Jtdylktuy

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 $Amathbf x=mathbf b$ 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 $Amathbf x=mathbf b$, then you have $A(beta_1-beta_2)=Abeta_1-Abeta_2=mathbf b-mathbf b = 0$, therefore $beta_1-beta_2$ is a solution to the homogeneous equation $Amathbf x=0$ 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 $Amathbf x=mathbf b$—so they are both solutions to the homogeneous equation, as you can verify by multiplying them by the matrix $A$.

On the other hand, $A{beta_1+beta_2over 2} = frac12(Abeta_1+Abeta_2)=frac12(mathbf b+mathbf b)=mathbf b$, 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 $Amathbf x=mathbf b$, 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 $Amathbf x=mathbf b$. Of course, you could have verified this directly by multiplying the expression by $A$ and simplifying.