# Recovering a matrix A from det(A)

I have made up this "fun" problem.

The first row of a 3x3 matrix is $(a_{11} a_{12} a_{13})$. The next row consists of the variables $h, g$ and $c$ in any order. The last row are distinct non-zero fixed scalars, such as $2, 1, 7$ or $-4$ etc. They could be any integer.

Given an expression for $det(A)$ for example:

$(-c -2h)a_{11} -3ca_{13} + 3ha_{12} + 2ga_{13} + ga_{12}$

What is the probability that a matrix you construct meeting the conditions given and with this determinate will be the original matrix A?

Ans: 1

If the last two rows are sets of distinct non-zero integers. What is the probability that a matrix you construct meeting the conditions given will be the original matrix A?

Example $det(A) = -71a_{11} +13a_{12} +11a_{13}$ how many matrices could produce this?

(This one seems to be limited to the number of solutions to a Diophantine system, and it's not I type I can solve, unless there is another way...)

If we let the scalars be any real number will the set of possible matrices become infinite?

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What about the (possible) special case when the last row is a multiple of the second row? In that case you have det(A) = 0, and the probability of finding the original matrix is 0, while the set of possible matrix that produce this result is infinite. Do you exclude this case? Or do you want us to take the probability of this case into account? –  Djaian Oct 27 '10 at 14:33
As written, that case is not excluded from the last question. So, it is a part of the solution, I guess. For the first question it's not possible h, g and c remain variables. –  a little don Oct 27 '10 at 14:35

Letting the entries in the other two rows be $a, b, c, d, e, f$, we are solving a system of Diophantine equations

$$bf - ce = P, cd - af = Q, ae - bd = R.$$

($P, Q, R$ are $-71, -13, 11$ in your example.) This is a system of three equations in six variables so we should expect lots of solutions.

Here's one way to write them down. Let $Q = \gcd(Q, R) Q', R = \gcd(Q, R) R'$ and fix $a = \gcd(Q, R), d = 2 \gcd(Q, R)$. Then the last two equations become $f = c - Q', e = R' + 2b$ and the first equation becomes

$$b(c - Q') - c(R' + b) = - bQ' - cR' = P.$$

since $\gcd(Q', R') = 1$, this always has a solution by Bezout's lemma. It can easily be found using the extended Euclidean algorithm. In fact, if it has one solution it has infinitely many, since if $(b, c)$ are a pair of solutions, so is $(b + Rt, c - Qt)$ for every $t \in \mathbb{Z}$. In particular, by choosing $t$ large enough you can arrange for $b$ to be very positive and $c$ to be very negative, which ensures that $a, b, c, d, e, f$ are distinct (although this is not a particularly natural restriction to me).

In other words, for any choice of $P, Q, R$ with at least two of them nonzero, you get infinitely many solutions, so this "probability" stuff doesn't make sense without more qualification. If two of them are equal to zero then the vectors $(a, d), (b, e), (c, f)$ are proportional and there are either zero or infinitely many solutions depending on whether the third is zero or nonzero.

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