Find a invertible matrix $Q$ such that $AQ$ = $B$ Hi I have calculate this matrices:
$$A = \begin{pmatrix} 1 & 0 & \frac{1}{3}\\ 0 & 0 & -\frac{1}{3} \\ 1 & 1 & \frac{2}{3}\end{pmatrix}$$
$$B = \begin{pmatrix} \frac{4}{3} & \frac{1}{3} & 1\\ -\frac{1}{3} & -\frac{1}{3} & 0 \\ \frac{5}{3} & \frac{8}{3} & 2\end{pmatrix}$$
And I'm trying to find a invertible matrix $Q$ such that $AQ$ = $B$, but I'm stuck.
Can you help me?
I have not seen determinats yet

Maybe your teacher want you to solve the equations one by one for
each column of $B$. For example let the first column of $Q$ be $q_{1}=\begin{pmatrix}x_{1}\\
x_{2}\\
x_{3}
\end{pmatrix}$ Then you have the equations $Aq_{1}=b_{1}$, that is
\begin{align*}
x_{1}+\frac{1}{3}x_{3} & =\frac{4}{3}\\
-\frac{x_{3}}{3} & =-\frac{1}{3}\\
x_{1}+x_{2}+\frac{2}{3}x_{3} & =\frac{5}{3}
\end{align*}
From the second line $x_{3}=1$, from the first line $x_{1}=1$ and
from the third line $x_{2}=0$. Hence the first column of $Q$ is
$$
q_{1}=\begin{pmatrix}1\\
0\\
1
\end{pmatrix}.
$$
You can find the other columns simply, in a similar way.
