Answer
See explanation.
Work Step by Step
1. Because $(−2, −1,1,1)$ is a solution to $Ax = 0$, so is any multiple $−2(−2, −1, 1, 1) = (4, 2, − 2, − 2)$ because the solution space is a subspace.
2. The solutions to $Ax = 0$ form a subspace, so any linear combination $2x_{1} − 3x_{2}$ of solutions $x_{1}$ and x_{2} is again a solution.
3. Let the first system be $Ax = b_1$. Because it is consistent, $b_1$ is in the column space of $A$. The second system is $Ax = b_{1}$, and $b_{2}$ is a multiple of $b_{1}$, so it is in the column space of $A$ as well. So, the second system is consistent.
4. $2x_{1} − 3x_{2}$ is not a solution (unless $b = 0$). The set of solutions to a nonhomogeneous system is not a subspace. If $Ax_{1} =b$ and $Ax_{2} =b$, then
$A(2x_{1} −3x_{2}) = 2Ax_{1} −3Ax_{2} =2b−3b= −b≠b$.
5. Yes, $b_1$ and $b_2$ are in the column space of $A$, therefore so
is $b_1 + b_2$.
6. If rank $A$ = rank $[A:b]$, then $b$ is in the column space of $A$ and the system is consistent. If rank $A <$ rank$[A:b]$, then $b$ is not in the column space and the system is inconsistent.
