Determinants Matrices Question 36
Question: Suppose $ \Delta = \begin{vmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \\ \end{vmatrix} $ and $ \Delta ‘= \begin{vmatrix} a_1+pb_1 & b_1+qc_1 & c_1+ra_1 \\ a_2+pb_2 & b_2+qc_2 & c_2+ra_2 \\ a_3+pb_3 & b_3+qc_3 & c_3+ra_3 \\ \end{vmatrix} $ . Then
Options:
A) $ \Delta ‘=\Delta $
B) $ \Delta ‘=\Delta (1-pqr) $
C) $ \Delta ‘=\Delta (1+p+q+r) $
D) $ \Delta ‘=\Delta (1+pqr) $
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Answer:
Correct Answer: D
Solution:
- [d] $ \Delta ‘= \begin{vmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \\ \end{vmatrix}+ \begin{vmatrix} pb_1 & qc_1 & ra_1 \\ pb_2 & qc_2 & ra_2 \\ pb_3 & qc_3 & ra_3 \\ \end{vmatrix} $ [All other determinants vanish] $ =\Delta +pqr\Delta =\Delta (1+pqr) $
BETA
