Given the equations [tex]15x + 7y = 10[/tex] and [tex]8x + 6y = -6[/tex],

1. Write the system of equations in [tex]AX = B[/tex] form:
[tex]\[
\begin{pmatrix}
15 & 7 \\
8 & 6 \\
\end{pmatrix}
\begin{pmatrix}
x \\
y \\
\end{pmatrix}
=
\begin{pmatrix}
10 \\
-6 \\
\end{pmatrix}
\][/tex]

2. Find [tex]A^{-1}[/tex]. Leave all answers in simplified fractional form.

3. Solve the system using the matrix equation. Write any answers as ordered pairs.



Answer :

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]