Determinants Matrices Question 77
Question: If $ a,b,c,d>0,x\in R $ and $ (a^{2}+b^{2}+c^{2})x^{2}-2(ab+bc+cd)x+b^{2}+c^{2}+d^{2}\le 0. $ Then, $ \begin{vmatrix} 33 & 14 & \log a \\ 65 & 27 & \log b \\ 97 & 40 & \log c \\ \end{vmatrix} $ is equal to
Options:
A) $ 1 $
B) $ -1 $
C) $ 2 $
D) $ 0 $
Show Answer
Answer:
Correct Answer: D
Solution:
- [d] Given, $ (a^{2}+b^{2}+c^{2})x^{2}-2(ab+bc+cd)x+b^{2}+c^{2} $
$ +d^{2}\le 0 $ .
$ \Rightarrow {{(ax-b)}^{2}}+{{(bx-c)}^{2}}+{{(cx-d)}^{2}}\le 0 $
$ \Rightarrow {{(ax-b)}^{2}}+{{(bx-c)}^{2}}+{{(cx-d)}^{2}}=0 $
$ \Rightarrow \frac{b}{a}=\frac{c}{b}=\frac{d}{c}=x $
$ \Rightarrow b^{2}=acor2\log b=\log a+\log c $
$ Now,\Delta = \begin{vmatrix} 33 & 14 & \log a \\ 65 & 27 & \log b \\ 97 & 40 & \log c \\ \end{vmatrix} $ Apply $ R_1\to R_1+R_3; $
$ \Delta \begin{vmatrix} 130 & 54 & \log a+\log c \\ 65 & 27 & \log b \\ 97 & 40 & \log c \\ \end{vmatrix}=0 $ Now, $ R_1\to R_1-2R_2; $
$ \Delta = \begin{vmatrix} 0 & 0 & 0 \\ 65 & 27 & \log b \\ 97 & 40 & \log c \\ \end{vmatrix}=0 $
BETA
