Solve the following system using Gauss's algorithm
(a)
x1 + 2x2 + 4x3 + 6x4 = 3
2x1 + x2 + 3x3 = -6
-2x1 + x2 + 6x3 + 4x4 = -11
2x1 + x2 + x3 = 0
so the matrix will be:
1 2 4 6 3
2 1 3 0 -6
-2 1 6 4 -11
2 1 1 0 0
numbering the lines 1-4,
#2 - 2*#1, #3 + 2*#1, #4 - 2*#1 gives
1 2 4 6 3
0 -3 -5 -12 -12
0 5 14 16 -5
0 -3 -7 -12 -6
Follow similar steps to place zeros in the other rows off the main diagonal, to end up with
so, (x1,x2,x3,x4) = 1/3 (-23,55,-9,-7)
A good calculator which shows the steps can be found at
www.gregthatcher.com/Mathematics/GaussJordan.aspx
awesome, thanks a lot Steve, that link will come in handy!
sure thing. Plus, I messed up my math. Solution is really (1,1,-3,2)
To solve this system using Gauss's algorithm, we will perform Gaussian elimination on the augmented matrix.
Step 1: Start with the given augmented matrix.
1 2 4 6 3
2 1 3 0 -6
-2 1 6 4 -11
2 1 1 0 0
Step 2: Perform row operations to eliminate the coefficients below the pivot in each column.
a) Row 2 becomes Row 2 - 2 * Row 1.
1 2 4 6 3
0 -3 -5 -12 -12
-2 1 6 4 -11
2 1 1 0 0
b) Row 3 becomes Row 3 + 2 * Row 1.
1 2 4 6 3
0 -3 -5 -12 -12
0 5 14 16 1
2 1 1 0 0
c) Row 4 becomes Row 4 - 2 * Row 1.
1 2 4 6 3
0 -3 -5 -12 -12
0 5 14 16 1
0 -3 -7 -12 -6
Step 3: Perform row operations to eliminate the coefficients below the pivot in each column.
a) Row 3 becomes Row 3 + (5/3) * Row 2.
1 2 4 6 3
0 -3 -5 -12 -12
0 0 1 -4 -7
0 -3 -7 -12 -6
b) Row 4 becomes Row 4 + (3/2) * Row 2.
1 2 4 6 3
0 -3 -5 -12 -12
0 0 1 -4 -7
0 0 1 -6 -15
Step 4: Perform row operations to eliminate the coefficients below the pivot in each column.
a) Row 4 becomes Row 4 - Row 3.
1 2 4 6 3
0 -3 -5 -12 -12
0 0 1 -4 -7
0 0 0 2 8
Step 5: Now, we have an upper triangular matrix. We can start back-substitution to find the values of x1, x2, x3, and x4.
From the last row, we can directly solve for x4:
2x4 = 8
x4 = 4
Then, from the third row, we can solve for x3 using the known value of x4:
x3 - 4(-4) = -7
x3 + 16 = -7
x3 = -7 - 16
x3 = -23
Similarly, from the second row, we can solve for x2 using the known values of x3 and x4:
-3x2 - 5(-23) - 12(4) = -12
-3x2 + 115 - 48 = -12
-3x2 + 67 = -12
-3x2 = -12 - 67
-3x2 = -79
x2 = -79 / -3
x2 = 79/3
Finally, from the first row, we can solve for x1 using the known values of x2, x3, and x4:
x1 + 2(79/3) + 4(-23) + 6(4) = 3
x1 + 158/3 - 92 + 24 = 3
x1 + 158/3 - 92 + 24 = 3
x1 + 158/3 - 68 = 3
x1 + 158/3 = 3 + 68
x1 + 158/3 = 71
x1 = 71 - 158/3
x1 = 213/3 - 158/3
x1 = 55/3
Therefore, the solution to the system of equations is:
x1 = 55/3
x2 = 79/3
x3 = -23
x4 = 4