For the given question, I am unclear about how special solutions are involved. I know what a special solution to a given system is and how to calculate it, but the role of special solutions in this question is too abstract for me to understand.
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Given that vectors $\mathbf{\beta_1, \beta_2}$ are distinct solutions to the system of equations $\mathbf{AX}= \mathbf{b}$, and that $\mathbf{\alpha_1, \alpha_2}$ forms the basis to the corresponding linear homogeneous system $\mathbf{AX}=\mathbf{0}$, and that $k_1, k_2$ are arbitrary constants, which of the following is always a general solution to the system $\mathbf{AX}= \mathbf{b}$ ?
A) $k_1 \mathbf{\alpha_1} + k_2(\mathbf{\alpha_1 + \alpha_2}) + \frac{\mathbf{\beta_1 - \beta_2}}{2}$
B) $k_1 \mathbf{\alpha_1} + k_2(\mathbf{\alpha_1 - \alpha_2}) + \frac{\mathbf{\beta_1 + \beta_2}}{2}$
C) $k_1 \mathbf{\alpha_1} + k_2(\mathbf{\beta_1 - \beta_2}) + \frac{\mathbf{\beta_1 - \beta_2}}{2}$
D ) $k_1 \mathbf{\alpha_1} + k_2(\mathbf{\beta_1 - \beta_2}) + \frac{\mathbf{\beta_1 + \beta_2}}{2}$
Given Solution: (B). Notice that (A) and (C) does not include special solutions to a non-homogeneous system of equations. In option (D), $\mathbf{\alpha_1}$ and $\mathbf{\beta_1 - \beta_2}$ are solutions to the homogeneous equation system but it is unknown whether they are linearly independent. For option (B), notice that $\mathbf{\alpha_1, \alpha_1 - \alpha_2}$ are basis vectors and that $\frac{\mathbf{\beta_1 + \beta_2}}{2}$ is a special solution. Therefore (B) is a general solution.
The general solution of the inhomogeneous equation $A\mathbf x=\mathbf b$ is usually given in the form $\mathbf v+N$, where $\mathbf v$ is any particular (special) solution to the equation and $N$ is the general solution of the homogeneous equation. Without the particular solution $\mathbf v$, though, you only have the solution to the homogeneous equation, so a particular solution of the inhomogeneous system must appear somewhere in any statement of the inhomogeneous equation’s general solution.
Now, if $\beta_1$ and $\beta_2$ are distinct solutions of $A\mathbf x=\mathbf b$, then you have $A(\beta_1-\beta_2)=A\beta_1-A\beta_2=\mathbf b-\mathbf b = 0$, therefore $\beta_1-\beta_2$ is a solution to the homogeneous equation $A\mathbf x=0$ as mentioned in the solution. Every term in expressions (A) and (C) is a solution to the homogeneous equation—there are not terms that solve $A\mathbf x=\mathbf b$—so they are both solutions to the homogeneous equation, as you can verify by multiplying them by the matrix $A$.
On the other hand, $A{\beta_1+\beta_2\over 2} = \frac12(A\beta_1+A\beta_2)=\frac12(\mathbf b+\mathbf b)=\mathbf b$, so $\frac12(\beta_1+\beta_2)$ is indeed a solution of the inhomogeneous equation. That gets you the required element in a general solution to $A\mathbf x=\mathbf b$, and it’s easily verified that $\alpha_1$ and $\alpha_1-\alpha_2$ form a basis for the solution of the homogeneous equation, so expression (B) qualifies as a general solution of $A\mathbf x=\mathbf b$. Of course, you could have verified this directly by multiplying the expression by $A$ and simplifying.