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The number of values of k for which the system of equations (k + 1) x + 8y = 4k, kx + (k + 3)y = 3k – 1 has infinitely many solutions, is
  • a)
    0
  • b)
    1
  • c)
    2
  • d)
    Infinite
Correct answer is option 'B'. Can you explain this answer?
Verified Answer
The number of values of k for which the system of equations (k + 1) x ...
(k + 1)x + 8y = 4k
kx + (k + 3)y = 3k – 1
The given system of linear equations has infinitely many solutions if ρ(A) = ρ(A|B) < n = 2
To get the rank less than 2, one row should be dependent on another.
nk = (k + 1), n(k + 3) = 8, n(3k – 1) = 4k
by solving the above equations, we get
n = 2, k = 1
Hence for k = 1 only the system has infinitely many solutions.
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The number of values of k for which the system of equations (k + 1) x + 8y = 4k, kx + (k + 3)y = 3k – 1 has infinitely many solutions, isa)0b)1c)2d)InfiniteCorrect answer is option 'B'. Can you explain this answer?
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