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

From linear independence of the rows we see that $A$ is invertible and so $Q = A^{-1}B$. You can use adjugate matrix and determinant to compute $A^{-1}$
$$
A^{-1} = \frac{{\rm adj} A}{\det A} = \frac{\begin{pmatrix}
1/3 & 1/3 & 0 \\
-1/3 & 1/3 & 1/3 \\
0 & -1 & 0 
\end{pmatrix}}{\frac{1}{3}} =
\begin{pmatrix}
1 & 1 & 0 \\
-1 & 1 & 1 \\
0 & -3 & 0 
\end{pmatrix}.
$$
Alternatively, you can find $A^{-1}$ using Gaussian elimination. Finally, we get
$$
Q = A^{-1} B  = \begin{pmatrix}
1 & 0 & 1 \\
0 & 2 & 1 \\
1 & 1 & 0
\end{pmatrix}.
$$
