Question:
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:
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:
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:
Show Hint
A homogeneous system has a non-zero solution iff the determinant of coefficients is zero; expand it and divide by abc.
Updated On: Oct 31, 2026
- \( {1 \over a},{1 \over b},{1 \over c} \) are in A.P.
- a + b + c = 0
- a, b, c are in G.P.
- a,b,c are in A.P.
Show Solution
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The Correct Option is A
Solution and Explanation
Non-zero solution means the determinant with rows (2, 2a, a), (2, 3b, b), (2, 4c, c) equals 0.
Expanding along the first row:
\(2(3bc-4bc)-2a(2c-2b)+a(8c-6b)=-2bc-4ac+4ab+8ac-6ab=-2bc+4ac-2ab\).
Setting it to zero: \(bc+ab=2ac\). Divide by abc: \(\dfrac1a+\dfrac1c=\dfrac2b\), so \(\dfrac1a,\dfrac1b,\dfrac1c\) are in A.P.
Answer: (A)
Expanding along the first row:
\(2(3bc-4bc)-2a(2c-2b)+a(8c-6b)=-2bc-4ac+4ab+8ac-6ab=-2bc+4ac-2ab\).
Setting it to zero: \(bc+ab=2ac\). Divide by abc: \(\dfrac1a+\dfrac1c=\dfrac2b\), so \(\dfrac1a,\dfrac1b,\dfrac1c\) are in A.P.
Answer: (A)
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