Using properties of determinants,prove the following: <img alt="\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 )." src="https://

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)

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