# Determinant of Matrix $S$ [closed]

Consider the matrix
$$S=\begin{pmatrix} s & 0 & 0 & \cdots & 0&a_{1}\\ -1 & s & 0 & \cdots &0& a_2\\ 0 & 0 & s & \cdots &0& a_{3}\\ \vdots & \ddots & \ddots & \ddots &\vdots&\vdots\\ 0 &0&\ddots&-1&s&a_{n-1}\\ 0 & 0 & \cdots &0& -1& s + a_n \end{pmatrix}$$ where $$s,a_1,a_2, \dots a_n \in F$$ such that $$s \neq 0$$.

How can I show that $$det(S) =s^{n} + a_ns^{n-1}+ \dots + a_2s +a_1$$ ?

## closed as off-topic by Brahadeesh, Delta-u, choco_addicted, Shailesh, ScientificaSep 26 '18 at 15:08

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• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Brahadeesh, Delta-u, choco_addicted, Shailesh, Scientifica
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• Try a Cofactor expansion + induction. – Morgan Rodgers Sep 26 '18 at 5:45
• @One Above All: Can I miss something in the edit? I don't know. If so, let me know or go ahead and edit again...! – Chinnapparaj R Sep 26 '18 at 5:48
• @oneaboveall Please show us your attempts and thoughts on this question. – paulplusx Sep 26 '18 at 8:24
• why isn't the answer accepted? – Viktor Glombik Feb 12 at 18:30

Base Case: Observe \begin{align} \begin{vmatrix} s & a_1\\ -1 & s+a_2 \end{vmatrix} = s^2+a_2s+a_1. \end{align}
Inductive Case: Suppose the statement holds for $$n=k$$, i.e. \begin{align} \begin{vmatrix} s & 0 & 0 & \cdots & 0 & a_1\\ -1 & s & 0 & \cdots & 0& a_2\\ 0 & -1 & s & \cdots & 0 & \vdots\\ 0 & 0 & \ddots & \ddots & 0 & \vdots\\ \vdots & \cdots & \cdots & -1 & s & a_{k-1}\\ 0 & \cdots & \cdots & \cdots & -1 & s+a_k \end{vmatrix} = s^k+a_k s^{k-1}+\cdots+ a_1 \end{align} Then we see that \begin{align} \begin{vmatrix} s & 0 & 0 & \cdots & 0 & a_1\\ -1 & s & 0 & \cdots & 0& a_2\\ 0 & -1 & s & \cdots & 0 & \vdots\\ 0 & 0 & \ddots & \ddots & 0 & \vdots\\ \vdots & \cdots & \cdots & -1 & s & a_{k}\\ 0 & \cdots & \cdots & \cdots & -1 & s+a_{k+1} \end{vmatrix} =& s\begin{vmatrix} s & 0 & 0 & \cdots & 0 & a_2\\ -1 & s & 0 & \cdots & 0& a_3\\ 0 & -1 & s & \cdots & 0 & \vdots\\ 0 & 0 & \ddots & \ddots & 0 & \vdots\\ \vdots & \cdots & \cdots & -1 & s & a_{k}\\ 0 & \cdots & \cdots & \cdots & -1 & s+a_{k+1} \end{vmatrix} + \begin{vmatrix} 0 & 0 & 0 & \cdots & 0 & a_1\\ -1 & s & 0 & \cdots & 0& a_3\\ 0 & -1 & s & \cdots & 0 & \vdots\\ 0 & 0 & \ddots & \ddots & 0 & \vdots\\ \vdots & \cdots & \cdots & -1 & s & a_{k-1}\\ 0 & \cdots & \cdots & \cdots & -1 & s+a_k \end{vmatrix}\\ =&\ s\left(s^k+a_{k+1}s^{k-1}+\cdots+a_2 \right) +(-1)^{k-1} a_1 (-1)^{k-1}\\ =& s^{k+1}+a_{k+1}s^k + \ldots +a_2s+ a_1 \end{align}