I am a bit in doubt if I am doing this correct, I am row reducing and trying to get as close to a reduced echelon form, but is that even what I want? I am not sure if I am seeking the right answers with my solution
1 Answer
$\begingroup$First, there's no such thing as the solution to a matrix. What you are actually solving is a system of equations - in this case, a system of two equations in three unknowns - and you are using a matrix to represent the system of equations, and using matrix operations to solve the system.
I don't know what that 6 is that's sitting on top of your first matrix, so I propose to ignore it.
Your first matrix represents the system $$\eqalign{x_1+3x_2+4x_3&=7\cr 3x_1+9x_2+7x_3&=6\cr}$$
Your next-to-last matrix (which, as Arturo notes in the comments, is in row-echelon form) is $$\pmatrix{1&3&4&7\cr0&0&1&3\cr}$$ which represents the system $$\eqalign{x_1+3x_2+4x_3&=7\cr x_3&=3\cr}$$ From this, you can read off the value of $x_3$, and you can then get a formula for $x_1$ in terms of $x_2$; you can treat $x_2$ as a free parameter, able to take on any real value.
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