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
 A: Suppose that we have $$A = \begin{pmatrix} 1 & 0 & \frac{1}{3}\\ 0 & 0 & -\frac{1}{3} \\ 1 & 1 & \frac{2}{3}\end{pmatrix}$$
$$B = \begin{pmatrix} \frac{4}{3} & \frac{1}{3} & 1\\ -\frac{1}{3} & -\frac{1}{3} & 0 \\ \frac{5}{3} & \frac{8}{3} & 2\end{pmatrix}$$
and we want $Q$ such that $AQ =B$. Let
$$ Q = \begin{pmatrix} a & b &c \\  d & e & f \\ g & h & i \end{pmatrix}$$
Then using matrix multiplication you'd see that
$$ AQ =  \begin{pmatrix} 1 & 0 & \frac{1}{3}\\ 0 & 0 & -\frac{1}{3} \\ 1 & 1 & \frac{2}{3}\end{pmatrix}   \begin{pmatrix} a & b &c \\  d & e & f \\ g & h & i \end{pmatrix} = \begin{pmatrix} a + \frac{g}{3}  & b + \frac{h}{3} & c + \frac{i}{3} \\ -\frac{g}{3} & \frac{-h}{3} & \frac{-i}{3} \\ a + d + \frac{2g}{3} & b + e + \frac{2h}{3} & c + f + \frac{2i}{3} \end{pmatrix} $$
now set it equal to $B$
$$ \begin{pmatrix} a + \frac{g}{3}  & b + \frac{h}{3} & c + \frac{i}{3} \\ -\frac{g}{3} & \frac{-h}{3} & \frac{-i}{3} \\ a + d + \frac{2g}{3} & b + e + \frac{2h}{3} & c + f + \frac{2i}{3} \end{pmatrix} =  \begin{pmatrix} \frac{4}{3} & \frac{1}{3} & 1\\ -\frac{1}{3} & -\frac{1}{3} & 0 \\ \frac{5}{3} & \frac{8}{3} & 2\end{pmatrix}$$
Multiply by $3$ on both sides and we get
$$ \begin{pmatrix} 3a + g  & 3b + h & 3c + i \\ -g & -h & -i \\ 3a + 3d + 2g & 3b + 3e 2h & 3c + 3f + 2i \end{pmatrix} =  \begin{pmatrix} 4 & 1 & 3 \\ -1 & -1 & 0 \\ 5 & 8 & 6\end{pmatrix}$$
Then you directly get $g,h,i$
$$ -g = -1 \implies g = 1 \\ -h = -1 \implies h = 1 \\ -i = 0 \implies i=0$$
Now substitute those into the top row
$$ 3a +1 =  4 \implies a =1 \\ 3b + 1 = 1 \implies b = 0 \\ 3c = 3 \implies c =1 $$
then continue and get the last three
A: 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
$$
Q = A^{-1} B  = \begin{pmatrix}
1 & 0 & 1 \\
0 & 2 & 1 \\
1 & 1 & 0
\end{pmatrix}.
$$
A: 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.
