Determinants Matrices Question 119
Question: Suppose the system of equations $ a_1x+b_1y+c_1z=d_1 $
$ a_2x+b_2y+c_2z=d_2 $
$ a_3x+b_3y+c_3z=d_3 $ has a unique solution $ (x_0,y_0,z_0) $ . If $ x_0=0, $ then which one of the following is correct-
Options:
A) $ \begin{vmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \\ \end{vmatrix}=0 $
B) $ \begin{vmatrix} d_1 & b_1 & c_1 \\ d_2 & b_2 & c_2 \\ d_3 & b_3 & c_3 \\ \end{vmatrix}=0 $
C) $ \begin{vmatrix} d_1 & a_1 & c_1 \\ d_2 & a_2 & c_2 \\ d_3 & a_3 & c_3 \\ \end{vmatrix}=0 $
D) $ \begin{vmatrix} d_1 & a_1 & b_1 \\ d_2 & a_2 & b_2 \\ d_3 & a_3 & b_3 \\ \end{vmatrix}=0 $
Show Answer
Answer:
Correct Answer: B
Solution:
- [b] The given system of equations is $ a_1x+b_1y+c_1z=d_1 $
$ a_2x+b_2y+c_2z=d_2 $ and $ a_3x+b_3y+c_3z=d_3 $ Let $ \Delta = \begin{vmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \\ \end{vmatrix} $ This system has a unique solution $ x_0,y_0,z_0 $ If $ \Delta \ne 0 $ and $ x_0=\frac{\Delta x}{\Delta }=0\Rightarrow \Delta x=0 $
$ \Rightarrow \begin{vmatrix} d_1 & b_1 & c_1 \\ d_2 & b_2 & c_2 \\ d_3 & b_3 & c_3 \\ \end{vmatrix}=0 $
BETA
