Using properties of determinants, show that

\begin{vmatrix} 3a &-a+b &-a+c \\ -b+a&3b -b+c & \\ -c+a& -c+b &3c \end{vmatrix}=3(a+b+c)(ab+bc+ca)

 

 

 

 
 
 
 
 

Answers (1)

\begin{vmatrix} 3a &-a+b &-a+c \\ -b+a&3b -b+c & \\ -c+a& -c+b &3c \end{vmatrix}=3(a+b+c)(ab+bc+ca)

LHS =RHS

 

LHS= \begin{vmatrix} 3a &-a+b & -a+c \\ -b+a&3b &-b+c \\ -c+a&-c+b &3c \end{vmatrix}

On applying operation C_1 \rightarrow C_1+C_2+C_3 we have,

\Rightarrow \begin{vmatrix} 3a-a+b-a+c &-a+b & -a+c\\ -b+a+3b-b+c& 3b &-b+c \\ -c+a-c+b+3c& -c+b &3c \end{vmatrix}

\Rightarrow \begin{vmatrix} a+b+c &-a+b & -a+c\\ a+b+c & 3b & -b+c \\ a+b+c & -c+b &3c \end{vmatrix}

\Rightarrow (a+b+c)\begin{vmatrix} 1 & -a+b &-a+c \\ 1 & 3b &-b+c \\ 1 & -c+b & 3c \end{vmatrix}

Again applying operations R_1\rightarrow R_1-R_3 and R_2\rightarrow R_2-R_3

\Rightarrow (a+b+c)\begin{vmatrix} 0 & -a+c & -a-2c\\ 0 & 2b+a & -b+a\\ 1 & -c+b & 3c \end{vmatrix}

On applying  R_2\rightarrow R_2-R_1, we get

\Rightarrow (a+b+c)\begin{vmatrix} 0 & -a+c & -a-2c\\ 0 & 2b+a & -b+a\\ 1 & -c+b & 3c \end{vmatrix}

Expanding about C_1, we get 

=(a+b+c)\begin{vmatrix} -a+c &-a-2c \\ 2b+a & -b+a \end{vmatrix}

\Rightarrow (a+b+c)\left [ (c-a)(a-b)+(a+2c)(2b+a) \right ]

=(a+b+c)\left [ ac-bc-a^2+ab+2ab+a^2 +4bc+2ac\right ]

=(a+b+c)\left [3ac+3bc+3ab\right ]

=(a+b+c)\times 3\left [ac+bc+ab\right ]

\Rightarrow 3(a+b+c)(ac+bc+ab)=RHS

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