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Another look at the last example

$\begin{array}{ccccccrrr} x&+&3y&+&z&=&5&\quad&(1)\\ 2x&+&7y&+&4z&=&17&&(2)\end{array}$
Find solutions of this system by applying operations
Aim to end up with a very simple sort of system where we can see the solutions easily.

$\begin{array}{ccccccrrr} x&+&3y&+&z&=&5&\quad&(1)\\ 2x&+&7y&+&4z&=&17&&(2)\end{array}$
Replace equation (2) with $(2)-2\times (1)$:
$\begin{array}{ccccccrrr} x&+&3y&+&z&=&5&\quad&(1)\\ &&y&+&2z&=&7&&(2)\end{array}$
Now replace equation (1) with $(1)-3\times (2)$
$\begin{array}{ccccccrrr} x&&&-&5z&=&-16&\quad&(1)\\ &&y&+&2z&=&7&&(2)\end{array}$

$\begin{array}{ccccccrrr} x&&&-&5z&=&-16&\quad&(1)\\ &&y&+&2z&=&7&&(2)\end{array}$
Can now easily rearrange (1) to find $x$ in terms of $z$, and we rearrange (2) to find $y$ in terms of $z$.
Since $z$ can take any value, write $z=t$ where $t$ is a “free parameter”
(which means $t$ can be any real number, or $t\in \mathbb{R}$).
Solution: \begin{align*} x&=-16+5t\\ y&=7-2t\\ z&=t,\qquad t\in \mathbb{R}\end{align*}

Solution: \begin{align*} x&=-16+5t\\ y&=7-2t\\ z&=t,\qquad t\in \mathbb{R}\end{align*}
Can also write this in “vector form”:
$\begin{bmatrix} x\\y\\z\end{bmatrix}=\begin{bmatrix} -16\\7\\0\end{bmatrix}+t\begin{bmatrix} 5\\-2\\1\end{bmatrix},\qquad t\in \mathbb{R}.$
This is the equation of the line where the two planes described by the original equations intersect.

$\begin{bmatrix} x\\y\\z\end{bmatrix}=\begin{bmatrix} -16\\7\\0\end{bmatrix}+t\begin{bmatrix} 5\\-2\\1\end{bmatrix},\qquad t\in \mathbb{R}$
For each value of $t$, we get a different solution (a different point on the line of intersection).
e.g. take $t=0$ to see that $(-16,7,0)$ is a solution
take $t=1.5$ to see that $(-16+1.5\times 5,7+1.5\times (-2),1.5) = (-8.5,4,1.5)$ is another solution
etc.
This works for any value $t\in\mathbb{R}$, and every solution may be written in this way

Observations

The operations we applied to the original linear system don't change the set of solutions. This is because each operation is reversible.
Writing out the variables $x,y,z$ each time is unnecessary. If we erase the variables from the system $$\begin{array}{ccccccrrr} x&+&3y&+&z&=&5&\quad&(1)\\ 2x&+&7y&+&4z&=&17&&(2)\end{array}$$ and write all the numbers in a grid, or a matrix, we get $\begin{bmatrix} 1&3&1&5\\2&7&4&17\end{bmatrix}$

$\begin{array}{ccccccrrr} x&+&3y&+&z&=&5&\quad&(1)\\ 2x&+&7y&+&4z&=&17&&(2)\end{array}$
$\begin{bmatrix} 1&3&1&5\\2&7&4&17\end{bmatrix}$
Each row corresponds to one equation.
The first column corresponds to the $x$ variable, the second to $y$, the third to $z$
Numbers in the final column are the right hand sides of the equations.
Instead of performing operations on equations, we can perform operations on the rows of this matrix.

\begin{align*} \begin{bmatrix} 1&3&1&5\\2&7&4&17\end{bmatrix} &\xrightarrow{R2\to R2-2\times R1} \begin{bmatrix} 1&3&1&5\\0&1&2&7\end{bmatrix} \\[6pt]&\xrightarrow{R1\to R1-3\times R1} \begin{bmatrix} 1&0&-5&-16\\0&1&2&7\end{bmatrix} \end{align*}

Translate back into equations and solve:
$\begin{array}{ccccccrrr} x&&&-&5z&=&-16&\quad&(1)\\ &&y&+&2z&=&7&&(2)\end{array}$
$\begin{bmatrix} x\\y\\z\end{bmatrix}=\begin{bmatrix} -16\\7\\0\end{bmatrix}+t\begin{bmatrix} 5\\-2\\1\end{bmatrix},\qquad t\in \mathbb{R}.$

This method always works:

take any system of linear equations
write down a corresponding matrix
perform certain reversible operations on the rows of this matrix to get a new matrix
write down a new system of linear equations with the same solutions as the original system.
If we do things in a sensible way then the new system will be easy to solve, so we'll be able to solve the original system (since the solutions are identical).
Let's give some terminology which will allow us to explain this clearly.

The augmented matrix and elementary operations

Definition

Given a system of linear equations: \begin{align*} a_{11}x_1+a_{12}x_2+\dots+a_{1m}x_m&=b_1\\ a_{21}x_1+a_{22}x_2+\dots+a_{2m}x_m&=b_2\\ \hphantom{a_{11}}\vdots \hphantom{x_1+a_{22}}\vdots\hphantom{x_2+\dots+{}a_{nn}} \vdots\ & \hphantom{{}={}\!} \vdots\\ a_{n1}x_1+a_{n2}x_2+\dots+a_{nm}x_m&=b_n \end{align*} its augmented matrix is \[ \begin{bmatrix} a_{11}&a_{12}&\dots &a_{1m}&b_1\\ a_{21}&a_{22}&\dots &a_{2m}&b_2\\ \vdots&\vdots& &\vdots&\vdots\\ a_{n1}&a_{n2}&\dots &a_{nm}&b_n \end{bmatrix}.\]

The numbers in this matrix are called its entries.

Example

Find the augmented matrix of the linear system\begin{align*}3x+4y+7z&=2\\x+3z&=0\\y-2z&=5\end{align*}
We can rewrite it as \begin{align*}3x+4y+7z&=2\\{\color{red}1}x+{\color{red}0y}+3z&=0\\{\color{red}0x}+{\color{red}1}y-2z&=5\end{align*}
So the augmented matrix is\[ \begin{bmatrix} 3&4&7&2\\1&0&3&0\\0&1&-2&5\end{bmatrix}.\]

Elementary operations on a system of linear equations

If we perform one of the following operations on a system of linear equations:

list the equations in a different order; or
multiply one of the equations by a non-zero real number; or
replace equation $j$ by “equation $j$ ${}+{}$ $c\times {}$ (equation $i$)”, where $c$ is a non-zero real number and $i\ne j$,

then the new system will have exactly the same solutions as the original system. These are called elementary operations on the linear system.

Why do elementary operations leave the solutions of systems unchanged?

we are doing the same thing to the left hand side and the right hand side of each equation, so any solution to the original system will also be a solution to the new system; and
these operations are reversible, using operations of the same type, so any solution to the new system will also be a solution to the original system.

Elementary row operations on a matrix

Recall that when we form the augmented matrix of a linear system, each equation in the system becomes a row of the matrix. So we can translate the elementary operations on the linear system into corresponding operations on the rows of the matrix. We get three different types:

change the order of the rows of the matrix;
multiply one of the rows of the matrix by a non-zero real number;
replace row $j$ by “row $j$ ${}+{}$ $c\times {}$ (row $i$)”, where $c$ is a non-zero real number and $i\ne j$.

The system of linear equations corresponding to these matrices will then have exactly the same solutions.

We call these operations elementary row operations or EROs on the matrix.

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