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Re: MTH603 Assignment 1 Solution and Discussion
Question #1: Find the root of the equation x^3+x^2+x-1 =0 correct to two decimal places by using bisection method.
Question #2: Solve the system of linear equations with the help of Gaussian elimination method.
2x + y + z = 9;3x −2y + 4z = 9;x +y-2z = 3
Solution File
@zareen said in MTH603 Assignment 1 Solution and Discussion:
Question #2: Solve the system of linear equations with the help of Gaussian elimination method.
2x + y + z = 9;3x −2y + 4z = 9;x +y-2z = 3
System of Linear Equations entered :
[1] 2x + y + z = 9
[2] 3x - 2y + 4z = 9
[3] x + y - 2z = 3
Solve by Substitution :
// Solve equation [3] for the variable y
[3] y = -x + 2z + 3
// Plug this in for variable y in equation [1]
[1] 2x + (-x +2z+3) + z = 9
[1] x + 3z = 6
// Plug this in for variable y in equation [2]
[2] 3x - 2•(-x +2z+3) + 4z = 9
[2] 5x = 15
// Solve equation [2] for the variable x
[2] 5x = 15
[2] x = 3
// Plug this in for variable x in equation [1]
[1] (3) + 3z = 6
[1] 3z = 3
// Solve equation [1] for the variable z
[1] 3z = 3
[1] z = 1
// By now we know this much :
x = 3
y = -x+2z+3
z = 1
// Use the x and z values to solve for y
y = -(3)+2(1)+3 = 2
Solution :
{x,y,z} = {3,2,1}
@zareen said in MTH603 Assignment 1 Solution and Discussion:
Question #1: Find the root of the equation x^3+x^2+x-1 =0 correct to two decimal places by using bisection method.
We now use the Bisection Method to approximate one of the solutions. The Bisection Method is an iterative procedure to approximate a root (Root is another name for a solution of an equation).
The function is F(x) = x3 + x2 + x - 1
At x= 0.00 F(x) is equal to -1.00
At x= 1.00 F(x) is equal to 2.00
Intuitively we feel, and justly so, that since F(x) is negative on one side of the interval, and positive on the other side then, somewhere inside this interval, F(x) is zero
Procedure :
(1) Find a point “Left” where F(Left) < 0
(2) Find a point ‘Right’ where F(Right) > 0
(3) Compute ‘Middle’ the middle point of the interval [Left,Right]
(4) Calculate Value = F(Middle)
(5) If Value is close enough to zero goto Step (7)
Else :
If Value < 0 then : Left <- Middle
If Value > 0 then : Right <- Middle
(6) Loop back to Step (3)
(7) Done!! The approximation found is Middle
Follow Middle movements to understand how it works :
Left | Value(Left) | Right | Value(Right) |
0.000000000 | -1.000000000 | 1.000000000 | 2.000000000 |
0.000000000 | -1.000000000 | 1.000000000 | 2.000000000 |
0.500000000 | -0.125000000 | 1.000000000 | 2.000000000 |
0.500000000 | -0.125000000 | 0.750000000 | 0.734375000 |
0.500000000 | -0.125000000 | 0.625000000 | 0.259765625 |
0.500000000 | -0.125000000 | 0.562500000 | 0.056884766 |
0.531250000 | -0.036590576 | 0.562500000 | 0.056884766 |
0.531250000 | -0.036590576 | 0.546875000 | 0.009502411 |
0.539062500 | -0.013703823 | 0.546875000 | 0.009502411 |
0.542968750 | -0.002140820 | 0.546875000 | 0.009502411 |
0.542968750 | -0.002140820 | 0.544921875 | 0.003670745 |
0.542968750 | -0.002140820 | 0.543945312 | 0.000762452 |
0.543457031 | -0.000689811 | 0.543945312 | 0.000762452 |
0.543457031 | -0.000689811 | 0.543701172 | 0.000036164 |
0.543579102 | -0.000326863 | 0.543701172 | 0.000036164 |
0.543640137 | -0.000145359 | 0.543701172 | 0.000036164 |
0.543670654 | -0.000054600 | 0.543701172 | 0.000036164 |
0.543685913 | -0.000009219 | 0.543701172 | 0.000036164 |
0.543685913 | -0.000009219 | 0.543693542 | 0.000013472 |
0.543685913 | -0.000009219 | 0.543689728 | 0.000002127 |
Next Middle will get us close enough to zero:
F( 0.543688774 ) is -0.000000710
The desired approximation of the solution is:
x ≓ 0.543688774
Note, ≓ is the approximation symbol
One solution was found :
x ≓ 0.543688774