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Joint Entrance Examination
Mathematics
Matrices and Determinants
Question 372
Question 372 of 472
Medium
MCQ
If the system of linear equations 2x + 2ay + az = 0 2x + 3by + bz = 0 2x + 4cy + cz = 0, where a, b, c \[ \in \] R are non-zero distinct; has a non-zero solution, then:
A
\[{1 \over a},{1 \over b},{1 \over c}\] are in A.P.
B
a + b + c = 0
C
a, b, c are in G.P.
D
a,b,c are in A.P.
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Official solution
For non-zero solution \[\left| {
(
2
2
a
a
2
3
b
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4
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c
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\begin{pmatrix} 2 & {2a} & a \\ 2 & {3b} & b \\ 2 & {4c} & c \\ \end{pmatrix}
2
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a
b
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} \right| = 0\] \[ \Rightarrow \] \[\left| {
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\begin{pmatrix} 1 & {2a} & a \\ 0 & {3b - 2a} & {b - a} \\ 0 & {4c - 2a} & {c - a} \\ \end{pmatrix}
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} \right| = 0\] \[ \Rightarrow \] (3b -- 2a) (c --a) -- (b -- a) (4c -- 2a) = 0 \[ \Rightarrow \] 2ac = bc + ab \[ \Rightarrow \] \[{2 \over b} = {1 \over a} + {1 \over c}\] \[ \therefore \] \[{1 \over a},{1 \over b},{1 \over c}\] are in A.P.
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