Question

please prove this determinant question:-

Answer 1

Hey there!

Solution:


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


Taking L.H.S


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



Applying $C_{1} => C_{1} + C_{2} + C_{3}$



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



Applying $R_{1} => R_{1} + R_{2} - R_{3}$


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


Taking ( a + b - c ) common from $R_{1}$


$= \: (a + b - c)\begin{vmatrix}1 &1& - 1 \\ b&b&c - a \\ c&b - a&c \end{vmatrix}$



Applying $C_{1} => C_{1} - C_{2}$


$= \: (a + b - c)\begin{vmatrix}0 &1& - 1 \\ 0&b&c - a \\ a - b + c&b - a&c \end{vmatrix}$


Taking ( a - b + c ) common from $C_{1}$


$= \: (a + b - c)(a - b + c)\begin{vmatrix}0 &1& - 1 \\ 0&b&c - a \\ 1&b - a&c \end{vmatrix}$


Expanding by $C_{1}$


∆ = ( a + b - c ) ( a - b + c ) [ 1 ( c - a + b ) ]


∆ = ( a + b - c ) ( a - b + c ) ( c - a + b )


L.H.S = R.H.S


Hence Proved.




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