Determinant of a certain matrix,

Question

I have been asked to solve the following problem:

The matrix $A$ is an $n \times n$ matrix. $A$'s elements consists of a set of the number one in the diagonal from lower left corner to the upper right corner and the rest are zeros. Determine the determinant of $A$.

Now, the problem I´m facing is that the determinant of A shifts between being 1 and -1, depending on how many columns (and rows) the matrix consists of, which is not certain (only that the rows and columns are the same is given). Now how do I show that determinant shifts without using any type of calculator?

I would appreciate any kind of assistance.

Answer 1

HINT

Use row or column flips to show that the determinant is $(-1)^{f(n)}$ where $f$ has fixed form you should discover.

To get this form think about having a matrix with $n$ rows in order $123\cdots n$. How many row flips do you need to move last element to first position to get $n123\cdots (n-1)$? hoe many row flips total to reverse it completely to get $n(n-1)\cdots 321$?

Answer 2

Let:

$$J_n=\begin{pmatrix}0&0&\cdots&0&1\\0&0&\cdots&1&0\\ \cdots&&\cdots&&\cdots\\0&1&\cdots&0&0\\1&0&\cdots&0&0 \end{pmatrix}$$ and $j_n=det(J_n).$

Expanding by Laplace rule $j_n$ by its last column, taking into account the checkerboard pattern of signs, we get:

$$j_n=(-1)^n . j_{n-1}$$

Thus, knowing that $j_0=1$, by an immediate recurrence, we have:

$$j_n=(-1)^n$$