[/tex] is:
[tex]\[ \text{det}(A) = ad - bc = (4 \cdot -2) - (5 \cdot 6) = -8 - 30 = -38 \][/tex]
Since the determinant is not zero, [tex]\( A \)[/tex] is invertible. The inverse is:
[tex]\[ A^{-1} = \frac{1}{-38} \begin{pmatrix} -2 & -5 \\ -6 & 4 \end{pmatrix} = \begin{pmatrix} \frac{2}{38} & \frac{5}{38} \\ \frac{6}{38} & \frac{-4}{38} \end{pmatrix} = \begin{pmatrix} \frac{1}{19} & \frac{5}{38} \\ \frac{3}{19} & \frac{-2}{19} \end{pmatrix} \][/tex]
Step 3: Multiply [tex]\( A^{-1} \)[/tex] by [tex]\( B \)[/tex] to find [tex]\( \mathbf{x} \)[/tex].
[tex]\[ \mathbf{x} = A^{-1} B \][/tex]
[tex]\[ \mathbf{x} = \begin{pmatrix} \frac{1}{19} & \frac{5}{38} \\ \frac{3}{19} & \frac{-2}{19} \end{pmatrix} \begin{pmatrix} 12 \\ 15 \end{pmatrix} \][/tex]
Simplifying the multiplication:
[tex]\[ \begin{pmatrix} \frac{1}{19} \times 12 + \frac{5}{38} \times 15 \\ \frac{3}{19} \times 12 + \frac{-2}{19} \times 15 \end{pmatrix} = \begin{pmatrix} 0.631578947368421 + 1.973684210526316 \\ 1.894736842105263 - 0.789473684210526 \end{pmatrix} = \begin{pmatrix} 2.6052631578947367 \\ 0.31578947368421073 \end{pmatrix} \][/tex]
Therefore, the solution to the system of equations is:
[tex]\[ x = 2.6052631578947367, \quad y = 0.31578947368421073 \][/tex]
So, the coordinates of the solution are approximately:
[tex]\[ (x, y) = \left(2.6052631578947367, 0.31578947368421073\right) \][/tex]
Consequently, the matrices are:
Matrix A:
[tex]\[ A = \begin{pmatrix} 4 & 5 \\ 6 & -2 \end{pmatrix} \][/tex]
Matrix B:
[tex]\[ B = \begin{pmatrix} 12 \\ 15 \end{pmatrix} \][/tex]
