If the system of equations $x + ay = 0,$ $az + y = 0$ and $ax + z = 0$ has infinite solutions,then the value of $a$ is

Let $\alpha_1, \alpha_2$ be two values of $\alpha$ for which the system $2\alpha x + y = 5$,$x - 6y = \alpha$,and $x + y = 2$ is consistent. Then $|2(\alpha_1 + \alpha_2)|$ is -

For a real number $\alpha$,if the system of linear equations $\begin{bmatrix} 1 & \alpha & \alpha^2 \\ \alpha & 1 & \alpha \\ \alpha^2 & \alpha & 1 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 1 \\ -1 \\ 1 \end{bmatrix}$ has infinitely many solutions,then $1+\alpha+\alpha^2=$

The values of $\lambda$ and $\mu$ such that the system of equations $x+y+z=6$,$3x+5y+5z=26$,and $x+2y+\lambda z=\mu$ has no solution are:

Examine the consistency of the system of equations: $x+y+z=1$,$2x+3y+2z=2$,and $ax+ay+2az=4$.

Let $f(x) = 2x^2 + 5x + 1$. If we write $f(x)$ as $f(x) = a(x+1)(x-2) + b(x-2)(x-1) + c(x-1)(x+1)$ for real numbers $a, b, c$, then: