Sure! Let's determine the solution for the system of linear equations:
[tex]\[
\begin{aligned}
&x_1 + 2x_2 = 11 \\
&3x_1 + 4x_2 = 25
\end{aligned}
\][/tex]
First, we write this system in the matrix form [tex]\(\mathbf{A}\mathbf{x} = \mathbf{b}\)[/tex], where:
[tex]\[
\mathbf{A} = \begin{pmatrix}
1 & 2 \\
3 & 4 \\
\end{pmatrix}
, \quad
\mathbf{x} = \begin{pmatrix}
x_1 \\
x_2 \\
\end{pmatrix}
, \quad
\mathbf{b} = \begin{pmatrix}
11 \\
25 \\
\end{pmatrix}
\][/tex]
Now, we need to solve for [tex]\(\mathbf{x} = \begin{pmatrix}
x_1 \\
x_2 \\
\end{pmatrix}
\)[/tex].
Given the matrix equation:
[tex]\[
\mathbf{A}\mathbf{x} = \mathbf{b}
\][/tex]
we see that:
[tex]\[
\begin{pmatrix}
1 & 2 \\
3 & 4 \\
\end{pmatrix}
\begin{pmatrix}
x_1 \\
x_2 \\
\end{pmatrix}
=
\begin{pmatrix}
11 \\
25 \\
\end{pmatrix}
\][/tex]
To find [tex]\(\mathbf{x}\)[/tex], we use the inverse of [tex]\(\mathbf{A}\)[/tex] (if it exists). The solution can be found by:
[tex]\[
\mathbf{x} = \mathbf{A}^{-1}\mathbf{b}
\][/tex]
Given the computations or solving by substitution/elimination, we determine the solutions for [tex]\( x_1 \)[/tex] and [tex]\( x_2 \)[/tex]. The solution provided is:
[tex]\[
x_1 = 3, \quad x_2 = 4
\][/tex]
So, the detailed solutions to the system of linear equations:
[tex]\[
\begin{aligned}
x_1 + 2x_2 &= 11 \\
3x_1 + 4x_2 &= 25
\end{aligned}
\][/tex]
are [tex]\( x_1 \approx 2.9999999999999987 \)[/tex] and [tex]\( x_2 \approx 4.000000000000001 \)[/tex].
