Using properties of determinants,prove that
\begin{vmatrix} b+c & a &a \\ b& c+a & b\\ c& c& a+b \end{vmatrix}= 4abc

 

 

 

 
 
 
 
 

Answers (1)

\begin{vmatrix} b+c & a &a \\ b& c+a & b\\ c& c & a+b \end{vmatrix}= 4\: abc
Let\: \Delta = \begin{vmatrix} b+c & a &a \\ b& c+a & b\\ c& c & a+b \end{vmatrix}
R_{1}\rightarrow R_{1}-R_{2}-R_{3}
\Delta = \begin{vmatrix} 0 & -2c &-2b \\ b& c+a &b \\ c & c& a+b \end{vmatrix}
Expanding R_{1}
\Delta = 0\begin{vmatrix} c+a & b\\ c & a+b \end{vmatrix}-\left ( -2c \right )\begin{vmatrix} b & b\\ c& a+b \end{vmatrix}+\left ( -2b \right )\begin{vmatrix} b &c+a \\ c& c \end{vmatrix}
    = 2c\left ( ab+b^{2} -bc\right )-2b\left ( bc-c^{2}-ac \right )
  = 2abc+2cb^{2}-2bc^{2}-2b^{2}c+2bc^{2}+2abc
 = 4\: abc

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