# How to find the final form of augmented matrix when solving for the inverse of this system of equations? 2x+3y=4 4x+5y=5

## A) $\left[\begin{matrix}1 & 0 \\ 0 & 1\end{matrix}\right] | \left[\begin{matrix}\frac{5}{2} & - \frac{3}{2} \\ - 3 & 4\end{matrix}\right]$ B) $\left[\begin{matrix}1 & 0 \\ 0 & 1\end{matrix}\right] | \left[\begin{matrix}- \frac{5}{2} & \frac{3}{2} \\ 2 & - 1\end{matrix}\right]$ C) $\left[\begin{matrix}1 & 0 \\ 0 & 1\end{matrix}\right] | \left[\begin{matrix}5 & - 3 \\ - 4 & 2\end{matrix}\right]$ D) $\left[\begin{matrix}1 & 0 \\ 0 & 1\end{matrix}\right] | \left[\begin{matrix}\frac{5}{2} & - \frac{3}{2} \\ - 2 & 1\end{matrix}\right]$ I have zero clue of what the question wants me to do and I've never heard of finding the inverse of an augmented matrix.... Plz halp :/

Jul 30, 2018

#### Answer:

The answer is $o p t i o n \left(B\right)$

#### Explanation:

You have $2$ equations with $2$ unknowns

$\left\{\begin{matrix}2 x + 3 y = 4 \\ 4 y + 5 y = 5\end{matrix}\right.$

The equations can be written as

$A X = B$

where $A$ is a matrix

$\left(\begin{matrix}2 & 3 \\ 4 & 5\end{matrix}\right) \left(\begin{matrix}x \\ y\end{matrix}\right) = \left(\begin{matrix}4 \\ 5\end{matrix}\right)$

The solution is

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

Where ${A}^{-} 1$ is the inverse of matrix $A$

To calculate the inverse, proceed as follows,

Write side by side matrix $A$ and the unit matrix $I$

Perform row operations until the unit matrix $I$ is on the left and the inverse on the RHS

$\left(\begin{matrix}2 & 3 \\ 4 & 5\end{matrix}\right) | \left(\begin{matrix}1 & 0 \\ 0 & 1\end{matrix}\right)$

$R 2 \leftarrow \left(R 2 - 2 R 1\right)$

$\left(\begin{matrix}2 & 3 \\ 0 & - 1\end{matrix}\right) | \left(\begin{matrix}1 & 0 \\ - 2 & 1\end{matrix}\right)$

$R 2 \leftarrow \left(R 2 \cdot - 1\right)$

$\left(\begin{matrix}2 & 3 \\ 0 & 1\end{matrix}\right) | \left(\begin{matrix}1 & 0 \\ 2 & - 1\end{matrix}\right)$

$R 1 \leftarrow R 1 - 3 R 2$

$\left(\begin{matrix}2 & 0 \\ 0 & 1\end{matrix}\right) | \left(\begin{matrix}- 5 & 3 \\ 2 & - 1\end{matrix}\right)$

$R 1 \leftarrow \left(\frac{R 1}{2}\right)$

$\left(\begin{matrix}1 & 0 \\ 0 & 1\end{matrix}\right) | \left(\begin{matrix}- \frac{5}{2} & \frac{3}{2} \\ 2 & - 1\end{matrix}\right)$

The answer is $o p t i o n \left(B\right)$