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

2. Learning Objective
To solve quadratic systems

3. 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.*

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

5. Augmented Matrix 1 3 8 0 −9 −15 0 0 48 22 −39 96 Algebra takes over!
−9 − −39 96
Augmented Matrix
Algebra takes over!
48𝑧=96
−9𝑦−15 2 =−39
𝑧=2
−9𝑦−30=−39
−9𝑦=−9
𝑦=1
𝑥 =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

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

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

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

9. 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):
D + 9E + F = 0
For (8, 7):
D + 7E + F = 0
For (–8, –1):
– 8D – E + F = 0
2𝐷+9𝐸+𝐹=−85 8𝐷+7𝐸+𝐹=−113 −8𝐷−𝐸+𝐹=−65

10. Augmented Matrix 2𝐷+9𝐸+𝐹=−85 8𝐷+7𝐸+𝐹=−113 −8𝐷−𝐸+𝐹=−65
 −8 −1 1 −85 −113 −65
–4R1 + R2
R2 + R3
−29 − − −178
−8 −36 − −113
6R2 + 29R3
8 7 1 −8 − −113 −65
−29 − − −3800
0 −174 − −5162

11. −29 − − −3800
Augmented Matrix
Algebra takes over!
40𝐹=−3800
−29𝐸−3 −95 =227
𝐹=−95
−29𝐸+285=227
−29𝐸=−58
𝐸=2
2𝐷 −95 =−85
2𝐷+18−95=−85
2𝐷−77=−85
2𝐷=−8
𝐷=−4
𝑥 2 + 𝑦 2 −4𝑥+2𝑦−95=0

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