3
$\begingroup$

Suppose we have matrix which has zeroes everywhere with the exception of bottom left corner, main diagonal and right above the main diagonal $$A_n=\begin{pmatrix} b_1 & c_1 & 0 & \ldots & \ldots & 0 & 0 \\ 0 & b_2 & c_2 & 0 & \ldots & \ldots & 0 \\ 0 & 0 & b_3 & \ddots & & & \vdots \\ \vdots & & & \ddots & \ddots & & \vdots \\ \vdots & & & & b_{n-2} & c_{n-2} & 0 \\ 0 & & & & 0 & b_{n-1} & c_{n-1} \\ c_n & 0 & \ldots & \ldots & 0 & 0 & b_n \\ \end{pmatrix}$$

If we look at Leibniz formula for determinants, we can see that there only two permutations where we have no zereos. (Namely the identity permutation and shift by one place.) From this we get for the determinant: $$D_n=\det(A_n) = b_1b_2\cdots b_n + (-1)^{n-1}c_1c_2\dots c_n$$ This is also the solution given in this question: How to find the determinant of this matrix

Are there some other interesting ways to derive this expression for the determinant?

$\endgroup$
1
$\begingroup$

It is quite straightforward (interesting?) to do the Laplace expansion along the first column or the last row since cofactors are determinants of triangular matrices, e.g. for the first column \begin{align} D_n&=b_1 \begin{vmatrix} b_2 &&&&\\ 0&b_3 &&&\\ 0 &0&b_4&\\ \vdots&\vdots&&\ddots &\\ 0&0&0&\ldots&b_n \end{vmatrix}+(-1)^{n-1}c_n \begin{vmatrix} c_1 &0&0&\ldots&0\\ &c_2 &0&\ldots&0\\ &&c_3&\\ &&&\ddots &\vdots\\ &&&&c_{n-1} \end{vmatrix}\\ &=b_1b_2\ldots b_n+(-1)^{n-1}c_1c_2\ldots c_n. \end{align}

$\endgroup$

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.