6.5 - Applications of Matrices and Determinants

# 6.5 - Applications of Matrices and Determinants -

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<?xml version="1.0" encoding="iso-8859-1"?> <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"> <html xmlns="http://www.w3.org/1999/xhtml"> <head> <title>6.5 - Applications of Matrices and Determinants</title> <meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1" /> <link href=". ./m116.css" rel="stylesheet" type="text/css" /> <link href="matrix.css" rel="stylesheet" type="text/css" /> </head> <body> <h1>6.5 - Applications of Matrices and Determinants</h1> <h2><img src="triangle.gif" alt="Area of a Triangle" width="232" height="214" class="imgrt" />Area of a Triangle</h2> <p>Consider a triangle with vertices at (x<sub>1</sub>,y<sub>1</sub>), (x<sub>2</sub>,y<sub>2</sub>), and (x<sub>3</sub>,y<sub>3</sub>). If the triangle was a right triangle, it would be pretty easy to compute the area of the triangle by finding one-half the product of the base and the height. </p> <p>However, when the triangle is not a right triangle, there are a couple of other ways that the area can be found.</p> <h3>Heron's Formula</h3> <p><img src="herons.gif" alt="Triangle with vertices at (-2,2), (1,5), and (6,-1)" width="368" height="273" class="imgrt" />If you know the lengths of the three sides of the triangle, you can use Heron's Formula to find the area of the triangle. </p> <p>In Heron's formula, s is the semi-perimeter (one-half the perimeter of the triangle). </p> <p>s = 1/2 ( a + b + c )<br /> Area = sqrt ( s ( s-a) ( s-b) ( s-c) )</p> <p>Consider the triangle with vertices at (-2,2), (1,5), and (6,1). </p> <p>Using the distance formulas, we can find that the lengths of the sides (arbitrarily assigning a, b, and c) are a = 3 sqrt(2), b = sqrt(61), and c = sqrt(73).</p> <p>Using those values gives us . ..</p> <p>s = 1/2 ( 3 sqrt(2) + sqrt(61) + sqrt(73) )<br /> s - a = 1/2 ( - 3 sqrt(2) + sqrt(61) + sqrt(73) )<br /> s - b = 1/2 ( 3 sqrt(2) - sqrt(61) + sqrt(73) )<br /> s - c = 1/2 ( 3 sqrt(2) + sqrt(61) - sqrt(73) )</p> <p>s ( s - a ) ( s - b ) ( s - c ) = 1089 / 4</p> <p>When you take the square root of that, you get 33/2, so the area of that triangle is 16.5.</p> <h4><strong>Problems with Heron's Formula include</strong> </h4> <ul> <li>Must know the lengths of the sides of the triangle. If you don't then you have to use the distance formula to find the lengths of the sides of the triangle.</li>

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<li>You have to compute the semi-perimeter, so chances are you will have fractions to work with.</li> <li>Lots of square roots are involved. For the lengths of the sides of the triangle and for the area of the triangle.</li> <li>It's not the easiest thing in the world to work with.</li> </ul> <h3>Geometric Technique</h3> <p><img src="geoarea.gif" alt="Triangle with vertices at (-2,2), (1,5), and (6,-1) bounded by rectangle" width="368" height="278" class="imgrt" />The triangle can be enclosed in a rectangle. The vertices of the triangle will intersect the rectangle in three places, forming three right triangles. These triangles are denoted A, B, and C in the
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6.5 - Applications of Matrices and Determinants -

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