The sum of three numbers is $6$. If we multiply the third number by $3$ and add the second number to it,we get $11$. By adding the first and third numbers,we get double the second number. Represent this algebraically and find the numbers using the matrix method.

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

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.

If the system of equations $x+y+z=6$,$2x+5y+\alpha z=\beta$,and $x+2y+3z=14$ has infinitely many solutions,then $\alpha+\beta$ is equal to.

Let $A=\left[\begin{array}{rr}2 & -1 \\ 3 & 4\end{array}\right], B=\left[\begin{array}{ll}5 & 2 \\ 7 & 4\end{array}\right], C=\left[\begin{array}{ll}2 & 5 \\ 3 & 8\end{array}\right]$. Find a matrix $D$ such that $CD-AB=O$.

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 -