Determinants Matrices Question 88
Question: $ A= \begin{vmatrix} 2a & 3r & x \\ 4b & 6s & 2y \\ -2c & -3t & -z \\ \end{vmatrix}=\lambda \begin{vmatrix} a & r & x \\ b & s & y \\ c & t & z \\ \end{vmatrix}, $ then what is the value of $ \lambda $ -
Options:
A) $ 12 $
B) $ -12 $
C) $ 7 $
D) $ -7 $
Show Answer
Answer:
Correct Answer: B
Solution:
- [b] Given, $ \begin{vmatrix} 2a & 3r & x \\ 4b & 6s & 2y \\ -2c & -3t & -z \\ \end{vmatrix}=\lambda \begin{vmatrix} a & r & x \\ b & s & y \\ c & t & z \\ \end{vmatrix} $ Taking 2 common from $ C_1 $ and 3 from $ C_2 $ in LHS
$ \therefore 2\times 3 \begin{vmatrix} a & r & x \\ 2b & 2s & 2y \\ -c & -t & -z \\ \end{vmatrix}=\lambda \begin{vmatrix} a & r & x \\ b & s & y \\ c & t & z \\ \end{vmatrix} $ Taking 2 common from $ R_2 $ and $ -1 $ from $ R_3 $ in LHS
$ \therefore -12 \begin{vmatrix} a & r & x \\ b & s & y \\ c & t & z \\ \end{vmatrix}=\lambda \begin{vmatrix} a & r & x \\ b & s & y \\ c & t & z \\ \end{vmatrix} $
$ \Rightarrow \lambda =-12 $
BETA
