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)}}$
Evaluate the determinants : $\left|\begin{array}{cc}2 & 4 \\ -5 & -1\end{array}\right|$
If $A = \left| {\,\begin{array}{*{20}{c}}1&1&1\\a&b&c\\{{a^3}}&{{b^3}}&{{c^3}}\end{array}\,} \right|,B = \left| {\,\begin{array}{*{20}{c}}1&1&1\\{{a^2}}&{{b^2}}&{{c^2}}\\{{a^3}}&{{b^3}}&{{c^3}}\end{array}\,} \right|,C = \left| {\,\begin{array}{*{20}{c}}a&b&c\\{{a^2}}&{{b^2}}&{{c^2}}\\{{a^3}}&{{b^3}}&{{c^3}}\end{array}\,} \right|,$ then which relation is correct
The following system of linear equations $2 x+3 y+2 z=9$ ; $3 x+2 y+2 z=9$ ;$x-y+4 z=8$