# Shortcut method for order 4 and above.

1. Aug 13, 2011

### median27

Is there shortcut methods for the solution of simultaneous equations when the given matrix is of order 4 and above? A more simplified technique other than Gaussian elimination and gauss-jordan method. Because when i solve orders ranging from 4 to 6, it takes me some time to finish it (30 mins and up) and barely get the right answers.

2. Aug 13, 2011

### Ben Niehoff

Cramer's rule works for any size matrices, but it's tedious. Gaussian elimination is usually the most efficient.

3. Aug 13, 2011

### AlephZero

Cramer's rule is hopelessly inefficient compared with Gaussian elimination and gets even more inefficient as the matrix order increases.

It is an important theoretical result, but no practical use for numerical work.

Unless your matrix has some special properties that you can use, Gaussian elimination is as good as it gets to find an "exact" solution. For larger matrices there are iterative methods which will can be much quicker to find an approximate numerical solution.

But in "real" life nobody would ever solve a 4x4 system of equation by hand. That's what computers are for - or even programmable calculators, for systems as small as 4x4.

4. Aug 13, 2011

### median27

I'm on my advanced engineering mathematics subject and we are practiced to perform the calculations manually. My calculator can do the calculation but it only allows 3x3 matrices and can only be used for checking (for our professor required us to include the solution).

I've been experimenting on reducing higher ordered matrices (for fast and accurate solving) and came up with eliminating one variable at a time and substituting it to the remaining equations until it ends up with a 3x3 matrix (or atleast 2x2) then solve it by gauss jordan. But when i did the checking, my answer is not consistent for the rest of the equations other than the first one. But sometimes, it gives the correct answer.

Do you find my reducing method applicable (and only needs to be polished) or not?

5. Aug 13, 2011

### gb7nash

You can do this (assuming that there exists a unique solution), but the chance for making arithmetic errors goes up, and you need to keep refining your matrix everytime you do this. I would stick with Gauss-Jordan elimination.