Question

do it............... ​

Answer 1

We need to find the value of,

$\longrightarrow \Delta=\left|\begin{array}{ccc}a+b+c&-c&-b\\-c&a+b+c&-a\\-b&-a&a+b+c\end{array}\right|$

Performing $R_1\to R_1+R_2,$

$\longrightarrow \Delta=\left|\begin{array}{ccc}a+b&a+b&-a-b\\-c&a+b+c&-a\\-b&-a&a+b+c\end{array}\right|$

Taking $a+b$ common from $R_1,$

$\longrightarrow \Delta=(a+b)\left|\begin{array}{ccc}1&1&-1\\-c&a+b+c&-a\\-b&-a&a+b+c\end{array}\right|$

Performing $R_2\to R_2+R_3,$

$\longrightarrow \Delta=(a+b)\left|\begin{array}{ccc}1&1&-1\\-b-c&b+c&b+c\\-b&-a&a+b+c\end{array}\right|$

Taking $b+c$ common from $R_2,$

$\longrightarrow \Delta=(a+b)(b+c)\left|\begin{array}{ccc}1&1&-1\\-1&1&1\\-b&-a&a+b+c\end{array}\right|$

Performing $C_3\to C_1+C_2+C_3,$

$\longrightarrow \Delta=(a+b)(b+c)\left|\begin{array}{ccc}1&1&1\\-1&1&1\\-b&-a&c\end{array}\right|$

Performing $R_1\to R_1-R_2,$

$\longrightarrow \Delta=(a+b)(b+c)\left|\begin{array}{ccc}2&0&0\\-1&1&1\\-b&-a&c\end{array}\right|$

Now, expanding the determinant along $R_1,$ we get,

$\longrightarrow\underline{\underline{\Delta=2(a+b)(b+c)(c+a)}}$

If $\Delta=0,$ atleast one among the following condition should be satisfied.

$a+b=0$

$b+c=0$

$c+a=0$

which means one among $a,\ b$ and $c$ should be additive inverse of the other.

Answer 2

☘️ See the attachment picture for Explanation.

$\Large\mathbb\pink{THANKS}$

$\Large\mathbb\green{HOPE\: IT'S\: HELPFUL}$