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I'm trying to solve this problem:

Assume that $(1, -2, 3)$ is a solution to a system $AX = b$ with $A$ a $4 \times 3$ matrix. Assume that $\{(3s, -s, 2s) | s \in \mathbb{R}\}$ is the solution space for $AX = 0$. Give the solution set for the system $AX = b$.

I start with trying to find (for some $T_A : \mathbb{R}^3 \to \mathbb{R}^4$) the $Im (T_A)$, since I know that for some $(b_1, b_2, b_3, b_4) \in Im (T_A) \Leftrightarrow AX = b$ has a solution with $b = \begin{bmatrix}b_1 & b_2 & b_3 & b_4\end{bmatrix}^T $

I know that from the given information you can infer that the reduced row echelon form $[A | 0]$ that solves the $AX = 0$ and gives the $ker(T_A) = \{(3s, -s, 2s) | s \in \mathbb{R}\}$ has to be

$ \begin{bmatrix} 1 & 3 & 0 & 0 \\ 0 & 2 & 1 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix} $

And based on $ker(T_A)$ there's 2 known results for $T_A$,

$T_A((-3, 1, -2)) = (0, 0, 0, 0)$ (and multiples of $(-3, 1, -2)$), and $T_A((1, -2, 3)) \neq (0, 0, 0, 0)$

I don't really know where I must go from here. I assume I must figure out $A$, the non reduced row-echelon form of it, somehow, and then I could easily determine the solution set for $AX = b$, but I'm not sure how I might do this.

Apologies for any inconsistent/strange vocabulary, I had to translate this problem.

fatCat9999
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    Note that if $Ax=b$ and $Ay=0$, then $A(x+y)=Ax+Ay=b+0=b$. Similarly, if $Ax=b$ and $Az=b$, then $x-z$ satisfies $A(x-z)=Ax-Az=b-b=0$. – plop Aug 16 '22 at 14:17
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    You can't say very much about the matrix form since it depends so much on the arbitrary $b$. Much easier, as the comment and answer so far have done, to just use properties of linear transformations without considering properties of the matrix representation. – aschepler Aug 16 '22 at 14:51
  • Thanks for pointing out the above explanation @user85667, I feel like your explanation is better than the one given in my book. – fatCat9999 Aug 16 '22 at 15:31
  • I see @aschepler, I overlooked an explanation on this property of the solution set in my book and got stuck on the matrix form rather than thinking about the solution set itself indeed. – fatCat9999 Aug 16 '22 at 15:31
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    @fatCat9999 See Introduction to Linear Algebra, $3$rd ed. by Gilbert Strang , chapter $2$ $\$ – Sourav Ghosh Aug 16 '22 at 15:38

1 Answers1

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The complete solution of $Ax=b$ is given by $X_c$.

Then $X_c= X_N+X_p$

Where $X_N$ be the null space solution i.e $X_N$ is the linear combination of null space basis and $X_p$ is a particular solution.

Given $X_N=s(3, -1, 2) $ and $X_p=(1, -2, 3) $.

Hence complete solution $X_c=s(3, -1, 2) +(1, -2, 3) $

Sourav Ghosh
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  • Thanks for this explanation! I had to go back to my question/book for a bit here since I'm mostly familiar with that long-form solution notation that was used in the question, but then I guess the correct way to write the long form is ${(1,−2,3) + s(3,−1,2) | s \in \mathbb{R}}$ for this problem. – fatCat9999 Aug 16 '22 at 15:38
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    @fatCat9999 Exactly. $\forall s\in\Bbb{R}$ gives the set of all solutions. – Sourav Ghosh Aug 16 '22 at 15:41
  • Hey Sourav, hope I am not disturbing but could you check this answer and say if there are any factual mistakes in it? I am confused what is in it. Thanks. – tryst with freedom Aug 19 '22 at 12:33