Question
Consider the system of linear equations
$$\eqalign{
& {a_1}x + {b_1}y + {c_1}z + {d_1} = 0, \cr
& {a_2}x + {b_2}y + {c_2}z + {d_2} = 0, \cr
& {a_3}x + {b_3}y + {c_3}z + {d_3} = 0, \cr} $$
Let us denote by $$\Delta \left( {a,b,c} \right)$$ 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 $$\Delta \left( {a,b,c} \right)\# \,0,$$ then the value of $$x$$ in the unique solution of the above equations is
A.
$$\frac{{\Delta \left( {b,c,d} \right)}}{{\Delta \left( {a,b,c} \right)}}$$
B.
$$\frac{{ - \Delta \left( {b,c,d} \right)}}{{\Delta \left( {a,b,c} \right)}}$$
C.
$$\frac{{\Delta \left( {a,c,d} \right)}}{{\Delta \left( {a,b,c} \right)}}$$
D.
$$ - \frac{{\Delta \left( {a,b,d} \right)}}{{\Delta \left( {a,b,c} \right)}}$$
Answer :
$$\frac{{\Delta \left( {b,c,d} \right)}}{{\Delta \left( {a,b,c} \right)}}$$
Solution :
From the given system of equations,
$$x = \frac{{{D_1}}}{D},y = \frac{{{D_2}}}{D},z = \frac{{{D_3}}}{D}$$
where, $$D = \Delta \left( {a,b,c} \right);\,\,\,{D_1} = \Delta \left( {d,b,c} \right)$$
$${D_2} = \Delta \left( {a,d,c} \right);\,{D_1} = \Delta \left( {a,b,d} \right)$$
Now, $$x = \frac{{\Delta \left( {d,b,c} \right)}}{{\Delta \left( {a,b,c} \right)}}$$
where, \[\Delta \left( {d,b,c} \right) = \left| {\begin{array}{*{20}{c}}
{ - {d_1}}&{{b_1}}&{{c_1}}\\
{ - {d_2}}&{{b_2}}&{{c_2}}\\
{ - {d_3}}&{{b_3}}&{{c_3}}
\end{array}} \right|\]
\[ = \, - \left| {\begin{array}{*{20}{c}}
{{b_1}}&{ - {d_1}}&{{c_1}}\\
{{b_2}}&{ - {d_2}}&{{c_2}}\\
{{b_3}}&{ - {d_3}}&{{c_3}}
\end{array}} \right|\, = + \left| {\begin{array}{*{20}{c}}
{{b_1}}&{{c_1}}&{ - {d_1}}\\
{{b_2}}&{{c_2}}&{ - {d_2}}\\
{{b_3}}&{{c_3}}&{ - {d_3}}
\end{array}} \right|\]
$$ = \,\Delta \left( {b,c,d} \right)\,\,\,{\text{Hence,}}\,\,x = \frac{{\Delta \left( {b,c,d} \right)}}{{\Delta \left( {a,b,c} \right)}}$$