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matrix_demo.html
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<!DOCTYPE html>
<html>
<head>
<meta charset="utf-8" />
<title>Matrices</title>
</head>
<body>
<script src="https://unpkg.com/[email protected]"></script>
<script src="https://cdn.rawgit.com/jasondavies/jsbn/v1.4/jsbn.js"></script>
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<style>
.katex { font-size: 1.0em; }
</style>
<script src="arithmetic.js"></script>
<script src="math.js"></script>
<script src="field_257.js"></script>
<script src="carryless.js"></script>
<script src="field_256.js"></script>
<script src="rational.js"></script>
<script src="matrix.js"></script>
<script src="demo_common.js"></script>
<script src="matrix_demo_common.js"></script>
<script src="cauchy_matrix_demo.js"></script>
<script src="row_reduce.js"></script>
<script src="matrix_inverse_demo.js"></script>
<style>
input:invalid { color: red; }
</style>
<div>
<h3>Example 1: Cauchy matrices</h3>
Working over the field of rational numbers, let
<span style="white-space: nowrap;">
<var>x</var> = [ 3, 4, 5 ]
</span>
and
<span style="white-space: nowrap;">
<var>y</var> = [ 0, 1, 2 ].
</span>
Then, the Cauchy matrix constructed from
<var>x</var> and <var>y</var> is
<pre>/ 1/3 1/2 1 \
| 1/4 1/3 1/2 |
\ 1/5 1/4 1/3 /,</pre>
which has inverse
<pre>/ 30 -180 180 \
| -36 192 -180 |
\ 9 -36 30 /.</pre>
</div>
<div id="cauchyMatrixDemo">
</div>
<script>
'use strict';
(function() {
const { h, render } = window.preact;
const root = document.getElementById('cauchyMatrixDemo');
render(h(CauchyMatrixDemo, {
initialX: '3, 4, 5', initialY: '0, 1, 2', initialFieldType: 'rational',
name: 'cauchyMatrixDemo',
header: h('h3', null, 'Interactive Example 1: Cauchy matrices'),
allowFieldTypeChanges: true,
}), root.parent, root);
})();
</script>
<style>
.swap-row-a { color: red; }
.swap-row-b { color: blue; }
.divide-row { color: red; }
.subtract-scaled-row-src { color: blue; }
.subtract-scaled-row-dest { color: red; }
</style>
<div>
<h3>Example 2: Matrix inversion via row reduction</h3>
Working over the field of rational numbers, let
<pre> / 0 2 2 \
M = | 3 4 5 |
\ 6 6 7 /.</pre>
The initial augmented matrix <var>A</var> is
<pre>/ 0 2 2 | 1 0 0 \
| 3 4 5 | 0 1 0 |
\ 6 6 7 | 0 0 1 /.</pre>
We need <var>A</var><sub>00</sub> to be non-zero, so swap rows <span class="swap-row-a">0</span> and <span class="swap-row-b">1</span>:
<pre>/ <span class="swap-row-a">0 2 2</span> | <span class="swap-row-a">1 0 0</span> \ / <span class="swap-row-b">3 4 5</span> | <span class="swap-row-b">0 1 0</span> \
| <span class="swap-row-b">3 4 5</span> | <span class="swap-row-b">0 1 0</span> | --> | <span class="swap-row-a">0 2 2</span> | <span class="swap-row-a">1 0 0</span> |
\ 6 6 7 | 0 0 1 / \ 6 6 7 | 0 0 1 /.</pre>
We need <var>A</var><sub>00</sub> to be 1, so divide row <span class="divide-row">0</span> by 3:
<pre>/ <span class="divide-row">3 4 5</span> | <span class="divide-row">0 1 0</span> \ / <span class="divide-row">1 4/3 5/3</span> | <span class="divide-row">0 1/3 0</span> \
| 0 2 2 | 1 0 0 | --> | 0 2 2 | 1 0 0 |
\ 6 6 7 | 0 0 1 / \ 6 6 7 | 0 0 1 /.</pre>
We need <var>A</var><sub>20</sub> to be 0, so subtract row <span class="subtract-scaled-row-src">0</span> scaled by 6 from row <span class="subtract-scaled-row-dest">2</span>:
<pre>/ <span class="subtract-scaled-row-src">1 4/3 5/3</span> | <span class="subtract-scaled-row-src">0 1/3 0</span> \ / 1 4/3 5/3 | 0 1/3 0 \
| 0 2 2 | 1 0 0 | --> | 0 2 2 | 1 0 0 |
\ <span class="subtract-scaled-row-dest">6 6 7</span> | <span class="subtract-scaled-row-dest">0 0 1</span> / \ <span class="subtract-scaled-row-dest">0 -2 -3</span> | <span class="subtract-scaled-row-dest">0 -2 1</span> /.</pre>
We need <var>A</var><sub>11</sub> to be 1, so divide row <span class="divide-row">1</span> by 2:
<pre>/ 1 4/3 5/3 | 0 1/3 0 \ / 1 4/3 5/3 | 0 1/3 0 \
| <span class="divide-row">0 2 2 </span> | <span class="divide-row"> 1 0 0</span> | --> | <span class="divide-row">0 1 1 </span> | <span class="divide-row">1/2 0 0</span> |
\ 0 -2 -3 | 0 -2 1 / \ 0 -2 -3 | 0 -2 1 /.</pre>
We need <var>A</var><sub>21</sub> to be 0, so subtract row <span class="subtract-scaled-row-src">1</span> scaled by −2 from row <span class="subtract-scaled-row-dest">2</span>:
<pre>/ 1 4/3 5/3 | 0 1/3 0 \ / 1 4/3 5/3 | 0 1/3 0 \
| <span class="subtract-scaled-row-src">0 1 1</span> | <span class="subtract-scaled-row-src">1/2 0 0</span> | --> | 0 1 1 | 1/2 0 0 |
\ <span class="subtract-scaled-row-dest">0 -2 -3</span> | <span class="subtract-scaled-row-dest">0 -2 1</span> / \ <span class="subtract-scaled-row-dest">0 0 -1</span> | <span class="subtract-scaled-row-dest">1 -2 1</span> /.</pre>
We need <var>A</var><sub>22</sub> to be 1, so divide row <span class="divide-row">2</span> by −1, which makes the left side of <var>A</var> a
unit upper triangular matrix:
<pre>/ 1 4/3 5/3 | 0 1/3 0 \ / 1 4/3 5/3 | 0 1/3 0 \
| 0 1 1 | 1/2 0 0 | --> | 0 1 1 | 1/2 0 0 |
\ <span class="divide-row">0 0 -1 </span> | <span class="divide-row"> 1 -2 1</span> / \ <span class="divide-row">0 0 1 </span> | <span class="divide-row">-1 2 -1</span> /.</pre>
We need <var>A</var><sub>12</sub> to be 0, so subtract row <span class="subtract-scaled-row-src">2</span> from row <span class="subtract-scaled-row-dest">1</span>:
<pre>/ 1 4/3 5/3 | 0 1/3 0 \ / 1 4/3 5/3 | 0 1/3 0 \
| <span class="subtract-scaled-row-dest">0 1 1</span> | <span class="subtract-scaled-row-dest">1/2 0 0</span> | --> | <span class="subtract-scaled-row-dest">0 1 0</span> | <span class="subtract-scaled-row-dest">3/2 -2 1</span> |
\ <span class="subtract-scaled-row-src">0 0 1</span> | <span class="subtract-scaled-row-src">-1 2 -1</span> / \ 0 0 1 | -1 2 -1 /.</pre>
We need <var>A</var><sub>02</sub> to be 0, so subtract row <span class="subtract-scaled-row-src">2</span> scaled by 5/3 from row <span class="subtract-scaled-row-dest">0</span>:
<pre>/ <span class="subtract-scaled-row-dest">1 4/3 5/3</span> | <span class="subtract-scaled-row-dest">0 1/3 0</span> \ / <span class="subtract-scaled-row-dest">1 4/3 0</span> | <span class="subtract-scaled-row-dest">5/3 -3 5/3</span> \
| 0 1 0 | 3/2 -2 1 | --> | 0 1 0 | 3/2 -2 1 |
\ <span class="subtract-scaled-row-src">0 0 1</span> | <span class="subtract-scaled-row-src">-1 2 -1</span> / \ 0 0 1 | -1 2 -1 /.</pre>
We need <var>A</var><sub>01</sub> to be 0, so subtract row <span class="subtract-scaled-row-src">1</span> scaled by 4/3 from row <span class="subtract-scaled-row-dest">0</span>, which makes the left side of <var>A</var> the identity matrix:
<pre>/ <span class="subtract-scaled-row-dest">1 4/3 0</span> | <span class="subtract-scaled-row-dest">5/3 -3 5/3</span> \ / <span class="subtract-scaled-row-dest">1 0 0</span> | <span class="subtract-scaled-row-dest">-1/3 -1/3 1/3</span> \
| <span class="subtract-scaled-row-src">0 1 0</span> | <span class="subtract-scaled-row-src">3/2 -2 1</span> | --> | 0 1 0 | 3/2 -2 1 |
\ 0 0 1 | -1 2 -1 / \ 0 0 1 | -1 2 -1 /.</pre>
Since the left side of <var>A</var> is the identity matrix, the right side of <var>A</var> is <var>M</var><sup>-1</sup>. Therefore,
<pre> / -1/3 -1/3 1/3 \
M^{-1} = | 3/2 -2 1 |
\ -1 2 -1 /.</pre>
</div>
<div id="matrixInverseDemo">
</div>
<script>
'use strict';
(function() {
const { h, render } = window.preact;
const root = document.getElementById('matrixInverseDemo');
render(h(MatrixInverseDemo, {
initialElements: '0, 2, 2, 3, 4, 5, 6, 6, 7', initialFieldType: 'rational',
name: 'matrixInverseDemo',
header: h('h3', null, 'Interactive Example 2: Matrix inversion via row reduction'),
allowFieldTypeChanges: true,
swapRowAColor: 'red',
swapRowBColor: 'blue',
divideRowColor: 'red',
subtractScaledRowSrcColor: 'blue',
subtractScaledRowDestColor: 'red',
}), root.parent, root);
})();
</script>
</body>
</html>