┌ ┐ │ 5 7 2 │ A = │ 6 -9 11 │ │ 1 1 -1 │ └ ┘ ┌ ┐ │ x │ X = │ y │ │ z │ └ ┘ ┌ ┐ │ 7 │ B = │ 3 │ │ 12 │ └ ┘

┌ ┐ │ 1 0 0 │ A = │ 0 1 0 │ │ 0 0 1 │ └ ┘ [code] Some problems have infinite/no solutions, so you won't always get it. But for something well-formed you [italic] should [italic]. An infinite solution problem will end up having columns with no leading ones - and therefore no solutions for the associated variable. A no solution problem will have a row full of zeros, yet the answer will be some non-zero. Anywho, the general process is to start at the top and: 1) Simplify the operating row so it has a leading one by row multiplication. 2) Now make all other elements below the leading one equal zero by multiplying your operating row, the adding. 3) Repeat. 4) Once at the bottom, work in reverse and use the leading zeros to remove those numbers above them (trailing behind the leading zeros of other rows). So, a simple gaussian elimination for a 4x4 matrix would look like this: [code] ┌ ┐ │ 4 8 │ 12 │ │ 3 9 │ 18 │ └ ┘ ┌ ┐ │ 1 2 │ 3 │ Multiply by (1/4) for leading one. │ 3 9 │ 18 │ └ ┘ ┌ ┐ │ 1 2 │ 3 │ │ 0 3 │ 9 │ Multiply the row above by -3, then add. └ ┘ ┌ ┐ │ 1 2 │ 3 │ │ 0 1 │ 3 │ Multiply by (1/3) for leading one. └ ┘ ┌ ┐ │ 1 0 │ -3 │ Multiply the row below by -2, then add. │ 0 1 │ 3 │ └ ┘