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Got the following problem where I can't find a way to solve:

Knowing $\begin{pmatrix}5\\ 3\\ 6\end{pmatrix}$ is the unique solution for the system $Ax=\begin{pmatrix}2\\1\\1\end{pmatrix}$, with $A \in \mathbb{R}^{3\times3}$

and $B=\begin{pmatrix} 1 & 2 & 1 & 2 \\ 1 & 0 & 4 & -1 \\ 1 & 3 & -3 & 6 \end{pmatrix}$

Find all solutions for $ABx=\begin{pmatrix}2\\1\\1\end{pmatrix}$

What I've tried:

  • The problem says that $Ax=b$ got unique solution, so I've tried by getting rid of $A$ by using the inverse matrix but it doesn't work sice I don't know $A$.
  • Since the constant matrix is $\begin{pmatrix}2\\1\\1\end{pmatrix}$ for both systems, I've tried $ABx = Ax$ but that also doesn't work for me.

Thanks in advance for your help and sorry for my bad English.

Lucas

Jacob
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Lucas
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1 Answers1

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As observed by @GerryMyerson, there is no need for $B$ to be a square matrix.
From

$$ Ax=b \implies x_0=A^{-1}b $$

and

$$ ABx=b \implies Bx=A^{-1}b $$

you get

$$ Bx=x_0 $$

where $x_0$ is the known solution of the first system.

Vincenzo Tibullo
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  • How can you tell that $A$ is invertiable ? – Belgi Sep 15 '12 at 17:27
  • @Belgi: in the question the OP says the first equation has a unique solution, and this is not the case if $\det A=0$. – Vincenzo Tibullo Sep 15 '12 at 17:32
  • ok, thanks for the reply! – Belgi Sep 15 '12 at 17:40
  • Hello @enzotib, first of all thanks for taking the time to answer. And yes, I'm sure B has four columns. – Lucas Sep 15 '12 at 17:41
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    @Lucas Since $A$ has 3 columns, this implies for $Ax$ to be defined that $x$ has 3 rows (I mean, you give an example of $x$ which is 3 by 1. Thus, for $ABx$ to be defined, $B$ must also be 3 by 3. – GeoffDS Sep 15 '12 at 20:59
  • Hello @Graphth. Thanks for your answer. I was sure B had 4 columns, but probably miscopied that since it seems there's no way to solve the problem. So I will check again and get back here. Thanks to all of you guys for your time! – Lucas Sep 15 '12 at 22:44
  • Should your first line be $Ax_0=b \implies x_0=A^{-1}b$? – Henry Sep 15 '12 at 23:48
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    There is no reason to think $B$ ought to be square. The solution by enzotib is correct, that is, all that's necessary is to find solution(s) $x$ of $Bx=x_0$. $x_0=(5,3,6)$, and $x$ will be a 4-tuple. – Gerry Myerson Sep 16 '12 at 01:05
  • Hello, I'm sorry I didn't comment this before. Just wanting to say the solution proposed by enzotib and clarification by Gerry Merson are correct. Thank you guys for your time. – Lucas Sep 24 '12 at 20:23