Question

How to prove determinants using properties ( 2bc-a^2 c^2 b^2 c^2 2ac-b^2 a^2 b^2 a^2 2ab-c^2 ) = ( a b c b c a c a



b.^2?

Answer 1

Answer:

$\left|\begin{array}{ccc}2bc-a^2&c^2&b^2\\c^2&2ac-b^2&a^2\\b^2&-a^2&2ab-c^2\end{array}\right|$

Step-by-step explanation:

$\left|\begin{array}{ccc}a&b&c\\b&c&a\\c&a&b\end{array}\right|^2$

$=\left|\begin{array}{ccc}a&b&c\\b&c&a\\c&a&b\end{array}\right|*\left|\begin{array}{ccc}a&b&c\\b&c&a\\c&a&b\end{array}\right|$

Interchanging Row2 and Row3 in second determinant we get

$=\left|\begin{array}{ccc}a&b&c\\b&c&a\\c&a&b\end{array}\right|*(-1)\left|\begin{array}{ccc}a&b&c\\c&a&b\\b&c&a\end{array}\right|$

$=\left|\begin{array}{ccc}a&b&c\\b&c&a\\c&a&b\end{array}\right|*\left|\begin{array}{ccc}-a&-b&-c\\c&a&b\\b&c&a\end{array}\right|$

By row by column multiplication

$=\left|\begin{array}{ccc}-a^2+bc+bc&-ab+ab+c^2&-ac+b^2+ac\\-ab+c^2+ab&-b^2+ac+ac&-bc+bc+a^2\\-ac+ac+b^2&-bc+a^2+bc&-c^2+ab+ab\end{array}\right|$

$=\left|\begin{array}{ccc}2bc-a^2&c^2&b^2\\c^2&2ac-b^2&a^2\\b^2&-a^2&2ab-c^2\end{array}\right|$

Answer 2

Step-by-step explanation:

this is solution...,,...........