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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.

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Do you know the determinant of an upper-triangular matrix? Can you get one from this matrix by swapping rows? Do you understand the effect on the determinant of swapping rows? –  Gerry Myerson May 11 '13 at 12:39
    
Or how about the sum over permutation form of the determinant? –  ronno May 11 '13 at 12:39

2 Answers 2

up vote 2 down vote accepted

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 $$ \det A=(a_{1n}a_{2,n-1}\dots a_{n1})(-1)^{n(n+3)/2}. $$

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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.

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