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↓ Slide 1

Another look at the last example

↓ Slide 2

Observations

  1. The operations we applied to the original linear system don't change the set of solutions. This is because each operation is reversible.
  2. Writing out the variables $x,y,z$ each time is unnecessary:

\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*}

This method always works:

→ Slide 3

The augmented matrix and elementary operations

↓ Slide 4

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.

↓ Slide 5

Example

↓ Slide 6

Elementary operations on a system of linear equations

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

  1. list the equations in a different order; or
  2. multiply one of the equations by a non-zero real number; or
  3. 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.

↓ Slide 7

Why do elementary operations leave the solutions of systems unchanged?

↓ Slide 8

Elementary row operations on a matrix

Translate elementary operations on the linear system into operations on the rows of the augmented matrix:

  1. change the order of the rows of the matrix;
  2. multiply one of the rows of the matrix by a non-zero real number;
  3. replace row $j$ by “row $j$ ${}+{}$ $c\times {}$ (row $i$)”, where $c$ is a non-zero real number and $i\ne j$.
  • These operations are called elementary row operations or EROs on the matrix.
  • The systems of linear equations corresponding to these matrices have exactly the same solutions.
↓ Slide 9

Example

Use EROs to find the intersection of the planes \begin{align*} 3x+4y+7z&=2\\x+3z&=0\\y-2z&=5\end{align*}

↓ Slide 10

Solution 1

\begin{align*} \def\go#1#2#3{\left[\begin{smallmatrix}#1\\#2\\#3\end{smallmatrix}\right]} \def\ar#1{\\\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*}

$\go{1&0&3&0}{0&1&-2&5}{0&0&1&-3}$

↓ Slide 11

Solution 2

We start in the same way, but by performing more EROs we make the algebra at the end simpler.

\begin{align*} \def\go#1#2#3{\left[\begin{smallmatrix}#1\\#2\\#3\end{smallmatrix}\right]} \def\ar#1{\\\xrightarrow{#1}&} &\go{3&4&7&2}{1&0&3&0}{0&1&-2&5} \ar{\ldots \text{same EROs as above}\ldots}\go{1&0&3&0}{0&1&-2&5}{0&0&1&-3} \ar{R2\to R2+2R3}\go{1&0&3&0}{0&1&0&-1}{0&0&1&-3} \ar{R1\to R1-3R3}\go{1&0&0&9}{0&1&0&-1}{0&0&1&-3} \end{align*}

\[ \go{1&0&0&9}{0&1&0&-1}{0&0&1&-3}\]

↓ Slide 12

Discussion

Both solutions use EROs to transform the augmented matrix.