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]
