3
$\begingroup$

Calculate the determinant of

$$ \left[ \begin{array}{cccc} 1 & 0 & 0 & 0 & \cdots & 1\\ 1 & a_1 & 0 & 0 & \cdots & 0 \\ 1 & 1 & a_2 & 0 & \cdots & 0 \\ 1 & 0 & 1 & a_3 & \cdots & 0 \\ \vdots & \vdots & \vdots & \vdots & \ddots & \vdots \\ 1 & 0 & 0 & \cdots& 1 & a_{n-1} \\ \end{array} \right]$$

I tried to develop at the first line, but got stuck. Any helps or hints appreciated.

$\endgroup$

1 Answer 1

3
$\begingroup$

Expanding from the first one will force a product along the leading diagonal; the one at the end of the first row will leave you with another determinant \begin{eqnarray*} \left[ \begin{array}{cccc} 1 & 0 & 0 & 0 & \cdots & 1\\ 1 & a_1 & 0 & 0 & \cdots & 0 \\ 1 & 1 & a_2 & 0 & \cdots & 0 \\ 1 & 0 & 1 & a_3 & \cdots & 0 \\ \vdots & \vdots & \vdots & \ddots & \ddots & \vdots \\ 1 & 0 & 0 & \cdots& 1 & a_{n-1} \\ \end{array} \right]=a_1 \cdots a_{n-1}+(-1)^{n+1}\left[ \begin{array}{cccc} 1 & a_1 & 0 & 0 & \cdots & 0 \\ 1 & 1 & a_2 & 0 & \cdots & 0 \\ 1 & 0 & 1 & a_3 & \cdots & 0 \\ \vdots & \vdots & \vdots & \ddots & \ddots & \vdots \\ 1 & 0 & 0 & \cdots& 1 & a_{n-2} \\ 1 & 0 & 0 & \cdots& 0 & 1 \\ \end{array} \right] \end{eqnarray*} Now take the last row of this determinant & make it the first row & shift all the other rows down by one; this will introduce another $(-1)^{n+1}$ but will be essentially the same determinant that you started with but of size one smaller \begin{eqnarray*} -(-1)^{n+1}(-1)^{n+1}\left[ \begin{array}{cccc} 1 & 0 & 0 & 0 & \cdots & 1\\ 1 & a_1 & 0 & 0 & \cdots & 0 \\ 1 & 1 & a_2 & 0 & \cdots & 0 \\ 1 & 0 & 1 & a_3 & \cdots & 0 \\ \vdots & \vdots & \vdots & \ddots & \ddots & \vdots \\ 1 & 0 & 0 & \cdots& 1 & a_{n-2} \\ \end{array} \right] \end{eqnarray*} So now we can use the above arguement recursively & we have \begin{eqnarray*} D=a_1 \cdots a_{n-1}-a_1 \cdots a_{n-2} +\cdots (-1)^{n-2}a_1+ (-1)^{n-1}. \end{eqnarray*} Tidy this expression up a little bit & we have the answer (exactly as Arden states in his comment) \begin{eqnarray*} D=\sum_{k=0}^{n-1}(-1)^{n-1-k}\prod_{i=1}^ka_i. \end{eqnarray*}

$\endgroup$
1
  • $\begingroup$ Yes, now it looks good ! Well done. $\endgroup$
    – Adren
    Feb 25, 2017 at 12:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.