Question

Which equation can be used to solve ? HELPPP

Answer 1

Option D:

$\left[\begin{array}{l}x_{1} \\x_{2}\end{array}\right]=\left[\begin{array}{cc}0.5 & -3 \\0 & 1\end{array}\right]\left[\begin{array}{c}2 \\-3\end{array}\right]$

Solution:

Given equation:

$\left[\begin{array}{ll}2 & 6 \\0 & 1\end{array}\right]\left[\begin{array}{l}x_{1} \\x_{2}\end{array}\right]=\left[\begin{array}{l}2 \\-3\end{array}\right]$

where

$A=\left[\begin{array}{ll}2 & 6 \\0 & 1\end{array}\right]$ ,   $X=\left[\begin{array}{l}x_{1} \\x_{2}\end{array}\right]$,   $B =\left[\begin{array}{l}2 \\-3\end{array}\right]$

This is in the form of AX = B.

To solve this equation.

Multiply by $A^{-1}$ on both sides.

$A^{-1}AX=A^{-1}B$

$X=A^{-1} B$

To find $A^{-1}$ using matrix formula:

$\left[\begin{array}{ll}a & b \\c & d\end{array}\right]^{-1}=\frac{1}{a d-b c}\left[\begin{array}{cc}d & -b \\-c & a\end{array}\right]$

$\left[\begin{array}{ll}2 & 6 \\0 & 1\end{array}\right]^{-1}=\frac{1}{2\times1- 6\times0}\left[\begin{array}{cc}1 & -6 \\0 & 2\end{array}\right]$

                 $=\frac{1}{2}\left[\begin{array}{cc}1 & -6 \\0 & 2\end{array}\right]$

Multiply $\frac{1}{2}$ into inside the matrix.

                 $=\left[\begin{array}{cc}\frac{1}{2} & \frac{-6}{2} \\\frac{0}{2} & \frac{2}{2} \end{array}\right]$

                 $=\left[\begin{array}{cc}0.5 & -3 \\0 & 1\end{array}\right]$

Substitute into $X=A^{-1} B$, we get

$\left[\begin{array}{l}x_{1} \\x_{2}\end{array}\right]=\left[\begin{array}{cc}0.5 & -3 \\0 & 1\end{array}\right]\left[\begin{array}{c}2 \\-3\end{array}\right]$

This equation can be used to solve the given matrix.

Option D is the correct answer.