To solve the following system of equations using the inverse matrix method:
[tex]\[ \frac{4}{x} + \frac{3}{y} = 1 \][/tex]
[tex]\[ \frac{3}{x} - \frac{2}{y} = \frac{1}{24} \][/tex]
First, let's rewrite these equations in a more standard linear system form. Suppose we introduce new variables for the reciprocals so that:
[tex]\[ u = \frac{1}{x} \quad \text{and} \quad v = \frac{1}{y} \][/tex]
In terms of [tex]\( u \)[/tex] and [tex]\( v \)[/tex], the system of equations is transformed to:
[tex]\[ 4u + 3v = 1 \][/tex]
[tex]\[ 3u - 2v = \frac{1}{24} \][/tex]
We can represent this linear system in matrix form [tex]\( A \mathbf{u} = \mathbf{b} \)[/tex], where:
[tex]\[ A = \begin{pmatrix}
4 & 3 \\
3 & -2
\end{pmatrix},
\quad \mathbf{u} = \begin{pmatrix}
u \\
v
\end{pmatrix},
\quad \mathbf{b} = \begin{pmatrix}
1 \\
\frac{1}{24}
\end{pmatrix} \][/tex]
To solve for [tex]\(\mathbf{u}\)[/tex], we need to find the inverse of matrix [tex]\(A\)[/tex] (denoted as [tex]\( A^{-1} \)[/tex]) and then multiply it by [tex]\( \mathbf{b} \)[/tex]. We have:
[tex]\[ \mathbf{u} = A^{-1} \mathbf{b} \][/tex]
The result of this calculation gives us the values of [tex]\( u \)[/tex] and [tex]\( v \)[/tex].
The solutions for [tex]\( u \)[/tex] and [tex]\( v \)[/tex] are:
[tex]\[ u = 0.125 \quad \text{and} \quad v = 0.16666667 \][/tex]
This means:
[tex]\[ \frac{1}{x} = 0.125 \quad \text{and} \quad \frac{1}{y} = 0.16666667 \][/tex]
Therefore, solving for [tex]\( x \)[/tex] and [tex]\( y \)[/tex], we get:
[tex]\[ x = \frac{1}{0.125} = 8 \][/tex]
[tex]\[ y = \frac{1}{0.16666667} \approx 6 \][/tex]
Hence, the solution to the system of equations is:
[tex]\[ x = 8 \][/tex]
[tex]\[ y = 6 \][/tex]
