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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

1 Answers 1

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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.

  • 0
    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.2012-09-24