To solve the system of linear equations given by the matrix equation
[tex]\[
\left[\begin{array}{cc}
2 & 6 \\
0 & 1
\end{array}\right]
\left[\begin{array}{c}
x_1 \\
x_2
\end{array}\right]
=
\left[\begin{array}{c}
2 \\
-3
\end{array}\right],
\][/tex]
we need to find the values of [tex]\( x_1 \)[/tex] and [tex]\( x_2 \)[/tex].
By inspecting this matrix equation, multiplying out the left side gives us:
[tex]\[
\left[\begin{array}{cc}
2 & 6 \\
0 & 1
\end{array}\right]
\left[\begin{array}{c}
x_1 \\
x_2
\end{array}\right]
=
\left[\begin{array}{c}
2x_1 + 6x_2 \\
x_2
\end{array}\right]
.
\][/tex]
This must equal the right-hand side:
[tex]\[
\left[\begin{array}{c}
2 \\
-3
\end{array}\right].
\][/tex]
Therefore, we get the following system of equations:
1. [tex]\( 2x_1 + 6x_2 = 2 \)[/tex]
2. [tex]\( x_2 = -3 \)[/tex]
Substitute [tex]\( x_2 = -3 \)[/tex] into the first equation:
[tex]\[
2x_1 + 6(-3) = 2.
\][/tex]
This simplifies to:
[tex]\[
2x_1 - 18 = 2,
\][/tex]
which further simplifies to:
[tex]\[
2x_1 = 20,
\][/tex]
so:
[tex]\[
x_1 = 10.
\][/tex]
Thus, the solution [tex]\( \left[\begin{array}{c}
x_1 \\
x_2
\end{array}\right] \)[/tex] is [tex]\( \left[\begin{array}{c}
10 \\
-3
\end{array}\right] \)[/tex].
Now, we need to identify which equation among the given choices correctly matches this solution process. The correct option is:
[tex]\[
\left[\begin{array}{l}
x_1 \\
x_2
\end{array}\right]
=
\left[\begin{array}{cc}
0.5 & -3 \\
0 & 1
\end{array}\right]
\left[\begin{array}{c}
2 \\
-3
\end{array}\right].
\][/tex]
Hence, the correct equation to solve the given problem is:
[tex]\[
\left[\begin{array}{l}
x_1 \\
x_2
\end{array}\right] =
\left[\begin{array}{cc}
0.5 & -3 \\
0 & 1
\end{array}\right]
\left[\begin{array}{c}
2 \\
-3
\end{array}\right].
\][/tex]
