Practicing Success
The value of $\left|\begin{array}{lll}1 & a & b+c \\ 1 & b & c+a \\ 1 & c & a+b\end{array}\right|$ is |
0 $(a+b+c)$ $(a-b)(b-c)(c-a)$ $a^2+b^2+c^2$ |
0 |
It $\Delta=\left|\begin{array}{lll}1 & a & b+c \\ 1 & b & a+c \\ 1 & c & a+b\end{array}\right|$ ⇒ $c_3 \rightarrow c_3+c_2 \quad \Delta=\left|\begin{array}{lll}1 & a & a+b+c \\ 1 & b & a+b+c \\ 1 & c & a+b+c\end{array}\right|$ taking (a + b + c) common factor out $\Rightarrow(a+b+c)\left|\begin{array}{lll}1 & a & 1 \\ 1 & b & 1 \\ 1 & c & 1\end{array}\right|=(a+b+c) \times 0=0$ (two columns have same values here value is zero) |