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Using properties of determinants, show that

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

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

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

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

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

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

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

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

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

Posted by

Ravindra Pindel

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