For the system of linear equations
$2x + 4y + 2az = b$
$x + 2y + 3z = 4$
$2x - 5y + 2z = 8$
which of the following is $NOT$ correct?

Consider the system of linear equations in $x, y, z$: $x+2y+tz=0, 6x+y+5tz=0, 3x+t^2y+z=0$. If this system has infinitely many solutions for all $t \in R$,then the determinant of the coefficient matrix must be zero for all $t$. Let $D(t)$ be the determinant of the coefficient matrix. If $D(t) = 0$ for all $t$,analyze the condition.

The positive value of $a$ for which the system of linear homogeneous equations $x+ay+z=0$,$ax+2y-z=0$,and $2x+3y+z=0$ has non-trivial solutions is

If $A = \begin{bmatrix} -1 & 2 \\ 2 & -1 \end{bmatrix}$ and $B = \begin{bmatrix} 3 \\ 1 \end{bmatrix}$,$AX = B$,then $X = $

The system of linear equations $(\sin \theta) x + y - 2z = 0$,$2x - y + (\cos \theta) z = 0$,and $-3x + (\sec \theta) y + 3z = 0$,where $\theta \neq (2n + 1) \frac{\pi}{2}$,has a non-trivial solution for:

The number of solutions of the system of equations $2x + y - z = 7$,$x - 3y + 2z = 1$,and $x + 4y - 3z = 5$ is: