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Compute the determinant of the following anti-triangular matrix. $$A = \begin{bmatrix} 0& 0&\dots & 0& a_{1\ n}\\ 0 & 0&\dots & a_{2\ n-1}& a_{2\ n}\\ \vdots & \vdots & \ddots & \vdots& \vdots\\ 0 & a_{n-1\ 2}& \dots & a_{n-1\ n-1}& a_{n-1\ n}\\ a_{n\ 1}& a_{n\ 2}& \dots & a_{n\ n-1}& a_{n\ n}\end{bmatrix}$$

Since any determinant is

$$\det A = (-1)^{1+1} \alpha_{1\ 1}+(-1)^{2+1}\alpha_{1\ 2}+\dots+(-1)^{n+1}\alpha_{1\ n}$$

where $\alpha_{ij}$ are the cofactors, I have correctly computed

$$\det A = a_{1\ 1} a_{2\ 2} \cdots a_{n\ n}$$

but the solution is that times a function of $(-1)^{f(n)}$ and I really don't know how to find that.

Rodrigo de Azevedo
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Aley20
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    Where in the matrix are the entries you give in the supposed answer? You’d better read and proofread very carefully. – Ted Shifrin May 24 '23 at 00:52
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    Expand along the first row. Expand the resulting determinant along its first row. Continue. – Gerry Myerson May 24 '23 at 01:54
  • Note: Your edit didn't significantly change the setup (EG We can permute rows and columns, and only change the determinant by a factor of $(-1)^n$), so everything that was said perviously still applies. – Calvin Lin May 24 '23 at 03:27

2 Answers2

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Right-multiplying by the $n \times n$ reversal matrix $\bf R$ (which is a permutation matrix and, thus, orthogonal), we reverse the order of the columns and obtain a lower triangular matrix whose determinant is easy to compute.

$$ \det ({\bf A} {\bf R}) = \det ({\bf A}) \underbrace{\det ({\bf R})}_{= \pm 1} = \prod_{i=1}^n a_{i,n+1-i}$$

Thus, what is left is determining the sign of $\det ({\bf R})$ as a function of $n$.

Rodrigo de Azevedo
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Hint Like you write, inductively applying the cofactor expansion to the top row (or leftmost column) of each lower-left square submatrix gives that $$\det A = (-1)^{n - 1} a_{1,n} \cdot (-1)^{n - 2} a_{2,n-1}\cdot \cdots \cdot (-1)^1 a_{n-1,2} \cdot (-1)^0 a_{n,1} = \pm \prod_{i = 1}^n a_{i, n + 1 - i}$$ for some sign $\pm$.

Collecting powers of $-1$ in the above expansion shows that the exponent of $-1$---whose parity determines $\pm$---is $(n - 1) + (n - 2) + \cdots + 2 + 1$.

This exponent coincides with ${n \choose 2}$. Note that since ${{n + 4} \choose 2} - {n \choose 2} = 4 n + 6$ is even for all $n$, the power $(-1)^{n \choose 2}$, hence the sign $\pm$, depends only on the residue class of $n$ modulo $4$.

Travis Willse
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  • but my question is how do you get the $n-1, n-2,\dots$ exponents? Due to the formula of a determinant that I have written in my post, it should be just the addition of the position of all the diagonal (from upper left to lower right), like for the position $a_{1,n}$ it should have $n+1$, for $a_{2,n-1}$ it should have $n-1+2=n+1$, isn't it? Or is my formula wrong? – Aley20 May 24 '23 at 10:29
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    When computing the determinant of the lower-left $(n - 1) \times (n - 1)$ submatrix $a'$ and expanding across its top row, the upper-right entry of $A'$ is the $(1, n - 1)$ entry of that matrix, not its $(2, n - 1)$ entry, so the cofactor expansion gives $\det A' = (-1)^{n - 1 + 1} a_{2, n - 1} \det A'' = (-1)^n a_{2, n - 1} \det A''$. My answer gives $n - 2$ for the power, but of course $(-1)^{n - 2} = (-1)^n$. – Travis Willse May 24 '23 at 13:14
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    (Note that if all of $n$ exponents of $-1$ were $n + 1$, the product of those powers would be $(-1)^{n (n + 1)} = 1$ for any $n$, whereas, for example, $\det\pmatrix{0&a_{12}\a_{21}&\ast} = -a_{12} a_{21}$.) – Travis Willse May 24 '23 at 13:17