If $x + y - z = 0,\,3x - \alpha y - 3z = 0,\,\,x - 3y + z = 0$ has non zero solution, then $\alpha = $
$-1$
$0$
$1$
$-3$
If $\alpha ,\beta \ne 0$ and $f\left( n \right) = {\alpha ^n} + {\beta ^n}$ and $\left| {\begin{array}{*{20}{c}}3&{1 + f\left( 1 \right)}&{1 + f\left( 2 \right)}\\{1 + f\left( 1 \right)}&{1 + f\left( 2 \right)}&{1 + f\left( 3 \right)}\\{1 + f\left( 2 \right)}&{1 + f\left( 3 \right)}&{1 + f\left( 4 \right)}\end{array}} \right|\; = K{\left( {1 - \alpha } \right)^2}$ ${\left( {1 - \beta } \right)^2}{\left( {\alpha - \beta } \right)^2}$ ,then $K=$ . . . . . .
Consider the system of equations
$ x-2 y+3 z=-1 $ ; $ -x+y-2 z=k $ ; $ x-3 y+4 z=1$
$STATEMENT -1$ : The system of equations has no solution for $\mathrm{k} \neq 3$. and
$STATEMENT - 2$ : The determinant $\left|\begin{array}{ccc}1 & 3 & -1 \\ -1 & -2 & \mathrm{k} \\ 1 & 4 & 1\end{array}\right| \neq 0$, for $\mathrm{k} \neq 3$.
Let $A = \left[ {\begin{array}{*{20}{c}}5&{5\alpha }&\alpha \\0&\alpha &{5\alpha }\\0&0&5\end{array}} \right]$, If ${\left| A \right|^2} = 25$, then $\left| \alpha \right|$ equals
If the system of equations $2 x+3 y-z=5$ ; $x+\alpha y+3 z=-4$ ; $3 x-y+\beta z=7$ has infinitely many solutions, then $13 \alpha \beta$ is equal to
If the system of equations $ax + y + z = 0 , x + by + z = 0 \, \& \, x + y + cz = 0$ $(a, b, c \ne 1)$ has a non-trivial solution, then the value of $\frac{1}{{1\, - \,a}}\,\, + \,\,\frac{1}{{1\, - \,b}}\,\, + \,\,\frac{1}{{1\, - \,c}}$ is :