ENGG2013 Unit 2 Linear Equations Jan, 2011.. Linear Equation in n variables a 1 x 1 + a 2 x 2 + …...
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Transcript of ENGG2013 Unit 2 Linear Equations Jan, 2011.. Linear Equation in n variables a 1 x 1 + a 2 x 2 + …...
![Page 1: ENGG2013 Unit 2 Linear Equations Jan, 2011.. Linear Equation in n variables a 1 x 1 + a 2 x 2 + … + a n x n = c – a 1, a 2, …, a n are called coefficients.](https://reader035.fdocuments.us/reader035/viewer/2022062715/56649d815503460f94a66048/html5/thumbnails/1.jpg)
ENGG2013Unit 2 Linear Equations
Jan, 2011.
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Linear Equation in n variables
• a1x1 + a2x2+ … + an xn = c– a1, a2, …, an are called coefficients (real numbers).
– x1, x2,…, xn are variables (or indeterminates).
– c is a constant term (real number).
• Example– 2x + 3y – 4z = 0.2
• Non-example– x2+y2=1
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Geometry of a linear equation
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Two variables: straight line
ax + by = c
Three variables: plane
ax + by + cz = d
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System of linear equations
• A system of linear equations (or linear system) is a collection of one or more linear equations.– for example:
• A solution is a list of numbers (s1, s2, …, sn) which satisfies all equalities after substituting xi by si, for i =1,2,…,n.
• The set of all solutions is called the solution set.
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Nutrition problem• Find a combination of food A, B, C and D in order to satisfy the nutrition requirement exactly.
• Let xA, xB, xC and xD be the amount of food A, B, C and D respectively.
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Food A Food B Food C Food D Requirement
Protein 9 8 3 3 5
Carbohydrate 15 11 1 4 5
Vitamin A 0.02 0.003 0.01 0.006 0.01
Vitamin C 0.01 0.01 0.005 0.05 0.01
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Formal notation
• Given a system of m linear equations in n variables
the solution set is defined as
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Doublesubscripts
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Review of set notation
• Set of Greek letters = {,,,,,,,,,, ,,,,,,,,,,,,,}
• Set of prime numbers = {2,3,5,7,11,13,17,23,29,31,37,41, …}
• Sphere with radius r centered at origin= {(x,y,z): x2+y2+z2=r2}
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finite
Countably infinite
Uncountably infinite
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Examples of solution sets
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{ (x,y): ax + by = c }
-1
-0.5
0
0.5
1
-1
-0.5
0
0.5
1-5
0
5
xy
z
x
y
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Consistency
• A linear system is called consistent if there is at least one solution, in other words, if the solution set is non-empty.
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x
yInconsistent,no solution
x
y
Consistent
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Classification
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Linear System
Inconsistent(no solution)
Consistent
Unique solution Infinitely many solutions
Tasks:Determine whether a linearsystem is consistent.If yes, find all solutions.
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Short-hand notation using matrix
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(2 rows, 4 columns)
(4 rows, 3 columns)Usually called the augmented matrix
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The nutrition example
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Elementary row operations
1. Interchange two rows2. Multiply a row by a non-zero constant3. Replace a row by the sum of itself and a
constant multiple of another row
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Facts: Elementary row operations do not change the solution(s). (There is no loss, and no gain, of information.)
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Illustration – row interchange
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Illustration – Multiply by constant
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2 2
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Illustration – Row replacement
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(1) (1) – (2) (1) (1) – (2)
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How to solve?
• Idea: Apply the three kinds of information-lossless elementary row operations, and transform the linear system into one which is easier to solve.
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Linear system in upper triangular matrix form can be easily solved
by backward substitution
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Carl Friedrich Gauss
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(1777~1855)
The old Deutsche 10-Mark note
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Gaussian elimination
• Step 0: Write the linear system in matrix format
• Step 1: Try to transform the matrix into upper triangular form
• Step 2: Solve for the variables one by one, in backward order
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![Page 20: ENGG2013 Unit 2 Linear Equations Jan, 2011.. Linear Equation in n variables a 1 x 1 + a 2 x 2 + … + a n x n = c – a 1, a 2, …, a n are called coefficients.](https://reader035.fdocuments.us/reader035/viewer/2022062715/56649d815503460f94a66048/html5/thumbnails/20.jpg)
Example 1 (row operations)
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(2) (2) – (1)
(3)
(2)
(1)
(3) (3) + (2)/2
Solve
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Example 1 (backward sub.)
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Upper triangular
(3) z = 7/3
(2) – 2y – (7/3) = 1 y = –5/3
(1) x+(–5/3)+(7/3) = 1 x = 1/3
Verify:x+y+z = 1/3 – 5/3 + 7/3 = 3/3 = 1x–y = (1/3) – (– 5/3) = 6/3 = 2y+2z = (– 5/3)+2(7/3) = 9/3 = 3
Solution: x=1/3, y = –5/3, z = 7/3(unique solution)
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Example 2 (row operations)
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(2) (2) – (1) (3) (3) – (1)
(3) (3) – 2 (2)
Solve
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Example 2 (backward sub.)
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z can be taken as a free variable.Let z to be any real number.
(1)
(2)
From (2), y = –1 – z
From (1), x +(–1 – z)+3z = 1 x = 2 – 2z
Solution: x= 2– 2z, y = –1–z, z = any real number.
Solution set = {(2 – 2z, –1–z, z): z is any real no.}(Infinitely many solutions)
Note: You can let y tobe the free variable as well,and obtain the solutions interms of y.
![Page 24: ENGG2013 Unit 2 Linear Equations Jan, 2011.. Linear Equation in n variables a 1 x 1 + a 2 x 2 + … + a n x n = c – a 1, a 2, …, a n are called coefficients.](https://reader035.fdocuments.us/reader035/viewer/2022062715/56649d815503460f94a66048/html5/thumbnails/24.jpg)
Example 2 (cont’d)
• Verification– x+y+3z
=(2 – 2z) + (– 1 – z) + 3z = 1
– x+2y+4z=(2 – 2z) + 2(– 1 – z) + 3z = 0
– x+3y+5z =(2 – 2z) + 3(– 1 – z) + 5z = – 1
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Solution: x= 2– 2z, y = –1–z, z = any real numberSolution set = {(2 – 2z, –1–z, z): z is any real no.}
-5
0
5
10
-50
510
-10
-8
-6
-4
-2
0
2
4
x
y
z
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Example 3 (row operations)
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Solve
(2) (2) – (1) (3) (3) – (1)
(3) (3) – 2(2)
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Example 3 (cont’d)
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Contradiction, cannot be true
Answer: the linear system is inconsistent
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Example 3 (picture)
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Cross-section
-5-4
-3-2
-10
1
-2
-1.8
-1.6
-1.4
-1.2
-1
0
0.5
1
1.5
2
2.5
x
y
z
No common intersection
An infinitely long triangular tubeis formed by the three planes
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Key concepts
• Three kinds of elementary row operations– The solution set is invariant under any elementary
row operation• Gaussian elimination
– Transform a linear system to upper triangular form– Backward substitution
• Three types of solutions– No solution– Unique solution– Infinitely many solutions
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