# Finding solutions for a linear system of equations

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.

Lucas

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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.

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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 $\det A=0$. –  enzotib Sep 15 '12 at 17:32
ok, thanks for the reply! –  Belgi Sep 15 '12 at 17:40
@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. –  Graphth Sep 15 '12 at 20:59
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 1:05