TS EAMCET · Maths · Determinants
The system of equations \(x+3 b y+b z=0, x+2 a y+a z=\) 0 and \(x+4 c y+c z=0\) has
- A only zero solution for any values of \(a, b, c\)
- B non-zero solution for any values of \(a, b, c\)
- C non-zero solution, whenever \(b(a+c)=2 a c\)
- D non-zero solution, whenever \(a+c=2 b\)
Answer & Solution
Correct Answer
(C) non-zero solution, whenever \(b(a+c)=2 a c\)
Step-by-step Solution
Detailed explanation
Given system : \(x+3 \mathrm{~b} y+\mathrm{b} z=0\) \(x+2 a y+a z=0 ; x+4 c y+c z=0\) has non-zero solution if \(\left|\begin{array}{lll}1 & 3 b & b \\ 1 & 2 a & a \\ 1 & 4 c & c\end{array}\right|=0\)…
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