How to calculate the determinant of a matrix using Laplace? How to calculate the determinant using Laplace?
$$
\det \begin{bmatrix}
       0 & \dots & 0 & 0 &   a_{1n}           \\[0.3em]
       0 & \dots & 0 & a_{2,n-1} &   a_{2n}   \\[0.3em]
       \dots & \dots & \dots & \dots & \dots             \\[0.3em]
       a_{n1} & \dots & a_{n,n-2} & a_{n,n-1} & a_{nn}  \\[0.3em]
     \end{bmatrix}
$$
I think it's something like:
$$
(a_{n1} * ... *a_{1n}) * (-1)^{n(n+1)}
$$
But I'm not sure about it.
 A: You can swap rows, or just develop with respect to the first row:
$$
\det A=
\det\, \begin{bmatrix}
       0 & \dots & 0 & 0 &   a_{1n}           \\[0.3em]
       0 & \dots & 0 & a_{2,n-1} &   a_{2n}   \\[0.3em]
       \dots & \dots & \dots & \dots & \dots             \\[0.3em]
       a_{n1} & \dots & a_{n,n-2} & a_{n,n-1} & a_{nn}  \\[0.3em]
     \end{bmatrix}
=(-1)^{1+n}a_{1n}
\det\, \begin{bmatrix}
       0 & \dots & 0 & a_{2,n-1} \\[0.3em]
       \dots & \dots & \dots & \dots \\[0.3em]
       a_{n1} & \dots & a_{n,n-2} & a_{n,n-1} \\[0.3em]
     \end{bmatrix}
$$
which has the same form; hence, in the end, you get
$$
\det A=(-1)^{n+1}(-1)^n(-1)^{n-1}\dots(-1)^2(a_{1n}a_{2,n-1}\dots a_{n1})
$$
and the final exponent of $-1$ is
$$
(n+1)+n+\dots+2=\frac{n(n+3)}{2}
$$
Thus
$$
\det A=(a_{1n}a_{2,n-1}\dots a_{n1})(-1)^{n(n+3)/2}.
$$
A: Interchanging two columns of a matrix changes the sign of determinant. Therefore
$$
\det \begin{bmatrix}
       0 & \dots & 0 & 0 &   a_{1n}           \\[0.3em]
       0 & \dots & 0 & a_{2,n-1} &   a_{2n}   \\[0.3em]
       \dots & \dots & \dots & \dots & \dots             \\[0.3em]
       a_{n1} & \dots & a_{n,n-2} & a_{n,n-1} & a_{nn}  \\[0.3em]
     \end{bmatrix}
=\\
=(-1)^{\frac{n(n-1)}2}
\det \begin{bmatrix}
       a_{1n} & 0 & 0 & \dots & 0 &              \\[0.3em]
       a_{2n} & a_{2,n-1} & 0 & \dots & 0        \\[0.3em]
       \dots & \dots & \dots & \dots & \dots             \\[0.3em]
        a_{nn} & a_{n,n-1} & a_{n,n-2} & \dots & a_{n1}     \\[0.3em]
     \end{bmatrix}
$$
(We have made switched the neighboring row $(n-1)$-times to get the last column to the first place. Then $(n-2)$-times for the column before. Altogether we need $1+2+\dots+(n-1)=\frac{n(n-1)}2$ transpositions.) 
Now we have a lower triangular matrix and the determinant is precisely the product of the elements on the diagonal. So the determinant is 
$$=(-1)^{\frac{n(n-1)}2} a_{1,n}a_{2,{n-1}}\cdots a_{n,1}.$$

This is the same result as the one given by egreg, since
the number $\frac{n(n+3)}2-\frac{n(n-1)}2=\frac{4n}2=\frac2n$ is even.
