Lesson 12 – 5 Augmented Matrix Pre-calculus Part 2 of 3

Learning Objective To solve quadratic systems

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.*

1. Solve the system using the augmented matrix method. 𝑥+3𝑦+8𝑧=22 2𝑥−3𝑦+𝑧=5 3𝑥−𝑦+2𝑧=12  1 3 8 2 −3 1 3 −1 2 22 5 12 want 0 here 1 3 8 0 −9 −15 3 −1 2 22 −39 12 –2R1 + R2 –3R1 + R3 0 here now 1 3 8 0 −9 −15 0 −10 −22 22 −39 −54 10R2 + (–9)R3 now 0 here 1 3 8 0 −9 −15 0 0 48 22 −39 96

Augmented Matrix 1 3 8 0 −9 −15 0 0 48 22 −39 96 Algebra takes over! 1 3 8 0 −9 −15 0 0 48 22 −39 96 Augmented Matrix Algebra takes over! 48𝑧=96 −9𝑦−15 2 =−39 𝑧=2 −9𝑦−30=−39 −9𝑦=−9 𝑦=1 𝑥+3 1 +8 2 =22 If you are adept enough, you can try the first 2 steps at the same time to speed up the process. 𝑥+3+16=22 𝑥+19=22 𝑥=3 𝑥, 𝑦, 𝑧  3, 1, 2

Lesson 12 – 5 Augmented Matrix Pre-calculus Part 2 of 3

2. Solve the system using the augmented matrix method. 2𝑟−3𝑠+3𝑡=−15 3𝑟+2𝑠−5𝑡=19 4𝑟−4𝑠−2𝑡=−2  2 −3 3 3 2 −5 4 −4 −2 −15 19 −2 6 −9 9 −6 −4 10 −45 −38 2 −3 3 0 −13 19 −15 −83 0 2 −8 28 3R1 + (–2)R2 –2R1 + R3 2 −3 3 0 −13 19 0 0 −66 −15 −83 198 −4 6 −6 4 −4 −2 30 −2 2R2 + 13R3 −13𝑠+19 −3 =−83 −66𝑡=198 𝑡=−3 𝑠=2 0 −26 38 0 26 −104 −166 364 2𝑟−3 2 +3 −3 =−15 𝑟, 𝑠, 𝑡 𝑟=0  0, 2, −3

Lesson 12 – 5 Augmented Matrix Pre-calculus Part 3 of 3

An application is to find the equation of a circle (in general form) knowing 3 of its points Augmented Matrix 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): 4 + 81 + 2D + 9E + F = 0 For (8, 7): 64 + 49 + 8D + 7E + F = 0 For (–8, –1): 64 + 1 – 8D – E + F = 0 2𝐷+9𝐸+𝐹=−85 8𝐷+7𝐸+𝐹=−113 −8𝐷−𝐸+𝐹=−65

Augmented Matrix 2𝐷+9𝐸+𝐹=−85 8𝐷+7𝐸+𝐹=−113 −8𝐷−𝐸+𝐹=−65  2 9 1 8 7 1 −8 −1 1 −85 −113 −65 –4R1 + R2 R2 + R3 2 9 1 0 −29 −3 0 6 2 −85 227 −178 −8 −36 −4 8 7 1 340 −113 6R2 + 29R3 8 7 1 −8 −1 1 −113 −65 2 9 1 0 −29 −3 0 0 40 −85 227 −3800 0 −174 −18 0 174 58 1362 −5162

2 9 1 0 −29 −3 0 0 40 −85 227 −3800 Augmented Matrix Algebra takes over! 40𝐹=−3800 −29𝐸−3 −95 =227 𝐹=−95 −29𝐸+285=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

Assignment Pg. 630 #1, 5, 11, 15, 17, 23, 27, 37