Determinants Matrices Question 16
Question: Let $ x<1, $ then value of $ \begin{vmatrix} x^{2}+2 & 2x+1 & 1 \\ 2x+1 & x+2 & 1 \\ 3 & 3 & 1 \\ \end{vmatrix} $ is
Options:
A) Non-negative
B) Non-positive
C) Negative
D) Positive
Show Answer
Answer:
Correct Answer: C
Solution:
- [c] $ R_1\to R_1-R_2, $
$ R_2\to R_2-R_3, $ (reduces the determinant to:) $ \begin{vmatrix} x^{2}-2x+1 & x-1 & 0 \\ 2x-2 & x-1 & 0 \\ 3 & 3 & 1 \\ \end{vmatrix} $
$ ={{(x-1)}^{3}}-2{{(x-1)}^{2}}={{(x-1)}^{2}}(x-1-2) $
$ ={{(x-1)}^{2}}(x-3), $ which is clearly negative for $ x<1 $
BETA
