Practicing Success
If the equations x = ay + z, y = az + x and z = ax + y are consistent having non-trivial solution, then |
a3 = 1 a3 + 1 = 0 a + 1 = 0 None of these |
None of these |
$\Delta=\left|\begin{array}{ccc}1 & -a & -1 \\ -1 & 1 & -a \\ -a & -1 & 1\end{array}\right|=0$ ⇒ a3 – 3a = 0 Hence (4) is the correct answer. |