Two lines a 1 + b 1 y + c 1 = 0 and a 2 x + b 0 y + c 2 = 0, if the denominator a 1 b 2 β a 2 b 1 β 0 then the given system of equations has unique solution (i.e. only one solution) and the solutions are said to be consistent.
β΄ a 1 b 2 β a 2 b 1 β 0
β
Question 1. Find the value of βPβ for which the given system of equations has only one solution (i.e. unique solution).
Px β y = 2 ....(i)
6x β 2y = 3 ....(ii)
Solution: a 1 = P, b 1 = β1, c 1 = β2
a 1 = 6 b 2 = β2, c 2 = β3.
Conditions for unique solution is
β
β΄ P can have all real values except 3.
Two lines a 1 x + b 1 y + c 1 = 0 and a 2 x + b 2 y + c 2 = 0, if the denominator a 1 b 2 β a 2 b 1 = 0 then the given system of equations has no solution and solutions are said to be consistent.
β΄
Question 1. Determine the value of k so that the following linear equations has no solution.
(3x + 1) x + 3y - 2 = 0
(k2 + 1) x + (k - 2) y β 5 = 0.
Sol. Here a 1 = 3k + 1, b 1 = 3 and c 1 = β2
a 2 = k 2 + 1, b 2 = k β 2 and c 2 = β5
For no solution, condition is
.
β (3k + 1) (k β 2) = 3(k 2 + 1)
β 3k 2 β 5k β 2 = 3k 2 + 3
β β5k β 2 = 3
β β5k = 5
β k = β1
Clearly, 3/k - 2 β 2/5 for k = β1.
Hence, the given system of equations will has no solution for k = - 1.
Two lines a
1
x + b
1
y + c
1
= 0 and a
2
x + b
2
y + c
2
= 0, if
then system of equations has many solution and solutions are said to be consistent.
Question 1. Find the value of k for which the system of linear equation
kx + 4y = k β 4
16x + ky = k has infinite solution.
Solution: a 1 = k, b 1 = 4, c 1 = β(k β 4)
a 2 = 16, b 2 = k, c2 = β k.
Here condition is
β k/16 = 4/k = (k - 4)/(k)
βΒ k/16 = 4/k also 4/k = k - 4/k
β k 2 = 64 β 4k = k 2 β 4k
β k = Β± 8 β k(k β 8) = 0.
k = 0 or k = 8 but k = 0 is not possible otherwise equation will be one variable.
β΄ k = 8 is correct value for infinite solution.