The value of $Δ=\begin{bmatrix}1 & a & b+c\\1 & b & c+a\\1 & c & a+b\end{bmatrix},$ is
Answer & explanation
Correct answer: option 1
The correct answer is option (4) : 0
Applying $C_2→C_2+C_3,$ we get
$Δ=\begin{bmatrix}1 & a+b+c & b+c\\1 & b+c+a & c+a\\1 & c+a+b & a+b\end{bmatrix}$
$⇒Δ= (a+b+c) \begin{bmatrix}1 & 1 & b+c\\1 & 1 & c+a\\1 & 1 & a+b\end{bmatrix}$ $[ \begin{bmatrix}\text{Taking out a+b + c}\\\text{common from} C_2\end{bmatrix}$
$⇒Δ=(a+b+c)×0=0$ $[∵ C_1 \, and \, C_2\, are \, identical]$