Consider the system of linear equations ${a_1}x + {b_1}y + {c_1}z + {d_1} = 0$, ${a_2}x + {b_2}y + {c_2}z + {d_2} = 0$ and ${a_3}x + {b_3}y + {c_3}z + {d_3} = 0$. Let us denote by $\Delta (a,b,c)$ the determinant $\left| {\,\begin{array}{*{20}{c}}{{a_1}}&{{b_1}}&{{c_1}}\\{{a_2}}&{{b_2}}&{{c_2}}\\{{a_3}}&{{b_3}}&{{c_3}}\end{array}\,} \right|$ if $\Delta (a,b,c) \ne 0$, then the value of $x$ in the unique solution of the above equations is
$\frac{{\Delta (bcd)}}{{\Delta (abc)}}$
$\frac{{ - \Delta (bcd)}}{{\Delta (abc)}}$
$\frac{{\Delta (acd)}}{{\Delta (abc)}}$
$ - \frac{{\Delta (abd)}}{{\Delta (abc)}}$
$\left| {\,\begin{array}{*{20}{c}}1&a&{{a^2} - bc}\\1&b&{{b^2} - ac}\\1&c&{{c^2} - ab}\end{array}\,} \right| = $
If the system of linear equations $2 x-3 y=\gamma+5$ ; $\alpha x+5 y=\beta+1$, where $\alpha, \beta, \gamma \in R$ has infinitely many solutions, then the value of $|9 \alpha+3 \beta+5 \gamma|$ is equal to
The system of equations $kx + y + z =1$ $x + ky + z = k$ and $x + y + zk = k ^{2}$ has no solution if $k$ is equal to
If $x = cy + bz,\,\,y = az + cx,\,\,z = bx + ay$ (where $x, y, z $ are not all zero) have a solution other than $x = 0$, $y = 0$, $z = 0$ then $a, b$ and $ c $ are connected by the relation
Let $S=\left\{A=\left(\begin{array}{lll}0 & 1 & c \\ 1 & a & d \\ 1 & b & e\end{array}\right): a, b, c, d, e \in\{0,1\}\right.$ and $\left.|A| \in\{-1,1\}\right\}$, where $|A|$ denotes the determinant of $A$. Then the number of elements in $S$ is. . . . .