Upload
merle
View
27
Download
0
Embed Size (px)
DESCRIPTION
Section 8.3 – Systems of Linear Equations - Determinants. Using Determinants to Solve Systems of Equations. A determinant is a value that is obtained from a square matrix. . Section 8.3 – Systems of Linear Equations - Determinants. - PowerPoint PPT Presentation
Citation preview
Section 8.3 – Systems of Linear Equations - DeterminantsUsing Determinants to Solve Systems of Equations
A determinant is a value that is obtained from a square matrix. 𝑎 𝑏𝑐 𝑑
2 57 1
𝐷=|2 57 1|¿2 ∙1−5 ∙7¿2−35¿−33
Section 8.3 – Systems of Linear Equations - Determinants
𝐷𝑥𝐷𝑒𝑡𝑒𝑟𝑚𝑖𝑛𝑎𝑛𝑡𝑜𝑓 𝑦 :𝐷𝑒𝑡𝑒𝑟𝑚𝑖𝑛𝑎𝑛𝑡 𝑜𝑓 𝑥 :
𝐷𝑦
𝑥=𝐷𝑥
𝐷
𝑦=𝐷𝑦
𝐷
Using Determinants to Solve Systems of Equations. (pg. 579 #18)
𝑆𝑜𝑙𝑢𝑡𝑖𝑜𝑛 :(−1 ,2)
Section 8.3 – Systems of Linear Equations - Determinants
{ 𝑥+3 𝑦=52𝑥−3 𝑦=−8
𝐷=|1 32 −3|¿1 ∙(−3)−3∙2¿−3−6 ¿−9
𝐷𝑥=| 5 3−8 −3|¿5 ∙(−3)−3 ∙(−8)¿−15+24¿9
𝐷𝑦=|1 52 −8|¿1 ∙(−8)−5∙2¿−8−10¿−18
𝑥=𝐷𝑥
𝐷¿9−9¿−1 𝑦=
𝐷𝑦
𝐷¿−18−9 ¿2
Using Determinants to Solve Systems of Equations
Determinant of a 3 x 3 Matrix
Section 8.3 – Systems of Linear Equations - Determinants
|𝑎11 𝑎12 𝑎13
𝑎21 𝑎22 𝑎23
𝑎31 𝑎32 𝑎33|
|𝑎11 𝑎12 𝑎13
𝑎21 𝑎22 𝑎23
𝑎31 𝑎32 𝑎33||𝑎11 𝑎12 𝑎13
𝑎21 𝑎22 𝑎23
𝑎31 𝑎32 𝑎33||𝑎11 𝑎12 𝑎13
𝑎21 𝑎22 𝑎23
𝑎31 𝑎32 𝑎33|
3 x 3 DeterminantsCalculate the determinant of the given matrix.
Section 8.3 – Systems of Linear Equations - Determinants
3 4 5−4 6 31 −4 3 | 3 4 5
−4 6 31 −4 3|
𝐷=¿3| 6 3−4 3|−4|−4 3
1 3|+5|−4 61 −4|
𝐷=¿3 (6 ∙3−3 ∙ (−4 ))−4 ((−4 )∙3−3∙1)+5 ((−4 )∙(−4 )−6 ∙1)
𝐷=90+6 0+5 0𝐷=200
Using Determinants to Solve Systems of EquationsSection 8.3 – Systems of Linear Equations - Determinants
𝐷≠0
Using Determinants to Solve Systems of EquationsSection 8.3 – Systems of Linear Equations - Determinants
−2 𝑥+ 𝑦−4 𝑧=45 𝑥−2 𝑦=18𝑥−5 𝑧=17
𝐷=¿−2 (−2 ∙−5−0 ∙0)−1(5∙ (−5)−0 ∙1)+(−4)(5 ∙0−(−2) ∙1)𝐷=−3
𝐷𝑥=¿4(−2 ∙(−5)−0 ∙0)−1(18 ∙(−5)−0 ∙17)+(−4)(18 ∙0−(−2) ∙17 )𝐷𝑥=−6
𝑥=𝐷𝑥
𝐷 =−6−3
=2
Using Determinants to Solve Systems of EquationsSection 8.3 – Systems of Linear Equations - Determinants
𝐷=−3
𝐷𝑦=¿−2 (18 ∙(−5)−0 ∙17)−4 (5 ∙(−5)−0∙1)+(−4)(5 ∙17−18 ∙1)𝐷𝑦=12
𝑦=𝐷𝑦
𝐷 = 12−3
=−4
𝐷𝑧=¿−2 ((−2)∙17−18 ∙0)−1(5∙17−18 ∙1)+4 (5 ∙0−(−2)∙1)𝐷𝑧=9
𝑧=𝐷𝑧
𝐷 = 9−3
=−3
Using Determinants to Solve Systems of EquationsSection 8.3 – Systems of Linear Equations - Determinants
−2 𝑥+ 𝑦−4 𝑧=45 𝑥−2 𝑦=18𝑥−5 𝑧=17
𝑆𝑜𝑙𝑢𝑡𝑖𝑜𝑛 : (2 ,−4 ,−3)
Properties of DeterminantsSection 8.3 – Systems of Linear Equations - Determinants
|1 2 31 2 34 5 6| 𝐷=¿1|2 3
5 6|−2|1 34 6|+3|1 2
4 5|𝐷=¿1(2 ∙6−3 ∙5)−2 (1∙6−3 ∙4)+3 (1 ∙5−2∙4 )¿−3+12−9¿0
Properties of DeterminantsSection 8.3 – Systems of Linear Equations - Determinants
𝑅1→−2𝑟2+𝑟1
Use the properties of determinants to solve the following problems.Section 8.3 – Systems of Linear Equations - Determinants
|𝑥 𝑦 𝑧𝑢 𝑣 𝑤1 2 3 |=4
𝑃𝑔 .579 ¿44
|𝑥 𝑦 𝑧𝑢 𝑣 𝑤2 4 6 |=¿2|𝑥 𝑦 𝑧
𝑢 𝑣 𝑤1 2 3 |=¿ 2(4)=¿8
𝑃𝑔 .579¿46
| 1 2 3𝑥−𝑢 𝑦−𝑣 𝑧−𝑤𝑢 𝑣 𝑤 |=¿
(−1 ) (−1 ) 4=¿4
|1 2 3𝑥 𝑦 𝑧𝑢 𝑣 𝑤|=¿
𝑅2→𝑟2+𝑟3 𝑅1↔𝑅2
(−1)|𝑥 𝑦 𝑧1 2 3𝑢 𝑣 𝑤|=¿
𝑅2↔𝑅3
(−1)(−1)|𝑥 𝑦 𝑧𝑢 𝑣 𝑤1 2 3 |=¿