Question

Using the property of determinants and without expanding, prove that: \(\begin{vmatrix}a-b&b-c&c-a\\b-c&c-a&a-b\\c-a&a-b&b-c\end{vmatrix}\)=0

Answer 1

\(\triangle\)= \(\begin{vmatrix}a-b&b-c&c-a\\b-c&c-a&a-b\\c-a&a-b&b-c\end{vmatrix}\)

Applying R₁\(\to\) R₁₊R₂,we have

\(\triangle\)= \(\begin{vmatrix}a-c&b-a&c-b\\b-c&c-a&a-b\\-(a-c)&-(b-a)&-(c-b)\end{vmatrix}\)

= \(-\begin{vmatrix}a-c&b-a&c-b\\b-c&c-a&a-b\\(a-c)&(b-a)&(c-b)\end{vmatrix}\)

Here, the two rows R₁ and R₃ are identical.
\(\triangle\) = 0.

Answer 2

Given :
\(\begin{vmatrix}a-b&b-c&c-a\\b-c&c-a&a-b\\c-a&a-b&b-c\end{vmatrix}\)
So, \(\begin{vmatrix}a&b&c\\b&c&a\\c&a&b\end{vmatrix}-\begin{vmatrix}b&c&a\\c&a&b\\a&b&c\end{vmatrix}\)
Therefore, by finding the determinant of each we get :
= a(cb - a²) - b(b² - ca) + c(ba - c²)
−b(ac - b²) + c(c² - ab) - a(cb - a²)
= abc - a³ - b³ + abc + abc - c³ - abc + b³ + c³ - abc - abc + a³
= 3abc - 3abc
= 0