Determinants Matrices Question 5
Question: Let $ \Delta = \begin{vmatrix} 1+x_1y_1 & 1+x_1y_2 & 1+x_1y_3 \\ 1+x_2y_1 & 1+x_2y_2 & 1+x_2y_3 \\ 1+x_3y_1 & 1+x_3y_2 & 1+x_3y_3 \\ \end{vmatrix} $ then value of $ \Delta $ is
Options:
A) $ x_1x_2x_3+y_1y_2y_3 $
B) $ x_1x_2x_3y_1y_2y_3 $
C) $ x_2x_3y_2y_3+x_3x_1y_3y_1+x_1x_2y_1y_2 $
D) 0
Show Answer
Answer:
Correct Answer: D
Solution:
- [d] We can write $ \Delta ={\Delta_1}+y_1{\Delta_2}, $ where $ {\Delta_1}= \begin{vmatrix} 1 & 1+x_1y_2 & 1+x_1y_3 \\ 1 & 1+x_2y_2 & 1+x_2y_3 \\ 1 & 1+x_3y_2 & 1+x_3y_3 \\ \end{vmatrix} $ and $ {\Delta_2}= \begin{vmatrix} x_1 & 1+x_1y_2 & 1+x_1y_3 \\ x_2 & 1+x_2y_2 & 1+x_2y_3 \\ x_3 & 1+x_3y_2 & 1+x_3y_3 \\ \end{vmatrix} $ In $ {\Delta_1}, $ use $ C_2\to C_2-C_1 $ and $ C_3\to C_3-C_1 $ so that, $ {\Delta_1}= \begin{vmatrix} 1 & x_1y_2 & x_1y_3 \\ 1 & x_2y_2 & x_2y_3 \\ 1 & x_3y_2 & x_3y_3 \\ \end{vmatrix}=0 $ [ $ \because C_2 $ and $ C_3 $ are proportional] In $ {\Delta_2}, $ us $ C_2\to C_2-y_2C_1 $ and $ C_3\to C_3-y_3C_1 $ to get $ {\Delta_2}= \begin{vmatrix} x_1 & 1 & 1 \\ x_2 & 1 & 1 \\ x_3 & 1 & 1 \\ \end{vmatrix}=0 $ [ $ \because C_2 $ and $ C_3 $ are identical]
$ \therefore \Delta =0 $
BETA
