12.5 Augmented Matrix Solutions

Another way to solve a system of equations uses an augmented matrix. In this method, we will create a “corner of zeros” and then let our algebra skills take over! *A lot of math is done in our heads, so be careful! Also, write good instructions to yourself to follow.* Let’s learn by doing! Ex 1) Solve the system using the augmented matrix method. –2R1 + R2 –3R1 + R310R2 + (–9)R3 want 0 here 0 here now now 0 here

Algebra takes over! 48z = 96 z = 2 –9y – 15(2) = –39 –9y – 30 = –39 –9y = –9 y = 1 x + 3(1) + 8(2) = 22 x = 22 x + 19 = 22 x = 3 (x, y, z)  (3, 1, 2) If you are adept enough, you can try the first 2 steps at the same time to speed up the process.

Ex 2) Solve 3R1 + (–2)R2 –2R1 + R3 2R2 + 13R3 *Try it with a partner from here! –66t = 198 t = –3 –13s + 19(–3) = –83 –13s – 57 = –83 –13s = –26 s = 2 2r – 3(2) + 3(–3) = –15 2r – 6 – 9 = –15 2r – 15 = –15 2r = 0 r = 0 (r, s, t)  (0, 2, –3)

An application is to find the equation of a circle (in general form) knowing 3 of its points. Ex 3) Determine the equation of the circle that passes through (2, 9), (8, 7), and (–8, –1). *remember a circle in general form is: x 2 + y 2 + Dx + Ey + F = 0 for (2, 9): D + 9E + F = 02D + 9E + F = –85 (8, 7): D + 7E + F = 08D + 7E + F = –113 (–8, –1): – 8D – E + F = 0–8D – E + F = –65 –4R1 + R2 R2 + R3

6R2 + 29R3 40F = –3800 F = –95 –29E – 3(–95) = 227 –29E = 227 –29E = –58 E = 2 2D + 9(2) + (–95) = –85 2D + 18 – 95 = –85 2D – 77 = –85 2D = –8 D = –4 x 2 + y 2 – 4x + 2y – 95 = 0

Homework #1205 Pg 630 #1, 5, 11, 15, 17, 23, 27, 37