Determinants Matrices Question 29
Question: If the system of linear equations $ x+2ay+az=0; $
$ x+3by+bz=0; $
$ x+4cy+cz=0 $ has a non - zero solution, then a, b, c.
Options:
A) Satisfy $ a+2b+3c=0 $
B) Are in A.P
C) Are in G.P
D) Are in H.P.
Show Answer
Answer:
Correct Answer: D
Solution:
- [d] For homogeneous system of equations to have non zero solution, $ \Delta =0 $
$ \begin{vmatrix} 1 & 2a & a \\ 1 & 3b & b \\ 1 & ac & c \\ \end{vmatrix}=0[\therefore C_2\to C_2-2C_3] $
$ \begin{vmatrix} 1 & 0 & a \\ 1 & b & b \\ 1 & 2c & c \\ \end{vmatrix}=0[R_3\to R_3-R_2,R_2\to R_2-R_1] $ On simplification, $ \frac{2}{b}=\frac{1}{a}+\frac{1}{c} $
$ \therefore $ a, b, c are in Harmonic Progression.
BETA
