# 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 ?

$$A^{-1}=\begin{pmatrix}2&2&0\\ 0&1&0\\ 0&\!-1&1\end{pmatrix}$$

Using the above, now you find $\;A\;$. How did I do the above? Knowing the determinant of $\;A^{-1}\;$ is $\;2\;$ and just adding zeros wherever possible.

• That is true, and apart from guessing the answer (which was cleaver) could i find an algorithm to find such matrix ? – Yariv Levy Jan 30 '18 at 12:08
• @YarivLevy Just the horrible, boring and annoying way you began to show in your question. Much better adding zeros as to get an easily calculable determinant for a $\;3\times3\;$ matrix. – DonAntonio Jan 30 '18 at 12:10

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.

You want that $A^{-1}=[x\ y\ z]$, where $y$ is your given column, and $\det(A^{-1})=2$.

Start completing $y$ to a basis of $\mathbb{R}^3$; call $x$ and $z'$ the vectors you get. Now, set $d=\det[x\ y\ z']$ and set $$z=\frac{2}{d}z'$$ Then you're done by taking $A=[x\ y\ z]^{-1}$.

In order to complete the basis, you can find the orthogonal complement, which is the null space of $[2\ 1\ {-1}]$, which is easily seen to have a basis consisting of $$\begin{bmatrix}-2\\1\\0\end{bmatrix} \qquad \begin{bmatrix}2\\0\\1\end{bmatrix}$$ Since $$\det\begin{bmatrix} -2 & 2 & 2 \\ 1 & 1 & 0 \\ 0 & -1 & 1 \end{bmatrix}=-6$$ With the above recipe, $$A^{-1}= \begin{bmatrix} -2 & 2 & -2/3 \\ 1 & 1 & 0 \\ 0 & -1 & -1/3 \end{bmatrix}$$ and so $$A= \begin{bmatrix} -1/6 & 2/3 & 1/3 \\ 1/6 & 1/3 & -1/3 \\ -1/2 & -1 & -2 \end{bmatrix}$$ No guessing here.

Note that any other completion of $y$ to a basis will give another matrix (and you can choose how to normalize the determinant in various other ways).

This method can be easily generalized to any set of given columns of $A$ (of any size).