system of linear equations - wordpress.com · 2020. 11. 10. · what is a system of linear equation...
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System of Linear Equations
What is a system of Linear equation ???
A system of equations is 2 or more equations which have the same variables.
Find one ( X, Y ) which solves both of these equations.
X + Y = 104X – Y = 0
This is a system of eq.
X + Y = 1……..(1)-X + Y = 1………(2)
How to Solve SLEs ??? Or
What type of Solutions we have for SLEs?
Example 1 Solve:
Elimination Method: If we simply add (1) & (2) then variable x from equation (2) is eliminated and we obtain by equation (2),
2Y = 2Y = 2/2 = 1
Therefore, by equation (1), X + Y = 1X + 1 = 1 X = 1 – 1 = 0
Graphical Method:
(2,0)
y
(0,0) (1,0) x
(0,1)
(-1,0)(0,-1) X + Y = 1-X + Y = 1
Therefore, Solution of the system is unique and it is (0,1).
X + Y = 1……..(1)X + Y = 2………(2)
Example 2 Solve:
Elimination Method: If we simply subtract (1) & (2) then variables x and y from equation (2) are eliminated and we obtain by equation (2),
0 = -1
Graphical Method:
(2,0)
y
(0,0) (1,0) x
(0,1)
(-1,0)(0,-1) X + Y = 1
X + Y = 2
Therefore, Solution of the system does not exist.
(0,2)
2X + 2Y = 2……..(1)X + Y = 1………(2)
Example 3 Solve:
Elimination Method: First we divide equation (1) by 2 to make coefficient of x in both the equation same, so we have
X + Y = 1……..(3)If we simply subtract (2) & (3) now then variables x and y from equation (2) are eliminated and we obtain by equation (2),
0 = 0
Graphical Method:
(2,0)
y
(0,0) (1,0) x
(0,1)
(-1,0)(0,-1) X + Y = 1
X + Y = 1Therefore, there are infinitely many Solutions of the system.
How to calculate all those solutions?
All such solutions are (1,0), (0,1), (1/2,1/2), (-1,2)…Y = k, k є R X = 1-k
(x, y) = {(1-k, k) / k є R}
What type of Solutions we have for SLEs?
• It may have Unique Solution• It may have Infinitely Many Solutions• Or Solution may not exist
Use of Matrix to Solve System of Linear Equations
SLEs to Matrix & Matrix to SLEs
1 1−1 1
𝑥𝑦 =
11
x + y = 1-x + y = 1
1x + 1y = 1-1x + 1y = 1
1 1−1 1
𝑥𝑦 =
11
1x + 1y = 1-1x + 1y = 1
x + y = 1-x + y = 1
x + y = 1-x + y = 1
Example 1 Solve using Matrix:
Therefore, Solution of the system is unique and it is (0,1).
1 1−1 1
𝑥𝑦 =
11
x + y = 1-x + y = 1
1x + 1y = 1-1x + 1y = 1
A X = B𝐴/𝐵 =
1 1 1−1 1 1
R1
R2
Augmented Matrix
R2 R2 + R1, 𝐴/𝐵 ≅1 1 10 2 2
1 10 2
𝑥𝑦 =
12
by row (2), 2Y = 2Y = 2/2 = 1
by row (1), X + Y = 1 X + 1 = 1 X = 1 – 1 = 0
X + Y = 1……..(1)-X + Y = 1………(2)
Example 1 Solve:
Elimination Method: If we simply add (1) & (2) then variable x from equation (2) is eliminated and we obtain by equation (2),
2Y = 2Y = 2/2 = 1
Therefore, by equation (1), X + Y = 1X + 1 = 1 X = 1 – 1 = 0
Graphical Method:
(2,0)
y
(0,0) (1,0) x
(0,1)
(-1,0)(0,-1) X + Y = 1-X + Y = 1
Therefore, Solution of the system is unique and it is (0,1).
Old Solution Methods
x + y = 1x + y = 2
Example 2 Solve using Matrix:
Therefore, Solution of the system does not exist.
1 11 1
𝑥𝑦 =
12
x + y = 1x + y = 2
1x + 1y = 11x + 1y = 2
A X = B𝐴/𝐵 =
1 1 11 1 2
R1
R2
Augmented Matrix
R2 R2 - R1, 𝐴/𝐵 ≅1 1 10 0 1
1 10 0
𝑥𝑦 =
11
by row (2), 0Y = 10 = 1 which Is not possible
X + Y = 1……..(1)X + Y = 2………(2)
Example 2 Solve:
Elimination Method: If we simply subtract (1) & (2) then variables x and y from equation (2) are eliminated and we obtain by equation (2),
0 = -1
Graphical Method:
(2,0)
y
(0,0) (1,0) x
(0,1)
(-1,0)(0,-1) X + Y = 1
X + Y = 2
Therefore, Solution of the system does not exist.
(0,2)
Old Solution Methods
2 x + 2y = 2x + y = 1
Example 3 Solve using Matrix:
Therefore, there are infinitely many solutions.
2 21 1
𝑥𝑦 =
21
2x + 2y = 2x + y = 1
2x + 2y = 21x + 1y = 1
A X = B𝐴/𝐵 =
2 2 21 1 1
R1
R2
Augmented Matrix
R1 R1/2 𝐴/𝐵 ≅1 1 11 1 1
1 10 0
𝑥𝑦 =
10
by row (2), 0Y = 00 = 0
R2 R2 – R1 𝐴/𝐵 ≅1 1 10 0 0
y = k, k є R, x = 1-k
(x, y) = {(1-k, k) / k є R}
2X + 2Y = 2……..(1)X + Y = 1………(2)
Example 3 Solve:
Elimination Method: First we divide equation (1) by 2 to make coefficient of x in both the equation same, so we have
X + Y = 1……..(3)If we simply subtract (2) & (3) now then variables x and y from equation (2) are eliminated and we obtain by equation (2),
0 = 0
Graphical Method:
(2,0)
y
(0,0) (1,0) x
(0,1)
(-1,0)(0,-1) X + Y = 1
X + Y = 1Therefore, there are infinitely many Solutions of the system.
How to calculate all those solutions?
All such solutions are (1,0), (0,1), (1/2,1/2), (-1,2)…Y = k, k є R X = 1-k
(x, y) = {(1-k, k) / k є R}
Old Solution Methods
a11x + a12y = b1a21x + a22y = b2
General Example for Solving 2x2 SLES using Matrix
𝑎11 𝑎12𝑎21 𝑎22
𝑥𝑦 =
𝑏1𝑏2
A X = B𝐴/𝐵 =
𝑎11 𝑎12 𝑏1𝑎21 𝑎22 𝑏2
R1 R1/a11,
𝐴/𝐵 ≅1 𝑎12/𝑎11 𝑏1/𝑎11𝑎21 𝑎22 𝑏2
1 𝑎12′0 𝑎22′
𝑥𝑦 = 𝑏1′
𝑏2′
R2 R2 – a21R1,
a11x + a12y = b1a21x + a22y = b2
𝐴/𝐵 ≅1 𝑎12/𝑎11 𝑏1/𝑎110 𝑎22 − 𝑎21𝑎12/𝑎11 𝑏2 − 𝑎21𝑏1/𝑎11
Case 1
𝐴 ≅ 1 𝑎12′0 𝑎22′
𝐴/𝐵 ≅ 1 𝑎12′ 𝑏1′0 𝑎22′ 𝑏2′
Case 2
Case 3
a22’ ≠ 0 Unique Solution
𝐴 ≅ 1 𝑎12′0 𝑎22′
𝐴/𝐵 ≅ 1 𝑎12′ 𝑏1′0 𝑎22′ 𝑏2′
a22’ = 0 b2’ ≠ 0 No Solution
𝐴 ≅ 1 𝑎12′0 𝑎22′
𝐴/𝐵 ≅ 1 𝑎12′ 𝑏1′0 𝑎22′ 𝑏2′
a22’ = 0 b2’ = 0 Infinitely Many Solutions
Next Lecture : Theories to Use Matrix Solution for SLEs