Given $A = \begin{bmatrix} 1 & 3 \\ 2 & 2 \end{bmatrix}$ and $I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$. If $A - \lambda I$ is a singular matrix,then:

The values of $p$ and $q$ so that the system of equations $2x + py + 6z = 8$,$x + 2y + qz = 5$ and $x + y + 3z = 4$ may have no solution are

If the system of equations $x + 2y + 3z = 4$,$x + py + 2z = 3$,and $x + 4y + \mu z = 3$ has an infinite number of solutions,then:

Consider the simultaneous linear equations $\beta x + \alpha y - z = -1$, $3x - \beta y + \alpha z = 0$, and $\alpha x + \beta y + z = 1$. In the usual notation used in Cramer's rule, given that $\frac{\Delta_1}{\Delta} = -1$, $\frac{\Delta_2}{\Delta} = 1$, and $\frac{\Delta_3}{\Delta} = 2$, then $(\alpha, \beta) = $

Let $\lambda, \mu \in R$. If the system of equations
$3x + 5y + \lambda z = 3$
$7x + 11y - 9z = 2$
$97x + 155y - 189z = \mu$
has infinitely many solutions,then $\mu + 2\lambda$ is equal to :

For the system $S$ of linear equations $x+y+z=3, 2x+2y-z=3, x+y+\lambda z=1$, the incorrect option among the following statements is: