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
$\begin{vmatrix}a_{1}&b_{1}&c_{1}\\ a_{2}&b_{2}&c_{2}\\ a_{3}&b_{3}&c_{3}\end{vmatrix},$ 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, $\Delta \left(d, b, c\right) = \begin{vmatrix}-d_{1}&b_{1}&c_{1}\\ -d_{2}&b_{2}&c_{2}\\ -d_{3}&b_{3}&c_{3}\end{vmatrix}$
$= -\begin{vmatrix}b_{1}&-d_{1}&c_{1}\\ b_{2}&-d_{2}&c_{2}\\ b_{3}&-d_{3}&c_{3}\end{vmatrix} = +\begin{vmatrix}b_{1}&c_{1}&-d_{1}\\ b_{2}&c_{2}&-d_{2}\\ b_{3}&c_{3}&-d_{3}\end{vmatrix}$
$= \Delta \left(b, c, d\right)$
Hence, $x = \frac{\Delta \left(b, c, d\right)}{\Delta \left(a, b, c\right)}$