If the system of equations $x + ay = 0,$ $az + y = 0$ and $ax + z = 0$ has infinite solutions, then the value of $a$ is
$-1$
$1$
$0$
No real values
The values of $\alpha$, for which $\left|\begin{array}{ccc}1 & \frac{3}{2} & \alpha+\frac{3}{2} \\ 1 & \frac{1}{3} & \alpha+\frac{1}{3} \\ 2 \alpha+3 & 3 \alpha+1 & 0\end{array}\right|=0$, lie in the interval
Let $\left| {\,\begin{array}{*{20}{c}}{6i}&{ - 3i}&1\\4&{3i}&{ - 1}\\{20}&3&i\end{array}\,} \right| = x + iy$, then
If $\left|\begin{array}{cc}x & 2 \\ 18 & x\end{array}\right|=\left|\begin{array}{cc}6 & 2 \\ 18 & 6\end{array}\right|,$ then $x$ is equal to
The system of linear equations $\lambda x+2 y+2 z=5$ ; $2 \lambda x+3 y+5 z=8$ ; $4 x+\lambda y+6 z=10$ has
$\left| {\,\begin{array}{*{20}{c}}1&1&1\\1&{1 + x}&1\\1&1&{1 + y}\end{array}\,} \right| = $