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)

$\frac{\Delta\left(b, c, d\right)}{\Delta\left(a, b, c\right)}$

Answer 2

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, $x = \frac{\Delta \left(b, c, d\right)}{\Delta \left(a, b, c\right)}$