Use hand row operation to find inverse

Question

$$\begin{cases} 2x+4y=-8\\ 4x-6y=-2 \end{cases}$$

I'm having trouble with hand row operations. Trying to teach myself. Could anyone help? Sorry for the bad format, I'm on my phone

Answer 1

You have the system $$\begin{cases} 2x+4y=-8\\ 4x-6y=-2 \end{cases} \iff \pmatrix{2 & 4 \\ 4 & -6}\pmatrix{x \\ y}=\pmatrix{-8 \\ -2}$$

I'm going to assume you want to solve this system by finding the inverse of the matrix $\pmatrix{2 & 4 \\ 4 & -6}$. That's doable.

There are formulas for the inverse matrices for $n\times n$ matrices for different sizes but let's just go through the most basic way of finding the inverse matrix: Gaussian elimination.

The idea here is to set of a block matrix of the form $\left[A\mid I\right]$ where $A$ is your matrix and $I$ is the identity matrix. Then perform elementary row operations until you get this augmented matrix into the form $\left[I\mid B\right]$. If you're successful (not every square matrix is invertible) then $B$ will be the inverse of $A$. So let's try it:

$$\begin{align}\left(\begin{array}{cc|cc} 2 & 4 & 1 & 0 \\ 4 & -6 & 0 & 1\end{array}\right) &\longrightarrow \left(\begin{array}{cc|cc} 2 & 4 & 1 & 0 \\ 0 & -14 & -2 & 1\end{array}\right) &\left(R_2\to R_2-2R_1\right)\\ &\longrightarrow \left(\begin{array}{cc|cc} 2 & 4 & 1 & 0 \\ 0 & 1 & \frac 17 & -\frac 1{14}\end{array}\right) &\left(R_2 \to -\frac 1{14}R_2\right)\\ &\longrightarrow \left(\begin{array}{cc|cc} 2 & 0 & \frac 37 & \frac 27 \\ 0 & 1 & \frac 17 & -\frac 1{14} \end{array}\right) &\left(R_1\to R_1-4R_2\right)\\ &\longrightarrow \left(\begin{array}{cc|cc} 1 & 0 & \frac 3{14} & \frac 17 \\ 0 & 1 & \frac 17 & -\frac 1{14} \end{array}\right) &\left(R_1\to \frac{1}{2}R_1\right)\end{align}$$

Thus $\frac 1{14}\pmatrix{3 & 2 \\ 2 & -1}$ should be the inverse matrix. Now just multiply both sides of your original matrix equation by this inverse matrix to get $$\pmatrix{x \\ y} = \frac 1{14}\pmatrix{3 & 2 \\ 2 & -1}\pmatrix{-8 \\ -2} = \pmatrix{-2 \\ -1}$$

Then if you plug in this pair $(x,y)$ into the original equation you'll find we (I) didn't make any mistakes. Yay!

Answer 2

I assume you are trying to solve the system to find a solution for $x$ and $y$. You have multiple options on how to proceed.

One of the simplest to learn is the method of substitution.

$\begin{cases}2x+4y=-8\\4x-6y=-2\end{cases}$

Move the $4y$ to the other side of the first equation.

$\begin{cases}2x=-8-4y\\4x-6y=-2\end{cases}$

Divide both equations by two.

$\begin{cases}x=-4-2y\\2x-3y=-1\end{cases}$

The first equation tells you what $x$ is equal to in terms of $y$ and constants. Use this information in the second equation.

$\begin{cases}x=-4-2y\\ 2(-4-2y)-3y=-1\end{cases}$

Now, the second equation is only in terms of $y$ and can be simplified using elementary methods.

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Another method to learn is the method of elimination. Try to combine the equations in such a way that you completely remove one of the variables.

$\begin{cases}2x+4y=-8\\4x-6y=-2\end{cases}$

Taking twice the first equation and subtracting the second equation, you are left with:

$\begin{cases}2x+4y=-8\\14y=-14\end{cases}$

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In either method, you will learn a value for $y$. Using this value for $y$, you can then use that information to learn what $x$ is.

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Both methods are generalized and can be expressed using matrices and the process known as Gaussian Elimination.

Following the algorithm described in the link exactly, we first describe our system using an augmented matrix:

$\left[\begin{array}{cc|c}2&4&-8\\4&-6&-2\end{array}\right]$

First, I make sure the top left entry is nonzero by swapping rows if necessary. (Already done). Next I make it the top left entry a one by multiplying (equivalently dividing) by an appropriate number. In this case, I divide by $2$.

$\left[\begin{array}{cc|c}1&2&-4\\4&-6&-2\end{array}\right]$

Next, I want to clear out all other entries in the first column. (exactly the same as elimination normally). I do so by subtracting four times the first row away from the second row.

$\left[\begin{array}{cc|c}1&2&-4\\0&-14&14\end{array}\right]$

I continue the process by shifting my focus down and over, now looking at the second row second column entry and trying to make that a one and clearing out the rest of its column until I arrive at something looking like this:

$\left[\begin{array}{cc|c}1&0&*\\0&1&**\end{array}\right]$

This can then be reinterpreted as your system of equations reading $x=*$ and $y=**$