Basis and general solution (homework assignement )

Question

If $\ X_1, X_2, X_3$ are linearly independent solutions with $\ n$ amount of rows to a set of linear equations $\ AX = B (B\neq0)$, where the rank of $\ A$, $\ r_A=n-2$ , find a basis for $\ AX=0$ and general solution set for $\ AX=B$

I got that if:

$\ AX_1=B$
$\ AX_2=B$
$\ AX_3=B$

then $\ A(X_1-X_2)=0$, however solution has also another equation $\ A(2X_1-X_2-X_3)=0$ which I don't really understand where it comes from. Can someone explain to me how this question should be solved?

Any help will be greatly apreciated !

Thanks a lot :)

Answer 1

Since rank of $A$ is $n-2$, we know that its null space must have dimension $2$. So you have one more independent solution to look for other than $X_1- X_2$. Verify that: $$A(2X_1 - X_2 - X_3) = 2AX_1 - AX_2 - AX_3 = 2B - B - B = 0$$ This solution and $X_1 - X_2$ are independent, since $X_3$ cannot be written in terms of $X_1$, $X_2$ (we are given that $X_1,X_2,X_3$ are independent). Of course, this is not the only possible choice of a second basis vector for the null space.