# Is the determinant of this matrix positive or negative?

$\left( \begin{array}{ccc} 1 & 1000 & 2 & 3 &4\\ 5 & 6 &7&1000 &8\\ 1000&9&8&7&6\\ 5 & 4&3&2&1000\\ 1&2&1000&3&4\\ \end{array} \right)$

When I compute the determinant online, I find that it is positive, but I'm supposed to "see" something about the matrix that allows me to know the determinant is positive. What properties does this specific matrix have that allow you to deduce the determinant will be positive?

• If you're supposed to 'see' something, the first thing that should pop right out is those massive 1000. So can you see something about their placement? – Graham Kemp Nov 26 '14 at 1:03
• Just for fun, if you replace $1000$ with $x$ you get $x^5-169 x^3+1508 x^2-4972 x+5232$. This function has three real roots, the largest one being $4.8639\cdots$. So, the determinant is positive for $x \ge 5$. You don't really need $x=1000$. But of course, none of this is "easy to see". – lhf Nov 26 '14 at 1:13

Expand the determinant as a sum of $120$ terms, each being a product of $5$ factors. (That is, imagine you expand it; don't write it out. . . )

All these terms will be positive; then $60$ of them will be added and $60$ will be subtracted. Just one of the terms will be $1000^5$, and by checking the permutation you can see that it will have a positive coefficient.

Every other term will be at most $1000^3\times8\times9$; since exactly $60$ of these terms will be subtracted, we have \eqalign{D &\ge1000^5-60\times1000^3\times8\times9\cr &>1000^5-100\times1000^3\times10\times10\cr &=990\times1000^4\ .\cr}

Using the position occupied by the terms $1000$ we can write the determinant in a nicer way

$$\left| \begin{array}{ccc} 1 & 1000 & 2 & 3 &4\\ 5 & 6 &7&1000 &8\\ 1000&9&8&7&6\\ 5 & 4&3&2&1000\\ 1&2&1000&3&4\\ \end{array} \right|=-\left| \begin{array}{ccc} 1000 & 1 & 2 & 3 &4\\ 6 & 5 &7&1000 &8\\ 9& 1000&8&7&6\\ 4 & 5&3&2&1000\\ 2&1&1000&3&4\\ \end{array} \right|\\=\left| \begin{array}{ccc} 1000 & 3 & 2 & 3 &4\\ 6 & 1000 &7&5 &8\\ 9& 7 &8&1000&6\\ 4 & 2&3&5&1000\\ 2&3&1000&1&4\\ \end{array} \right|=-\left| \begin{array}{ccc} 1000 & 3 & 3 & 2 &4\\ 6 & 1000 &5&7 &8\\ 9& 7 &1000&8&6\\ 4 & 2&5&3&1000\\ 2&3&1&1000&4\\ \end{array} \right|\\=\left| \begin{array}{ccc} 1000 & 3 & 3 & 4 &2\\ 6 & 1000 &5&8 &7\\ 9& 7 &1000&6&8\\ 4 & 2&5&1000&3\\ 2&3&1&4&1000\\ \end{array} \right|=10^{15}+x.$$

Now, having in mind that in a determinant of a matrix of order $5$ there are $120$ terms and one of them is $10^{15}$ we can bound $x.$ Indeed, we have that

$$x>-119 \cdot 1000^3\cdot 9\cdot 8= -8568\cdot 10^9.$$

So,

$$10^{15}+x>10^{15}-8568\cdot 10^9>0.$$

By definition, the determinant is the sum of $60$ positive and $60$ negative terms, one of which is $\mbox{sgn}{\left( \begin{array}{ccccc} 1 & 2 & 3 & 4 & 5 \\ 3 & 1 & 5 & 2 & 4 \\ \end{array} \right)}(1000)^5=\mbox{sgn}\left((13542)\right)(1000)^5=+(1000)^5$.

Each of the other terms has absolute value at most $9\cdot1000^4$, so the sum of the 60 negative terms is greater than or equal to $-60\cdot9\cdot1000^4$, and the determinant is greater than or equal to $(1000)^5-60\cdot9\cdot1000^4$, which is positive.

With an even number of adjacent row swaps, you can put the $1000$s on the diagonal. That determinant must be positive. Using the algorithm which looks at minors, the largest term is always positive.

• ...and the important thing is that the largest term is very large compared to all the others. Obviously it's at least $1000/9 \approx 111$ times larger than any other term, but there are $119$ other terms so a small amount of subtlety is needed. But if you think about it, it's actually at least $1000^2/9^2$ times larger than any other term, which is plenty large enough to dominate the sum. – Erick Wong Nov 26 '14 at 1:01
• Alternatively, note that only 60 of those 119 terms are negative. – Steve Kass Nov 26 '14 at 2:07

You can use 1000 in every row to eliminate the other element to get an upper triangular matrix, and in the process for 1000 is large enough, you can reckon the element in other column won't change. Then you can get the elements on the diagonal all are1000, which shows the determinant is positive.`