Question

Consider the system of linear equations
$a_{1}x+b_{1}y+c_{1}z+d_1=0,$
$a_{2}x+b_{2}y+c_{2}z+d_2=0,$
$a_{3}x+b_{3}y+c_{3}z+d_3=0$,
Let us denote by $\Delta \left(a,b,c\right)$ the determinant
$\left|\begin{array}{ccc}{a}_1& b_1& c_1\\ a_2& b_2& c_2\\ a_3& b_3& c_3\end{array}\right|,$ 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. $-\frac{\Delta \left(b,c,d\right)}{\Delta \left(a,b,c\right)}$

Answer 1

Answer: (A)

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, $\Delta \left(d,b,c\right)=\left|\begin{array}{ccc}-d_1& b_1& c_1\\ -d_2& b_2& c_2\\ -d_3& b_3& c_3\end{array}\right|$
$=-\left|\begin{array}{ccc}{b}_1& -d_1& c_1\\ b_2& -d_2& c_2\\ b_3& -d_3& c_3\end{array}\right|=+\left|\begin{array}{ccc}{b}_1& c_1& -d_1\\ b_2& c_2& -d_2\\ b_3& c_3& -d_3\end{array}\right|$
$=\Delta \left(b,c,d\right)$
Hence, $x=\frac{\Delta \left(b,c,d\right)}{\Delta \left(a,b,c\right)}$