Interesting determinant

Number Theory Level 2

Let $A$ be a $n \times n$ matrix such that $A_{i,j} = min\{i,j\}$ for $1 \le i,j \le n$ .

Compute $det(A)$ .

(Bonus: Prove that it doesn't matter what $n$ is)

The answer is 1.

2 solutions

Daniel Yarochewsky
Nov 27, 2016

Following the algorithm given, let's build the rows of A...

row 1: a11 = min{1,1} = 1, a12 = min{1,2} = 1, a13 = min{1,3} = 1 , .... , a1n = min{1,n} = 1 row 2: a21 = min{2,1} = 1, a22= min{2,2} = 2, a23 = min{2,3} = 2, ...., a2n = min{2,n} = 2 . . . row n : a1n = min{1,n} = 1, a2n = min{2,n} = 2, a3n = min{3,n} = 3, ...., ann = min{n, n} = n

So, A loos like this:

|1 1 1 . . . 1 | |1 2 2 . . . 2 | |1 2 3 . . . 3 | |. . . . . . . | |1 2 3 4 ... n|

By row reducing A, we can cancel out the rows in the following fashion:

row(1) * (-1) + row(2) row(1) * (-1) + row(3) ... and do this with all rows, until arriving:

|1 1 .... 1| |0 1 .... 1| |0 0 1 ..1| |............| |0 0 0 ..1|

At this point, we know 2 facts:

(1) row reducing A with the described operations doesn't change the determinant of A. (2) the determinant of the row reduced echelon form of A given above is just the product of the diagonal entries, because it's an upper triangular matrix.

Therefore, det(A) = 1.

For any dimension n, you will arrive at a similar row reduced triangular form with determinant = 1.

Denton Young
Jan 15, 2020

Since it doesn't matter what $n$ is, take $n = 1$ .

Then we have a 1 by 1 matrix with a single entry, namely 1, and its determinant is therefore 1.

Interesting determinant

The answer is 1.

2 solutions

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