Table of Contents
Let's look at the example from the end of Lecture 2 more closely: $$\begin{array}{ccccccrrr} x&+&3y&+&z&=&5&\quad&(1)\\ 2x&+&7y&+&4z&=&17&&(2)\end{array}$$ We find the solutions of this system by applying operations to the system to make a new system, aiming to end up with a very simple sort of system where we can see the solutions easily.
First replace equation (2) with $(2)-2\times (1)$. We'll call the resulting equations (1) and (2) again, although of course we end up with a different system of linear equations: $$\begin{array}{ccccccrrr} x&+&3y&+&z&=&5&\quad&(1)\\ &&y&+&2z&=&7&&(2)\end{array}$$ Now replace equation (1) with $(1)-3\times (2)$: $$\begin{array}{ccccccrrr} x&&&-&5z&=&-16&\quad&(1)\\ &&y&+&2z&=&7&&(2)\end{array}$$ Notice that we can now easily rearrange (1) to find $x$ in terms of $z$, and we can rearrange (2) to find $y$ in terms of $z$. Since $z$ can take any value, we write $z=t$ where $t$ is a “free parameter” (which means $t$ can be any real number, or $t\in \mathbb{R}$). \begin{align*} x&=-16+5t\\ y&=7-2t\\ z&=t,\qquad t\in \mathbb{R}\end{align*} We can also write this in so-called “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 (1) and (2) intersect.
Note for each different value of $t$, we get a different solutions (that is, a different point on the line of intersection). For example, setting $t=0$ we see that $(-16,7,0)$ is a solution; setting $t=1.5$, we see that $(-16+1.5\times 5,7+1.5\times (-2),1.5) = (-8.5,4,1.5)$ is another solution, and so on. 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:
\[ \begin{bmatrix} 1&3&1&5\\2&7&4&17\end{bmatrix}\] Notice that the first column corresponds to the $x$ variable, the second to $y$, the third to $z$ and the numbers in the final column are the right hand sides of the equations. Each row corresponds to one equation. So 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} \\[6pt]\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*} Now we translate this back into equations to solve: $$\begin{array}{ccccccrrr} x&&&-&5z&=&-16&\quad&(1)\\ &&y&+&2z&=&7&&(2)\end{array}$$ so \[ \begin{bmatrix} x\\y\\z\end{bmatrix}=\begin{bmatrix} -16\\7\\0\end{bmatrix}+t\begin{bmatrix} 5\\-2\\1\end{bmatrix},\qquad t\in \mathbb{R}.\]
This sort of thing works in general: we can 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, and then 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 solution set is the same).
Let's give some terminology which will allow us to make this process clear.
The augmented matrix of a system of linear equations
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 the entries of the matrix. We can be a bit more precise: the number in row $i$ and column $j$ is called the $(i,j)$ entry of the matrix.
Example
To find the augmented matrix of the linear system \begin{align*} 3x+4y+7z&=2\\x+3z&=0\\y-2z&=5 \end{align*} notice that 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 \[ \begin{bmatrix} 3&4&7&2\\1&0&3&0\\0&1&-2&5\end{bmatrix}.\]
- the $(2,3)$ entry of this matrix is $3$;
- the $(3,2)$ entry is $1$;
- the $(1,4)$ entry is $2$;
- the $(4,1)$ entry is undefined (since this matrix does not have a $4$th row).
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,
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.
Example
Use EROs to find the intersection of the planes \begin{align*} 3x+4y+7z&=2\\x+3z&=0\\y-2z&=5\end{align*}
Solution 1
\begin{align*} \def\go#1#2#3{\begin{bmatrix}#1\\#2\\#3\end{bmatrix}} \def\ar#1{\\[6pt]\xrightarrow{#1}&} &\go{3&4&7&2}{1&0&3&0}{0&1&-2&5} \ar{\text{reorder rows}}\go{1&0&3&0}{0&1&-2&5}{3&4&7&2} \ar{R3\to R3-3R1}\go{1&0&3&0}{0&1&-2&5}{0&4&-2&2} \ar{R3\to R3-4R2}\go{1&0&3&0}{0&1&-2&5}{0&0&6&-18} \ar{R3\to \tfrac16 R3}\go{1&0&3&0}{0&1&-2&5}{0&0&1&-3} \end{align*}
So
- from the last row, we get $z=-3$
- from the second row, we get $y-2z=5$, so $y-2(-3)=5$, so $y=-1$
- from the first row, we get $x+3z=0$, so $x+3(-3)=0$, so $x=9$
The conclusion is that \[ \begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}9\\-1\\-3\end{bmatrix}\] is the only solution.