calculus – ii gauss elimination by dr. eman saad & dr. shorouk ossama
TRANSCRIPT
• Gauss Elimination:
We need write the equations in form of
augmented matrix for the system. The fourth
column consists of the constant on the right-
hand sides.
• The elementary operations referred to
previously become elementary row operations
on the matrix. The final matrix is said to be in
echelon form: that is, it has zeros below the
diagonal elements starting from the top left.
We can solve the equations by back
substation.
Example: Find the solution by using gauss elimination
2 R1 + R2 = R`2
- R1 + R3 = R`3
R1 - R3 = R`1
-(1/7) R3 + R2 = R`2
-(2/7) R2 + R3 = R`3 (-2) R2 + R1 = R`1
- (7/23) R3 = R`3
- (1/7) R2 = R`2
by back substation x3 = 2 x2 = -1 x1 = 1
1
2
3
4
5
Example: Using gauss elimination & back substitution,
solve the set of equations
-2 R1 + R2 = R`2 (-2/3) R3 + R4 = R`4
R2↔R3(-1/3) R3 = R`3
(3/5) R4 = R`4
R4 – R2= R`4
x4 = 3 x3 = 2 x2 = 1
x1 = -1
2
3
4
5
6
0 1 -1 2 5
0 1 -1 2 5
0 1 -1 2 5
0 0 -2 1 -1
0 0 0 5/3 5
0 0 0 1 3
X1 X2 X3
=
back substation
0 + 0 + 0 + X4 = 3
0 + 0 + X3 + 1/3 X4 = 3
0 + X2 + X3 + X4 = 6
X1 + X2 + 2X3 + 0 = 4
0 0 0 1 3
X4
• For two equations in two unknowns the
occurrences of unique solutions are easy to
detect.
• For higher-order sets of equations these
possibilities are not so obvious.
Consider the set of equations:
det A = = zero
• So the Cramer’s Rule will fail, although there
still may be solutions.
• We can determine whether solutions exist
more readily by using Gaussian Elimination.
• In this case the application of row operations
on the augmented matrix leads to:
-2 R1 + R2 = R`2
- R1 + R3 = R`3
By Back Substation
Row 3 is impossible to be satisfied
0 = 1
No Solution
(-1/2) R2 + R3 = R`3
1 1 -1 3 0 -4 1 -40 -2 3 -1
1
2
4
1 1 -1 33 -1 3 51 -1 2 2
1 1 -1 3 0 -4 6 -40 0 0 1
6
-2 R1 + R2 = R`2
- R1 + R3 = R`3By Back Substation
Row 3 is consistentRow 2 is -4y+6z=2
Hence, y = -1/4 ( 2 – 6z )
From Row 1, X = 1 – y + z = 3/2 – ½ zThus, z can take any value, say t
So the full solution set is:
(-1/2) R2 + R3 = R`3
1 1 -1 1 0 -4 6 2 0 0 0 0
1
2
1 1 -1 13 -1 3 51 -1 2 2
Infinite Number of Solution
Advantage of using gauss elimination more than cramer’s rule when the number of equations differs from the number of unknowns.
-3 R1 + R2 = R`1
-R1 + R3 = R`3
- R1 + R4 = R`4
Row 4 is consistent, while row 3 is implies z=3.
Then y and x can be found by back substitution row2 and 1
So: y= 4 , x=0
(-1/2) R2 + R3 = R`3
1 1 -1 1 0 -4 6 2
0 -2 3 0 0 -1 2 3/2
1
2
1 1 -1 13 -1 3 51 -1 2 2 1 0 1 3
1 1 -1 10 -4 6 2
0 0 -2 -6 0 0 0 0
Show that the following equations are inconsistent:
- R1 + R2 = R`2
-3 R1 + R3 = R`3
Row 3, 0 ≠ 2The system is
inconsistent
-2 R2 + R3 = R`3
1
2
1 1 -1 2 11 -2 3 -1 43 -3 7 0 7
1 1 1 2 10 -3 2 -3 30 -6 4 -6 4
1 1 12 1 10 -3 2 -3 30 0 0 0 -2
Determine the complete sets of values for a and b which make the equations :
-2 R1 + R2 = R`2
- R1 + R3 = R`3
1. System has unique solution if:
a ≠ -1, then z= (b-1/a+1)
2. System is inconsistency if:a = -1 and b ≠ 1
3. System has infinite solutions
If: a = -1 and b = 1
- R2 + R3 = R`3
1
2
1 -2 3 22 -1 2 31 1 a b
1 -2 3 10 3 -4 -1
0 0 a+1 b-1
Homogeneous Linear Systems:
A system of linear equations is said to be homogeneous if the
constant terms are all zero; that is, the system has the form:
Every homogeneous system of linear equations is consistent
because all such systems have x1=0,x2=0,…,xn=0 as a solution.
This solution is called the trivial solution; if there are other
solutions, they are called non-trivial solutions.
Because a homogeneous linear system always has the trivial solution, there are only two possibilities for its solutions:
• The system has only the trivial solution.• The system has infinitely many solutions in addition to the trivial solution.
In the special case of a homogeneous linear system of two equations in two unknowns, say:
the graphs of the equations are lines through the origin, and the trivial solution corresponds to the point of intersection at
the origin.
After solving the system, the corresponding system of equations is:
Solving for the leading variables we obtain:
Note that the trivial solution results when r = s = t = 0
• Any set of equations AX=0 is known as a
homogeneous set; it is a set of linear
equations with zero right-hand sides. So the
equations always have trivial solution x=o, but
there may exist non-trivial solutions. What are
the conditions for their existence.
- R1 + R2 = R`2
- R1 + R3 = R`3
System is non-trivial solutions if and if only:
a = 5, put z= tBy back substitution
Y – Z = 0Y = z = t
So, x = -y –z = -2tFor any t
the solution is non-trivial
if t ≠ 0
4 R2 + R3 = R`3
1
2
1 1 1 01 2 0 01 -3 a 0
1 1 1 00 1 -1 0
0 0 a-5 0
We therefore have the following test:
For Homogeneous equations AX = 0, where A
is square:
• If det A = 0, there is an infinite number of
non-trivial solutions.
• det A ≠ 0, No solution ( trivial solution ).
Non-Homogeneous System AX = b
The system might has:1. One Solution = Unique Solution
2. Infinite Solutions = non-trivial Solution
3. No Solution = Trivial Solution (det A ≠ 0)
.. .. .. ..
.. .. .. ..
0 0
.. .. .. ..
.. .. .. ..
0 0 0 0
Consistent
.. .. .. ..
.. .. .. ..
0 0 0In Consistent
Homogeneous System AX = 0
The system might has:1. No Solution = Trivial Solution
at det A ≠ 0
2. Infinite Solutions = non-trivial Solutionat det A = 0
Note: If the matrix is non- square (n ≠ m) It is non-trivial Solution and used Gauss
Elimination