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While preparing an exam, I found the following question:

What is the largest possible number of zero entries in any $5 \times 5$ matrix with a non-zero determinant?

  1. 25
  2. 15
  3. 16
  4. 20

I know that the answer is 20, but why? I can find the formula or the theorem that makes that answer true. Any help please?

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4 Answers 4

Nytram12 already gave the lower bound: the identity matrix shows that a $5 \times 5$ matrix with 20 zeroes can have determinant 1, by choosing the identity matrix. To show that no more than 20 zeroes are possible, note that if a matrix has non-zero determinant, it has a non-zero entry in every row and every column: If matrix $A$ has a column consists of only zeroes, we can multiply that row by 0 and get the same matrix. Thus, in this case, $0 \cdot \det A = \det A$ by the multilinearity of the determinant, i.e., $\det A = 0$.

This generalizes to larger matrices: A $n\times n$ matrix with determinant $\neq 0$ can have at most $n^2-n$ zeroes, and the identity matrix is an example of such a matrix.

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Surely it's just because the Identity matrix would be the one where you have the most zeroes but a non-zero determinant?

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Using Leibniz formula for calculating determinants you can see that if you have more than $20$ zeros the determinant is zero.
Or, if you have more than $20$ zeros show, using pigeonhole principle, that necessarily a row must be zero.

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Clearly we can have a diagonal matrix with a non-zero determinant. It follows that we must show that there does not exist an invertible matrix with $4$ or less non-zero entries.

This is true because we have $5$ rows and $4$ or less non-zero entries. Therefore at most $4$ of the $5$ rows can be non-zero. In particular, this means that there must exist an all zero row, which is bad news for our determinant.

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