My professor went through an example in class that said :
Let A be the augmented matrix of a system of linear equations. Assume that every entry in the last column of A is 0. Explain why the system of equations must have at least one solution. Provide an example of such a system with a unique solution. Provide an example of such a system with infinitely many solutions.
The answer they gave was:
"The zero vector is always a solution. Therefore the system must have at least one solution. The system
x +y = 0, x- y = 0
has only one solution, x=y=0.
whereas the system
x+y=0, and 3x+3y = 0
has infinitely many solutions, since one of the equations is a scalar multiple of the other one"
I am confused as to how they concluded the matrix had an infinite number of solutions.
Best Answer
If the last column of the augmented matrix is all zeroes, then you are attempting to solve $Ax = 0$. Essentially, you are finding all vectors that produce the zero vector through multiplication with $A$. Naturally, the zero vector satisfies this equation, so you must have at least one solution.
For the system with infinitely many solutions given, all solutions will satisfy y = -x. You have the choice of what x (or y) is such that this equation is satisfied. You have infinitely many choices as you are working with an infinite set (assuming the real numbers). When the equations are linear combinations of each other, you'll end up with the freedom to choose some values.