Determinants and Matrices — Maths STD 11 Science — Question

Applying $\mathrm{C}_3 \rightarrow \mathrm{C}_3+\mathrm{C}_2$, we get.

$\mathrm{D}=\left|\begin{array}{ccc}1 & a & a+b+c \\ 1 & b & a+b+c \\ 1 & c & a+b+c\end{array}\right|$

Taking $(a+b+c)$ common from $C_3$, we get

$D=(a+b+c)\left|\begin{array}{lll}1 & a & 1 \\ 1 & b & 1 \\ 1 & c & 1\end{array}\right|$

= (a + b + c)(0)

$\ldots\left[\because C_1\right.$ and $C_3$ are identical $]$

= 0