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 Post subject: Determinant
PostPosted: Tue, 10 Mar 2009 10:37:05 UTC 
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Posts: 182
Compute the determinant of the symmetric matrix
\left(\begin{array}{ccccc}
a_1b_1 & a_1b_2 & a_1b_3 & \cdots & a_1b_n\\
a_1b_2 & a_2b_2 & a_2b_3 & \cdots & a_2b_n\\
a_1b_3 & a_2b_3 & a_3b_3 & \cdots & a_3b_n\\
\vdots & \vdots & \vdots & \ddots & \vdots \\
a_1b_n & a_2b_n & a_3b_n & \cdots & a_nb_n\\
\end{array}\right)

Any hint ?


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 Post subject: Re: Determinant
PostPosted: Thu, 12 Mar 2009 17:18:22 UTC 
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Joined: Mon, 10 Jan 2005 18:23:46 UTC
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D=\left|\begin{array}{ccccc}
a_1b_1 & a_1b_2 & a_1b_3 & \cdots & a_1b_n\\
a_1b_2 & a_2b_2 & a_2b_3 & \cdots & a_2b_n\\
a_1b_3 & a_2b_3 & a_3b_3 & \cdots & a_3b_n\\
\vdots & \vdots & \vdots & \ddots & \vdots \\
a_1b_n & a_2b_n & a_3b_n & \cdots & a_nb_n\\
\end{array}\right|=a_1b_n \left|\begin{array}{ccccc}
b_1 & b_2 & b_3 & \cdots & b_n\\
a_1b_2 & a_2b_2 & a_2b_3 & \cdots & a_2b_n\\
a_1b_3 & a_2b_3 & a_3b_3 & \cdots & a_3b_n\\
\vdots & \vdots & \vdots & \ddots & \vdots \\
a_1 & a_2 & a_3 & \cdots & a_n\\
\end{array}\right|

Now subtract b_k \cdot R_n (bottom row) from R_k (k-th row) for each k>1.
Expand that by minors (choose the row with all zeros but one) to get:
Spoiler:
$ D= a_1b_n \prod_{k=1}^{n-1} \left|\begin{array}{cc} 
b_k & b_{k+1} \\ a_k & a_{k+1} \\
\end{array}\right|

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