Question

$\Large\textsf{~Demonstre:}\\\\\\\\\Large\begin{array}{l}\begin{vmatrix} \mathsf{a~~a~~a~~a} \\ \mathsf{a~~b~~b~~b} \\ \mathsf{a~~b~~c~~c} \\ \mathsf{a~~b~~c~~d} \end{vmatrix}}\end{array}\mathsf{=a(b-a)(c-b)(d-c)}$


*Matrizes, propriedades, regra de Chió*

Answer 1

Olá!

Por triangularização, devemos deixar todos os elementos da diagonal principal valendo 1 e o elementos abaixo dela deverão ser nulos. E, para que não alteremos o determinante, devemos multiplicá-lo pelo mesmo valor que o dividimos (na linha).
 
 Segue,

$\\ \mathsf{\Delta} = \begin{vmatrix} a & a & a & a \\ a & b & b & b \\ a & b & c & c \\ a & b & c & d \end{vmatrix} \\\\\\ \mathsf{\Delta} = \begin{vmatrix} a & a & a & a \\ a & b & b & b \\ a & b & c & c \\ a & b & c & d \end{vmatrix} \quad \mathsf{L_1 \leftarrow \frac{1}{a} \cdot L_1}$

$\\ \mathsf{\Delta = a \cdot} \begin{vmatrix} 1 & 1 & 1 & 1 \\ a & b & b & b \\ a & b & c & c \\ a & b & c & d \end{vmatrix} \quad \mathsf{L_2 \leftarrow L_2 - a \cdot L_1 \ \wedge \ L_3 \leftarrow L_3 - a \cdot L_1 \ \wedge \ L_4 \leftarrow L_4 - a \cdot L_1} \\\\\\ \mathsf{\Delta = a \cdot} \begin{vmatrix} 1 & 1 & 1 & 1 \\ 0 & b - a & b - a & b - a \\ 0 & b - a & c - a & c - a \\ 0 & b - a & c - a & d - a \end{vmatrix} \quad \mathsf{L_2 \leftarrow \frac{1}{(b - a)} \cdot L_2}$

$\\ \mathsf{\Delta = a \cdot (b - a) \cdot} \begin{vmatrix} 1 & 1 & 1 & 1 \\ 0 & 1 & 1 & 1 \\ 0 & b - a & c - a & c - a \\ 0 & b - a & c - a & d - a \end{vmatrix} \\\\\\ \mathsf{\Delta = a \cdot (b - a) \cdot} \begin{vmatrix} 1 & 1 & 1 & 1 \\ 0 & 1 & 1 & 1 \\ 0 & b - a & c - a & c - a \\ 0 & b - a & c - a & d - a \end{vmatrix} \quad \mathsf{L_3 \leftarrow L_3 - (b - a) \cdot L_2 \ \wedge \ L_4 \leftarrow L_4 - (b - a) \cdot L_2}$

$\\ \mathsf{\Delta = a \cdot (b - a) \cdot} \begin{vmatrix} 1 & 1 & 1 & 1 \\ 0 & 1 & 1 & 1 \\ 0 & 0 & c - a - (b - a) & c - a - (b - a) \\ 0 & 0 & c - a - (b - a) & d - a - (b - a) \end{vmatrix} \\\\\\ \mathsf{\Delta = a \cdot (b - a) \cdot} \begin{vmatrix} 1 & 1 & 1 & 1 \\ 0 & 1 & 1 & 1 \\ 0 & 0 & c - b & c - b \\ 0 & 0 & c - b & d - b \end{vmatrix} \quad \mathsf{L_3 \leftarrow \frac{1}{c - b} \cdot L_3}$

$\\ \mathsf{\Delta = a \cdot (b - a) \cdot (c - b) \cdot } \begin{vmatrix} 1 & 1 & 1 & 1 \\ 0 & 1 & 1 & 1 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & c -b & d - b \end{vmatrix} \quad \mathsf{L_4 \leftarrow L_4 - (c - b) \cdot L_3} \\\\\\ \mathsf{\Delta = a \cdot (b - a) \cdot (c - b) \cdot} \begin{vmatrix} 1 & 1 & 1 & 1 \\ 0 & 1 & 1 & 1 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & d - b - (c - b) \end{vmatrix}$
 
$\\ \mathsf{\Delta = a \cdot (b - a) \cdot (c - b) \cdot } \begin{vmatrix} 1 & 1 & 1 & 1 \\ 0 & 1 & 1 & 1 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & d - c \end{vmatrix} \quad \mathsf{L_4 \leftarrow \frac{1}{(d - c)} \cdot L_4} \\\\\\ \mathsf{\Delta = a \cdot (b - a) \cdot (c - b) \cdot (d - c) \cdot} \begin{vmatrix} 1 & 1 & 1 & 1 \\ 0 & 1 & 1 & 1 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 1 \end{vmatrix}$


 Por fim, efectuamos as multiplicações, veja:

$\\ \mathsf{\Delta = a \cdot (b - a) \cdot (c - b) \cdot (d - c) \cdot 1 \cdot 1 \cdot 1 \cdot 1} \\\\ \boxed{\boxed{\mathsf{\Delta = a(b - a)(c - b)(d - c)}}}$