x-y = 1
3x - 3y = k
1 -1 ¦ 1
3 -3 ¦ k
1 -1 ¦ 1
0 0 ¦ k-3
so when k doesnt equal 3 there is no solutions
k = 3 infinite solutions
but i cant do the one solution, is there one? or have done it wrong?
Thanks
x-y = 1
3x - 3y = k
1 -1 ¦ 1
3 -3 ¦ k
1 -1 ¦ 1
0 0 ¦ k-3
so when k doesnt equal 3 there is no solutions
k = 3 infinite solutions
but i cant do the one solution, is there one? or have done it wrong?
Thanks
It looks like you've already answered your question. The problem is, the matrix you're dealing with is singular. That implies automatically, I think, that the existence of a unique solution is impossible. In other words, you've found all the possibilities. Just say there are no values of k for which the system has exactly one solution.
It's probably more accurate to say that the lines, represented by the system of equations you have there, are either the same line, or non-intersecting, corresponding to infinite solutions and zero solutions, respectively. Does that make sense?