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Using properties of determinants,prove the following:
\begin{vmatrix} a+b+c & -c &-b \\ -c& a+b+c & -a\\ -b&-a & a+b+c \end{vmatrix}= 2\left ( a+b \right )\left ( b+c \right )\left ( c+a \right ).

 

 

 

 
 
 
 
 

Answers (1)
R Ravindra Pindel

\begin{vmatrix} a+b+c & -c &-b \\ -c& a+b+c & -a\\ -b&-a & a+b+c \end{vmatrix}= 2\left ( a+b \right )\left ( b+c \right )\left ( c+a \right )
Taking LHS
               = \begin{vmatrix} a+b+c & -c &-b \\ -c& a+b+c & -a\\ -b&-a & a+b+c \end{vmatrix}
  Applying  R_{2} \rightarrow R_{2}+R_{1} and R_{3}\rightarrow R_{3}+R_{1}, we get
= \begin{vmatrix} a+b+c & -c &-b \\ a+b& a+b& -\left ( a+b \right )\\ a+c&-\left ( a+c \right ) & a+c \end{vmatrix}
Taking \left ( a+b \right ) comman \: from\: R_{2} \: and \left ( a+c \right )
comman from R_{3}, we get
= \left ( a+b \right )\left ( a+c \right )\begin{vmatrix} a+b+c & -c &-b \\ 1 &1 &-1 \\ 1&-1 &1 \end{vmatrix}
now, Applying c_{1}\rightarrow c_{2}+c_{1}
= \left ( a+b \right )\left ( a+c \right )\begin{vmatrix} \left ( a+b \right )& -\ c\right ) &-b \\ 2& 1&-1 \\ 0& -1&1 \end{vmatrix}
Applying c_{2}\rightarrow c_{2}+c_{3}
= \left ( a+b \right )\left (a+c \right )\: \begin{vmatrix} \left ( a+b \right ) & \left -( b+c \right ) &-b \\ 2& 0&-1 \\ 0 & 0 & 1 \end{vmatrix}
Now,Expanding along R_{3} , weget
= \left ( a+b \right )\left ( a+c \right )\left [ 1\left ( 0+2 \right \left ( b+c \right) ) \right ]
= 2\left ( a+b \right )\left ( a+c \right )\left [ \left [ b+c \right ] \right ]
\Rightarrow \: 2\left ( a+b \right )\left ( a+c \right ) \left ( b+c \right ) = RHS
hence proved
 

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