To solve the given system of linear equations:
[tex]\[
\begin{aligned}
4x + 8y = 7, \\
2x - 3y = 0,
\end{aligned}
\][/tex]
we can represent it in matrix form as:
[tex]\[
A \mathbf{x} = \mathbf{B},
\][/tex]
where [tex]\(A\)[/tex] is the coefficient matrix, [tex]\(\mathbf{x}\)[/tex] is the column vector of variables, and [tex]\(\mathbf{B}\)[/tex] is the column vector of constants.
[tex]\[
A = \begin{pmatrix}
4 & 8 \\
2 & -3
\end{pmatrix}, \quad
\mathbf{x} = \begin{pmatrix}
x \\
y
\end{pmatrix}, \quad
\mathbf{B} = \begin{pmatrix}
7 \\
0
\end{pmatrix}.
\][/tex]
To solve for [tex]\(\mathbf{x}\)[/tex], we need to find the inverse of matrix [tex]\(A\)[/tex] (denoted as [tex]\(A^{-1}\)[/tex]) and multiply it with [tex]\(\mathbf{B}\)[/tex]:
[tex]\[
\mathbf{x} = A^{-1} \mathbf{B}.
\][/tex]
So let's check which given option matches the correct representation of [tex]\( A^{-1} \mathbf{B} \)[/tex]:
Given options:
a.
[tex]\[
M = \frac{1}{4}
\begin{pmatrix}
3 & 8 \\
2 & -4
\end{pmatrix}
\begin{pmatrix}
0 \\
7
\end{pmatrix}
\][/tex]
b.
[tex]\[
M = \frac{-1}{28}
\begin{pmatrix}
-3 & -8 \\
-2 & 4
\end{pmatrix}
\begin{pmatrix}
7 \\
0
\end{pmatrix}
\][/tex]
c.
[tex]\[
M = \frac{1}{28}
\begin{pmatrix}
3 & -2 \\
-8 & -4
\end{pmatrix}
\begin{pmatrix}
-7 \\
0
\end{pmatrix}
\][/tex]
d.
[tex]\[
M = \frac{1}{4}
\begin{pmatrix}
4 & 2 \\
8 & -3
\end{pmatrix}
\begin{pmatrix}
7 \\
0
\end{pmatrix}
\][/tex]
To determine the correct choice, we apply the steps involved in finding the inverse of matrix [tex]\(A\)[/tex]:
The inverse of a 2x2 matrix [tex]\(A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}\)[/tex] is:
[tex]\[
A^{-1} = \frac{1}{ad - bc} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix}.
\][/tex]
For our matrix [tex]\(A = \begin{pmatrix} 4 & 8 \\ 2 & -3 \end{pmatrix}\)[/tex]:
Determinant: [tex]\( \text{det}(A) = (4 \times -3) - (8 \times 2) = -12 - 16 = -28 \)[/tex].
So,
[tex]\[
A^{-1} = \frac{1}{-28} \begin{pmatrix} -3 & -8 \\ -2 & 4 \end{pmatrix}.
\][/tex]
Now multiply [tex]\(A^{-1}\)[/tex] with [tex]\(\mathbf{B}\)[/tex]:
[tex]\[
A^{-1} \mathbf{B} = \frac{1}{-28} \begin{pmatrix} -3 & -8 \\ -2 & 4 \end{pmatrix} \begin{pmatrix} 7 \\ 0 \end{pmatrix}.
\][/tex]
This matches option b:
[tex]\[
M = \frac{-1}{28} \begin{pmatrix} -3 & -8 \\ -2 & 4 \end{pmatrix} \begin{pmatrix} 7 \\ 0 \end{pmatrix}.
\][/tex]
Thus, the correct option is:
[tex]\[
\boxed{b}
\][/tex]
