# 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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• 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

## 1 Answer

Prove by induction.

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}

• My bad but i didn't get the logic of the last part can you explain why. – One Above All Sep 26 '18 at 13:02