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2. Linear Equations2. Linear Equations
Objectives:
1. Introduction to Gaussian Elimination.
2. Using multiple row operations.
3. Exercise - let’s see if you can do it.
Refs: B & Z 4.2, 4.3.
Example 1 (revisited):
constant
€
3 −1 −2
1 1 6
⎡
⎣ ⎢
⎤
⎦ ⎥
x y
€
R1 → R2
€
R2 → R1
Use operation (A)
€
R2 − 3R1 → R2
Use operation (C)
€
−1 4 R2 → R2
Use operation (B)€
1 1 6
3 −1 −2
⎡
⎣ ⎢
⎤
⎦ ⎥~
€
1 1 6
0 −4 −20
⎡
⎣ ⎢
⎤
⎦ ⎥~
€
1 1 6
0 1 5
⎡
⎣ ⎢
⎤
⎦ ⎥~
€
1 0 1
0 1 5
⎡
⎣ ⎢
⎤
⎦ ⎥~
€
R1 − R2 → R1
Use operation (C)
Bingo! Read off solution.
x y constant
Solution: x=1 y=5
We have just used a procedure known as Gaussian elimination (or row reduction) which transforms a matrix
The procedure also applies to larger matrices.
€
a b e
c d g
⎡
⎣ ⎢
⎤
⎦ ⎥
into
€
1 0 h
0 1 i
⎡
⎣ ⎢
⎤
⎦ ⎥
Example 2:
Solve for x, y and z:
€
2x + y − 4z = 3
x − 2y + 3z = 4
−3x + 4y − z = −2.
The first step is to construct the augmented matrix:
€
2 1 −4 3
1 −2 3 4
−3 4 −1 −2
⎡
⎣
⎢ ⎢ ⎢
⎤
⎦
⎥ ⎥ ⎥€
← equation 1
€
← equation 2
€
← equation 3
x coefficientsconstant terms
z coefficientsy coefficients
Our aim is to produce an equivalent augmented matrix which has 1’s on the diagonal and zeroes elsewhere (onthe LHS).
€
2 1 −4 3
1 −2 3 4
−3 4 −1 −2
⎡
⎣
⎢ ⎢ ⎢
⎤
⎦
⎥ ⎥ ⎥
€
1 −2 3 4
0 5 −10 −5
−3 4 −1 −2
⎡
⎣
⎢ ⎢ ⎢
⎤
⎦
⎥ ⎥ ⎥
~
~
€
1 −2 3 4
2 1 −4 3
−3 4 −1 −2
⎡
⎣
⎢ ⎢ ⎢
⎤
⎦
⎥ ⎥ ⎥
2
€
R1 → R2
€
R2 → R1
Use (A) to get a 1 in the top left corner
€
R2 − 2R1 → R2
Use (C) to get a 0 in the position indicated
€
R3 + 3R1 → R3 Use (C) to get a 0 in the bottom left position
€
R1 + 2R2 → R1
Use (C) to get a 0 in the positionindicated
€
R3 + 2R2 → R3
Use (C) to get a 0 in the position indicated
€
14 R3 → R3 Use (B) to get a 1 in the
bottom right corner
~
€
1 0 −1 2
0 1 −2 −1
0 −2 8 10
⎡
⎣
⎢ ⎢ ⎢
⎤
⎦
⎥ ⎥ ⎥-2
~
€
1 0 −1 2
0 1 −2 −1
0 0 4 8
⎡
⎣
⎢ ⎢ ⎢
⎤
⎦
⎥ ⎥ ⎥4
Use (B) to get a 1 in the centre
€
15 R2 → R2
€
1 −2 3 4
0 1 −2 −1
0 −2 8 10
⎡
⎣
⎢ ⎢ ⎢
⎤
⎦
⎥ ⎥ ⎥
-2~€
1 −2 3 4
0 5 −10 −5
0 −2 8 10
⎡
⎣
⎢ ⎢ ⎢
⎤
⎦
⎥ ⎥ ⎥
5~
Solution:
€
x = 4, y = 3, z = 2.
Always substitute these values back into ALL of your equations to check your solution.
Note: It is very easy to make algebraic mistakes!!!
€
1 0 0 4
0 1 −2 −1
0 0 1 2
⎡
⎣
⎢ ⎢ ⎢
⎤
⎦
⎥ ⎥ ⎥
-2~
€
1 0 0 4
0 1 0 3
0 0 1 2
⎡
⎣
⎢ ⎢ ⎢
⎤
⎦
⎥ ⎥ ⎥
~
€
R2 + 2R3 → R2Use (C) to get a 0 in the position indicated
We now have the required form
x constantzy
€
1 0 −1 2
0 1 −2 −1
0 0 1 2
⎡
⎣
⎢ ⎢ ⎢
⎤
⎦
⎥ ⎥ ⎥
~
€
R1 + R3 → R1Use (C) to get a 0 in the Top right corner
Check:
€
2x + y − 4z = 3
x − 2y + 3z = 4
−3x + 4y − z = −2,
x = 4, y = 3, z = 2?
€
1. 2(4)+ (3) − 4(2)
= 8 + 3− 8 = 3
€
2. 4 − 2(3)+ 3(2)
= 4 − 6 + 6 = 4
€
3. − 3(4)+ 4(3) − 2
= −12 +12 − 2 = −2.
Using multiple operationsWe can alter more than one row at a time to speed up the Gaussian elimination procedure.
Example 3:
€
4 2 8
9 3 6
⎡
⎣ ⎢
⎤
⎦ ⎥
€
4 2 8
9 3 6
⎡
⎣ ⎢
⎤
⎦ ⎥
€
2 1 4
9 3 6
⎡
⎣ ⎢
⎤
⎦ ⎥~
€
2 1 4
3 1 2
⎡
⎣ ⎢
⎤
⎦ ⎥~
€
2 1 4
3 1 2
⎡
⎣ ⎢
⎤
⎦ ⎥~
€
12 R1 → R1
€
13 R2 → R2
€
12 R1 → R1
€
13 R2 → R2
No problem - we have saved some time.(Here we are using multiple (B)Operations)
Example 4:
We can perform multiple (C ) operations provided at least one row is kept constant and only multiples of it are used to perform the other operations.
€
1 2 −2 4
3 1 0 −1
2 2 −1 0
⎡
⎣
⎢ ⎢ ⎢
⎤
⎦
⎥ ⎥ ⎥
€
1 2 −2 4
0 −5 6 −13
2 2 −1 0
⎡
⎣
⎢ ⎢ ⎢
⎤
⎦
⎥ ⎥ ⎥
~
€
1 2 −2 4
0 −5 6 −13
0 −2 3 −8
⎡
⎣
⎢ ⎢ ⎢
⎤
⎦
⎥ ⎥ ⎥
~
€
R2 − 3R1 → R2
€
R3 − 2R1 → R3
Obviously, performing multiple (A) type operations causes no problem.
Exercise 1: Solve the following system of simultaneous equations:
€
2x + 4y − z = −3
x − 3y + 2z =11
4x − 2y + 5z = 21.
Example 4 (continued):
€
1 2 −2 4
3 1 0 −1
2 2 −1 0
⎡
⎣
⎢ ⎢ ⎢
⎤
⎦
⎥ ⎥ ⎥
€
R3 − 2R1 → R3
€
R2 − 3R1 → R2
€
1 2 −2 4
0 −5 6 −13
0 −2 3 −8
⎡
⎣
⎢ ⎢ ⎢
⎤
⎦
⎥ ⎥ ⎥
~No problem -We kept R1 constantAnd used it to get R2 and R3
Solution to Exercise
€
2 4 −1 −3
1 −3 2 11
⎡
⎣
⎢ ⎢ ⎢
⎤
⎦
⎥ ⎥ ⎥
€
⎡
⎣
⎢ ⎢ ⎢
⎤
⎦
⎥ ⎥ ⎥
€
2 4 −1 −3 ⎡
⎣
⎢ ⎢ ⎢
⎤
⎦
⎥ ⎥ ⎥
€
2 4 −1 −3
1 −3 2 11
4 −2 5 21
⎡
⎣
⎢ ⎢ ⎢
⎤
⎦
⎥ ⎥ ⎥€
R1 → R2
€
R2 → R1
€
R2 − 2R1 → R2
€
R3 − 4R1 → R3
€
R2 10 → R2
€
R3 2 → R3
€
R3 −10R2 → R3€
R1 + 3R2 → R1
€
€
R1 −1 2 R3 → R1
€
R2 +1 2 R3 → R2
€
1 −3 2 11
2 4 −1 −3
4 −2 5 21
⎡
⎣
⎢ ⎢ ⎢
⎤
⎦
⎥ ⎥ ⎥
~
€
1 −3 2 11
0 10 −5 −25
0 10 −3 −23
⎡
⎣
⎢ ⎢ ⎢
⎤
⎦
⎥ ⎥ ⎥
~
€
1 0 1 2 7 2
0 1 −1 2 −5 2
0 0 1 1
⎡
⎣
⎢ ⎢ ⎢
⎤
⎦
⎥ ⎥ ⎥
~€
1 0 1 2 7 2
0 1 −1 2 −5 2
0 0 2 2
⎡
⎣
⎢ ⎢ ⎢
⎤
⎦
⎥ ⎥ ⎥
~€
1 −3 2 11
0 1 −1 2 −5 2
0 10 −3 −23
⎡
⎣
⎢ ⎢ ⎢
⎤
⎦
⎥ ⎥ ⎥
~