Option D:
[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]
Solution:
Given equation:
[tex]\left[\begin{array}{ll}2 & 6 \\0 & 1\end{array}\right]\left[\begin{array}{l}x_{1} \\x_{2}\end{array}\right]=\left[\begin{array}{l}2 \\-3\end{array}\right][/tex]
where
[tex]A=\left[\begin{array}{ll}2 & 6 \\0 & 1\end{array}\right][/tex] , [tex]X=\left[\begin{array}{l}x_{1} \\x_{2}\end{array}\right][/tex], [tex]B =\left[\begin{array}{l}2 \\-3\end{array}\right][/tex]
This is in the form of AX = B.
To solve this equation.
Multiply by [tex]A^{-1}[/tex] on both sides.
[tex]A^{-1}AX=A^{-1}B[/tex]
[tex]X=A^{-1} B[/tex]
To find [tex]A^{-1}[/tex] using matrix formula:
[tex]$\left[\begin{array}{ll}a & b \\c & d\end{array}\right]^{-1}=\frac{1}{a d-b c}\left[\begin{array}{cc}d & -b \\-c & a\end{array}\right][/tex]
[tex]$\left[\begin{array}{ll}2 & 6 \\0 & 1\end{array}\right]^{-1}=\frac{1}{2\times1- 6\times0}\left[\begin{array}{cc}1 & -6 \\0 & 2\end{array}\right][/tex]
[tex]$=\frac{1}{2}\left[\begin{array}{cc}1 & -6 \\0 & 2\end{array}\right][/tex]
Multiply [tex]\frac{1}{2}[/tex] into inside the matrix.
[tex]$=\left[\begin{array}{cc}\frac{1}{2} & \frac{-6}{2} \\\frac{0}{2} & \frac{2}{2} \end{array}\right][/tex]
[tex]=\left[\begin{array}{cc}0.5 & -3 \\0 & 1\end{array}\right][/tex]
Substitute into [tex]X=A^{-1} B[/tex], we get
[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]
This equation can be used to solve the given matrix.
Option D is the correct answer.
