$\left|\begin{array}{ccc} a+b & a+2 b & a+3 b \\ a+2 b & a+3 b & a+4 b \\ a+4 b & a+5 b & a+6 b \end{array}\right|=$ |
$a^2+b^2+c^2-3 a b c$ 3ab 3a + 5b 0 |
0 |
$\left|\begin{array}{ccc}a+b & a+2 b & a+3 b \\ a+2 b & a+3 b & a+4 b \\ a+4 b & a+5 b & a+6 b\end{array}\right|=\left|\begin{array}{ccc}a+b & a+2 b & a+3 b \\ b & b & b \\ 3 b & 3 b & 3 b\end{array}\right|=0$ by applying R2 → R2 – R1 R3 → R3 – R1 Hence (4) is the correct answer. |