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

The following system of equations $3x - 2y + z = 0$,$\lambda x - 14y + 15z = 0$,$x + 2y - 3z = 0$ has a solution other than $x = y = z = 0$ for $\lambda$ equal to

If $A=\begin{bmatrix} 1 & 5 & 3 \\ 2 & 4 & 0 \\ 3 & -1 & -5 \end{bmatrix}$,$B=\begin{bmatrix} -1 \\ -2 \\ 4 \end{bmatrix}$ and $[x \ y \ z] A^{T}=B^{T}$,then $x+y+z=$

Let $A$ be a $3 \times 3$ real matrix such that $A \begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix} = \begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix}$,$A \begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix} = \begin{bmatrix} -1 \\ 0 \\ 1 \end{bmatrix}$,and $A \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix} = \begin{bmatrix} 1 \\ 1 \\ 2 \end{bmatrix}$. If $X = (x_1, x_2, x_3)^T$ and $I$ is an identity matrix of order $3$,then the system $(A - 2I)X = \begin{bmatrix} 4 \\ 1 \\ 1 \end{bmatrix}$ has:

If the system of equations $x+y+z=1$, $x+2y+4z=k$ and $x+4y+10z=k^2$ is consistent, then $k$ is equal to

The number of $3 \times 3$ matrices $A$,whose entries are either $1$ or $-1$ and for which the system $A\begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 1 \\ -1 \\ 0 \end{bmatrix}$ has exactly three distinct solutions,is