i must demostrate the next property for the determinant of the next matrix
$\begin{vmatrix} t & 0 & 0 & \dots & 0 & 0 & 0 & a_0 \\ -1 & t & & \dots & 0 & 0 & 0 & a_1 \\ 0 & -1 & t & \dots & 0 & 0 & 0 & a_2 \\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots & \vdots & \vdots \\ 0 & 0 & 0 & \dots & -1 & t & 0 & a_{n-3} \\ 0 & 0 & 0 & \dots & 0 & -1 & t & a_{n-2} \\ 0 & 0 & 0 & \dots & 0 & 0 & -1 & t+a_{n-1} \end{vmatrix}= t^n+a_{n-1}t^{n-1}+...+a_1t+a_0$
i had demostrate the property for a 1x1, 2x2 and 3x3. then i supose its true till an (n-1)x(n-1) and i want demostrate for an nxn
i soposse the property
$\begin{vmatrix} t&0&0&...&0&0&0&a_0 \\ -1&t&&0...&0&0&0&a_1 \\ 0&-1&t&...&0&0&0&a_2 \\ ...&...&...&...&...&...&...&... \\ 0&0&0&...&-1&t&0&a_{n-4} \\ 0&0&0&...&0&-1&t&a_{n-3} \\ 0&0&0&...&0&0&-1&t+a_{n-2} \\ \end{vmatrix}= t^{n-1}+a_{n-2}t^{n-2}+...+a_1t+a_0$
now comes the crazy thing, whithout using de inductive Hypotesis i can just work the nxn matrix like
\begin{vmatrix} t&0&0&...&0&0&0&a_0 \\ -1&t&0&...&0&0&0&a_1 \\ 0&-1&t&...&0&0&0&a_2 \\ ...&...&...&...&...&...&...&... \\ 0&0&0&...&-1&t&0&a_{n-3} \\ 0&0&0&...&0&-1&t&a_{n-2} \\ 0&0&0&...&0&0&-1&t+a_{n-1} \\ \end{vmatrix}
\begin{vmatrix} t&0&0&...&0&0&0&a_0 \\ 0&t&0&...&0&0&0&a_1+a_o \frac{1}{t} \\ 0&-1&t&...&0&0&0&a_2 \\ ...&...&...&...&...&...&...&... \\ 0&0&0&...&-1&t&0&a_{n-3} \\ 0&0&0&...&0&-1&t&a_{n-2} \\ 0&0&0&...&0&0&-1&t+a_{n-1} \\ \end{vmatrix}
\begin{vmatrix} t&0&0&...&0&0&0&a_0 \\ 0&t&0&...&0&0&0&a_2+a_o \frac{1}{t} \\ 0&0&t&...&0&0&0&a_1+a_1 \frac{1}{t}+a_0 \frac{1}{t^2} \\ ...&...&...&...&...&...&...&... \\ 0&0&0&...&-1&t&0&a_{n-3} \\ 0&0&0&...&0&-1&t&a_{n-2} \\ 0&0&0&...&0&0&-1&t+a_{n-1} \\ \end{vmatrix}
\begin{vmatrix} t&0&0&...&0&0&0&a_0 \\ 0&t&0&...&0&0&0&a_1+a_o \frac{1}{t} \\ 0&0&t&...&0&0&0&a_2+a_1 \frac{1}{t}+a_0 \frac{1}{t^2} \\ ...&...&...&...&...&...&...&... \\ 0&0&0&...&0&t&0&a_{n-3}+a_{n-4} \frac{1}{t}+a_{n-5}\frac{1}{t^2}...+a_0\frac{1}{t^{n-3}} \\ 0&0&0&...&0&-1&t&a_{n-2} \\ 0&0&0&...&0&0&-1&t+a_{n-1} \\ \end{vmatrix}
\begin{vmatrix} t&0&0&...&0&0&0&a_0 \\ 0&t&0&...&0&0&0&a_1+a_o \frac{1}{t} \\ 0&0&t&...&0&0&0&a_2+a_1 \frac{1}{t}+a_0 \frac{1}{t^2} \\ ...&...&...&...&...&...&...&... \\ 0&0&0&...&0&t&0&a_{n-3}+a_{n-4} \frac{1}{t}+a_{n-5}\frac{1}{t^2}...+a_0\frac{1}{t^{n-3}} \\ 0&0&0&...&0&0&t&a_{n-2}+a_{n-3} \frac{1}{t}+...+a_0\frac{1}{t^{n-2}} \\ 0&0&0&...&0&0&0&t+a_{n-1}+a_{n-2} \frac{1}{t}+...+a_0 \frac{1}{t^{n-1}} \\ \end{vmatrix}
and now i just know that thas the miltiplication of the pivots and that determinant gives me $t^n+a_{n-1}t^{n-1}+...+a_1t+a_0$ but i didnt used the inductive hypotesis so i dont know if i jus didnt did it by induction cause the base case and the inductive hypotesis dont realy care for the resolution, help me plz to know if i did it right.
Yes! You are correct that you did not use induction, since you just used Gaussian elimination and reduced the entire matrix. However, there's an easy fix. Remember that when you try to do induction, you want to "actively" look for your induction hypothesis. Here, we simply note that after doing the first row reduction (when you do row 2 = row 2 + row 1 / t), the bottom right corner is exactly what your induction hypothesis looks like! So instead of continuing to expand, you can just use your hypothesis and substitute.