Do you mean that the matrix has infinitely many solutions? also how did you get your answer? also why did you not choose the method of reducing the matrix to a row echelon form? and how can you solve for and if nor is part of that equation?
Thanks again for your continued support
This is a system of the form .
I could see that there were a lot of 0's in A, so just by using matrix multiplication, I could multiply A by x and get a vector that is equal to b.
This system is not only unsolvable, it's nonsense. and can not be solved, and has to equal two different things at once in order for the system to make any sense at all.
Are you sure you copied the matrix A down correctly?
Oh my god you are absolutely correct... that matrix was for another question sorry to have wasted your time but this is the true matrix A:
Now, i've checked and double checked and so far this matrix A is correctly copied. also, x and b still are the same.
Sorry to have caused such trouble
Because there isn't an equation we can back-substitute into that can give us JUST .
When we back substitute, we get an equation that has both and .
So the best we can do is write one of them in terms of the other.
In other words, we can let be whatever we like, it won't change the system. The only thing that will change is , because it depends on .