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so I've come across this question in preparation for an exam:

Let $A$ be a $4\times 4$ matrix where $rank(A)=3$. The vectors $(1,2,0,-1),(0,2,1,1)$ are solutions to the system $A\underline{x}=\underline{b}$ .

  1. Find the solution space to the homogeneous system of equations $A\underline{x}=\underline{0}$

  2. Find the general solution to the system $A\underline{x}=\underline{b}$

I solved the first part without too much of a problem:

Let $P(A)$ be the space of all solutions to the system $A\underline{x}=\underline{0}$. We know that $dimP(A)=n-rank(A)$, so $dimP(A)=4-3=1$ . We also know that the difference between two solutions to a non-homogeneous system will give us a solution to the corresponding homogeneous system - we'll call that solution vector $\underline{v}$ .

So $ \underline{v}=\begin{pmatrix}1\\2\\0\\-1\end{pmatrix}-\begin{pmatrix}0\\2\\1\\1\end{pmatrix}=\begin{pmatrix}1\\0\\-1\\-2\end{pmatrix}$

Since $dimP(A)=1$, $P(A)=Sp\{\underline{v}\}=Sp\{(1,0,-1,-2)\}$

The second part is what got me. How can I find the general solution to the system given only two specific solutions? In one of the course's early lectures the professor told us the following:

If $M$ is a system of equations that has a solution and $O$ is its corresponding homogeneous system, then:

General solution of $M$ = general solution of $O$ + specific solution of $M$

And I have no idea what to make of that. It doesn't seem logical to me to just add the two solutions I have to the general solution of the homogeneous system using vector addition - can anyone help me out here?

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It $u_0$ and $u_1$ are the two given solutions of $Ax=b$, $u_1-u_0$ is a non-zero solution of the homogeneous equation, hence a generator of the subspace of its solutions, since this subspace has dimension $1$.

Thus the affine subspace of solutions of $Ax=b$ is $$ \bigl\{ u_0+t(u_1-u_0)\mid t\in\mathbf R\bigr\},$$ which constitutes a parametric representation of the solutions.

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This idea has a lot of relations to differential equations.

So, let $AX=0$ and $AY=B$ for some particular $X,Y$.

Note that $cX$ is a general solution to $AX=0$ since any multiple of $X$ should also be in the nullspace. $$A(cX+Y)=A(cX)+AY=cAX+AY=c\quad\!\!\!\!\!\!\cdot 0+AY=B$$

So it follows that $cX+Y$ is the general solution to $M$.

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  • $\begingroup$ Okay... I understand the logic but I don't really understand how to apply it... If the general solution to $A\underline{x}=\underline{0}$ is $\{c(1,0,-1,-2)\ |\ c\in\mathbb{R}\}$ then the general solution to $A\underline{x}=\underline{b}$ would be... what, exactly? $\endgroup$ – Or Bairey-Sehayek Feb 9 '16 at 0:29

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