Determinants Matrices Question 39
Question: If $ a\ne b\ne c $ are all positive, then the value of the determinant $ \begin{vmatrix} a & b & c \\ b & c & a \\ c & a & b \\ \end{vmatrix} $ is
Options:
A) Non-negative
B) Non-positive
C) Negative
D) Positive
Show Answer
Answer:
Correct Answer: C
Solution:
- [c] $ \begin{vmatrix} a & b & c \\ b & c & a \\ c & a & b \\ \end{vmatrix}= \begin{vmatrix} a+b+c & b & c \\ a+b+c & c & a \\ a+b+c & a & b \\ \end{vmatrix} $
$ (\because C_1\to C_1+C_2+C_3) $
$ =(a+b+c) \begin{vmatrix} 1 & b & c \\ 1 & c & a \\ 1 & a & b \\ \end{vmatrix} $ [on taking (a+b+c) common from $ C_1 $ ] $ =(a+b+c)[1(bc-a^{2})-b(b-a)+c(a-c)] $
$ =(a+b+c)[bc-a^{2}-b^{2}+ab+ac-c^{2}] $
$ =(a+b+c)[-(a^{2}+b^{2}+c^{2}-ab-bc-ca)] $
$ =-\frac{1}{2}(a+b+c)[{{(a-b)}^{2}}+{{(b-c)}^{2}}+{{(c-a)}^{2}}] $ Hence, the determinant is negative value
BETA
