Twos and minus ones · Problem 120

s Shown after a correct answer. Revealing it first marks the problem as seen, not solved.

Idea. Expanding along the first row turns the determinant into a short recurrence, which has a very simple solution.

Step 1: the recurrence. Expanding $D_{n}$ along the first row: the entry $2$ multiplies $D_{n - 1}$ . The entry $- 1$ multiplies a minor whose first column has a single nonzero entry $- 1$ ; expanding that column leaves $D_{n - 2}$ . With the signs, $D_{n} = 2 D_{n - 1} - D_{n - 2} .$

Step 2: start. $D_{1} = 2$ and $D_{2} = 4 - 1 = 3$ . Then $D_{3} = 6 - 2 = 4$ , $D_{4} = 5$ , and so on.

Step 3: solve. The recurrence says $D_{n} - D_{n - 1} = D_{n - 1} - D_{n - 2}$ : the differences are constant, equal to 1. So $D_{n} = n + 1$ , and $D_{10} = 11.$

Another way. Gaussian elimination produces pivots $2, \frac{3}{2}, \frac{4}{3}, \dots, \frac{n + 1}{n}$ , whose product telescopes to $n + 1$ . This matrix is the discrete version of $- \frac{d ^{2}}{d x ^{2}}$ ; its eigenvalues are $4 sin^{2} \frac{k π}{2 ( n + 1 )}$ , $k = 1, \dots, n$ .