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I was reading this Transversal of Primes, and the solution shown for an $11 \times 11$ grid. Made me think of an identity matrix.

First, have each $a_{ij}$ be either $1$ for a prime number or $0$ otherwise. then the $11$ primes will "map" to the identity matrix. (for $GL_{11}(\mathbb{R})$)?

I am asking to help clarify my own knowledge, I have started learning a bit of Abstract Algebra and reviewing Linear Algebra (because my TI-89 made class to easy).

Also, are there any other Prime Identity matrices, which can "map" to an identity matrix? I haven't found a good way of computing the distance between each prime number corresponding to the location in the matrix, still working on that.

*Please edit to make better, I admit this may be an awkward question.

EDIT to make the question more self-contained: a "transversal of primes" is obtained by writing the numbers from $1$ to $p^2$ ($p$ a prime) in their natural order in a $p\times p$ grid and then choosing primes in such a way that you choose exactly one number from each row and from each column. The link gives an example for $p=11$ and asks whether there is such a transversal for every prime $p$. For $p=3$, we're talking about the grid $\matrix{1&2&3\cr4&5&6\cr7&8&9\cr}$ and $3, 5, 7$ is a transversal of primes.

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    @anon Not sure if I am using the correct usage of "maps", but if you replace all the Primes that fit the correct location, would that then map, aka be mathematically the same as the identity matrix? That is why I am asking, because I want to make sure I have the correct idea.2012-06-13

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No, replacing the prime entries by $1$ and others by $0$ will never produce the identity matrix. For one thing, the upper left corner has $1$, which is not prime. But even if we considered $1$ as a prime number for purposes of this problem, there is another issue: the bottom right corner has $p^2$, which is definitely not prime. Besides, there are prime numbers outside of the main diagonal, such as $p$ in the upper right corner.