Suppose $A$ is a $7\times 7$ matrix with all entries less than $1$ in magnitude. Prove that $\det (100I+A) > 0$.
This is my first post on the mathematics stack exchange, so forgive me if it’s not in the right format or if it’s been asked before (I couldn’t find anything like it, though).
My elementary linear algebra professor posted this question in the beginning of class this morning and left it up to us to discuss independently.
I can gather that the determinant of $100 I_7$ is $100^7$ and every element in $A$ is in between $-1$ and $1$ (given) but other than that, I’ve got no clue where to start.
Thank you!
The product of elements in the main diagonal is $100^7$. Estimate the sum of all other summands in the determinant: notice, that every other permutation contains $100$ at most five times. How many permutations are there all together? It is $7!=5040$. But $100^7>100^5\cdot 5040$, which means that all the other permutations cannot be as negative as how big that one permutation is alone.
Edit: Thank you user1551 for pointing out the small mistake: we only know that the product of elemnts in the main diagonal is at least $99^7$, while when bounding the negative part, we should use $101$, not $100$. But the proof still works as $99^7>101^5\cdot 5040$.