On the Question on Arrays

Previously, I posted a question on arrays and asked for a solution.

\[ \begin{bmatrix} 2&0&1&0&2&0\\ 0&2&0&1&2&0\\ 1&0&2&0&2&0\\ 0&1&0&2&2&0\\ 1&1&1&1&2&0\\ 0&0&0&0&0&0 \end{bmatrix} \] In the above \( 6 \times 6 \) array, one can choose any \( k \times k \) subarray, with \( 1 < k \leq 6 \) and add \( 1 \) to all its entries. Is it possible to perform the operation a finite number of times such that all entries in the array are multiples of 3?

This was because I did not understand the solution given. This was the answer from the Singapore Mathematical Olympiad 2013 ( Senior )

The answer is no. Let the original array be A A . Consider the following array M=[011110101101110011110011101101011110].M= \begin{bmatrix} 0&1&1&-1&-1&0\\ -1&0&1&-1&0&1\\ -1&-1&0&0&1&1\\ 1&1&0&0&-1&-1\\ 1&0&-1&1&0&-1\\ 0&-1&-1&1&1&0 \end{bmatrix}. Multiply the corresponding elements of the two arrays and compute the sum modulo 3. It's easy to verify that this sum is invariant under the given operation. Since the original sum is 2, one can never obtain an array where all the entries are multiples of 3.

What does this mean? I think that the answer given is pretty vague, and I do not understand the reason or purpose of multiplying the elements of A A and M M , or even how M was derived in the first place. What does this have to do with the original operation in the question?

Please help! I really need it. Thank you.

Note by Timothy Wan
5 years, 1 month ago

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The idea is that you want to find a property that is invariant under the allowed operations. Then, if can calculate the value of the property for matrix A, and see how that compares with the value of the property for "all entries are multiples of 3".

Calculating the sum modulo 3 makes a lot of sense since we want them to be multiples of 3. Other possibilities could be modulo 6 or 9.

The "multiply the corresponding elements" simply means "weight the elements by this extent" IE the contribution of the first row of A is equal to 2×0+0×1+1×1+0×1+2×1+0×0 2 \times 0 + 0 \times 1 + 1 \times 1 + 0 \times -1 + 2 \times -1 + 0 \times 0 .

Coming up with the matrix M is much harder. It is not immediately apparent why that would work. It takes some practice and ideas to figure out how to approach the problem.

Calvin Lin Staff - 5 years, 1 month ago

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