<s>[INST] Finding a matrix with conditions on $A^{-1}$ and $\det(A)=0.5$ 
Find matrix $A$ (without guessing) if $A^{-1}$ $2^{\mkern1mu\text{nd}}$ column is\begin{pmatrix} 2 \\ 1 \\-1 \end{pmatrix} 
And $\det(A) = 0.5$

Attempt -
I could make a progress knowing that $\det(A^{-1}) = 2$ obviously.
I could also notify that if $A = $
\begin{pmatrix} a_1 & a_2 & a_3\\ b_1 & b_2 & b_3 \\c_1 & c_2 & c_3 \end{pmatrix}
Then - \begin{pmatrix} 2a_1 + a_2 - a_3 = 0  \\ 2b_1 +b_2 -b_3 = 1 \\2c_1 +c_2 -c_3 = 0\end{pmatrix} 
I don't know how to procced, Any help ?
 [/INST] I would proceed like this $A^{-1}=\begin{pmatrix} a & 2 & d\\ b & 1 & e \\ c & -1 & f\end{pmatrix}$
The only additional constraint we have is $\det(A^{-1})=af+ae-bd-2bf+2ce-cd=2$
Since we are required to find an instance of $A$ that works and not all possible matrices, then just select some values of $(a,b,c,d,e,f)$ that fit nicely. 
For instance 


*

*let start with $a=0$, it remains $-bd-2bf+2ce-cd=2$

*let choose $b=0$, it remains $2ce-cd=2$

*a trivial choice is then $d=0,f=0,c=e=1$


And we get $A^{-1}=\begin{pmatrix} 0 & 2 & 0\\ 0 & 1 & 1 \\ 1 & -1 & 0\end{pmatrix}$
But I could have started 


*

*let choose $f=0$, it remains $ae-bd+2ce-cd=2$

*continue with $e=0$, it remains $-bd-cd=2$

*a trivial choice could then be $d=1,b=0,c=-2,a=0$


And we get $A^{-1}=\begin{pmatrix} 0 & 2 & 1\\ 0 & 1 & 0 \\ -2 & -1 & 0\end{pmatrix}$
But you can as well set five of the six variables arbitrarily 


*

*set $a=1,b=2,c=3,d=4,e=5$, it remains $-3f+15=2$

*$f=3/13$ is forced.


And we get $A^{-1}=\begin{pmatrix} 1 & 2 & 4\\ 2 & 1 & 5 \\ 3 & -1 & \frac 3{13}\end{pmatrix}$
It is just up to you to select whichever numbers you desire, until at some point, the equation relative to $\det(A^{-1})=2$ constrains the remaining last free variable.
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