Is there any programme for solving a system of 5 linear equations in five variables?

Question

The system of equation that I have has coefficients of variables in decimals, therfore it is very difficult to solve it by elimination.

Answer 1

Yes, there are programs for very large matrices with 100 or more equations.

Look for matrix inversion. What sort of program did you want? Let me know and I can steer you in the right direction.

+ I just looked in the index of my book for the TI-83. It says that it can handle up to 99 rows or columns!

++ I looked at your posts and I see that you know something about statistics. The R program will easily handle matrices this large. The program is also free. It will also do other types of matrix manipulation such as eigenvalue decomposition...

You could also check to see if Minitab will help you.

If you have a TI-83 or equivalent, then you can do the job.

Answer 2

Other than successive substitutions, there's nothing I can recommend, as there is really no shortcuts (that I'm aware of). If you know matrices, then putting them in a matrix and reducing them to reduced row echelon form would solve the problem, but would still require a considerable amount of work.

Follow these steps:
1) Take the first equation and solve for one of the variables in terms of the rest. It may look something like this:
x = 2y - 5z + 4w + 8a
But the key thing to know is that x is written with 4 variables.

2) Substitute this value into your other four equations.

3) Expand, reduce, collect like terms, in all 4 equations. What this leaves you with is now a set of 4 equations, 4 unknowns.

4) Repeat the process; in the first equation, solve for one variable in terms of the others.

5) Plug this into the other three equations.

6) Expand, reduce, collect like terms, in all three equations. This will now leave you with 3 equations, three unknowns.

7) Repeat the process again, with the three equations. Solve for one variable in terms of the others.

8) Plug this value into the other 2 equations. This will give you two equations, two unknowns.

9) Solve the system of two equations and two unknowns. This will get you TWO values.

10) Use these two values and look back at your three equation, three unknowns problem, to get the third value.

11) Use these 3 values and look back at our four equations, four unknowns problem, to get the fourth value.

12) Finally, use the 4 values you now have to get the 5th value, by plugging these values into one of your original 5 equations.

13) TEST TEST TEST! There are so many things that can go wrong in such a large number of system of equations, that you should plug all of the values you obtained and see if they all work.

Answer 3

2x - y + z = -4 .........1st equation 3x + y - 2z = 0.......2d equation 3x - y = -4........third equation you initiate with information from taking the third equation and expressing x in words of y, you get: 3x=y-4 or, x=(y-4)/3 ......4th equation change 4th in 1st you get: (2y-8)/3-y+z = -4 or, 2y-8-3y+3z = -12 or, -y + 3z = -4 ...............fifth equation also, change equation 4 in 2, you get: y - 4 + y - 2z = 0 or, 2y - 2z - 4 = 0...............sixth equation Now be certain equations 5 and six 2(- y + 3z = - 4) 2y - 2z = 4 or, -2y + 6z = - 8 2y - 2z= 4 ___________ 4z = - 4 or z = -a million change z contained in the above equation(2y - 2 * (-a million)=4), you get y = a million change y in 4th equation and also you get, x = a million - 4/3 = -a million for this reason, the answer is x = -a million, y = a million and z = -a million The question is to unravel a equipment of three equations with 3 variables in them. those equations are solvable because you want a minimum of three equations to unravel for 3 variables. in case you had 2 equations and three variables then you won't be able to be certain it.

Answer 4

You just need something that can handle 5 x 5 matrices.

Something like Mathematica any good to you?