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Value of Parameters for Infinitely Many Solutions in System of Equations

If the system of linear equations :

x+y+z=6,x+2y+5z=10,2x+3y+λz=μ\begin{array}{c} x + y + z = 6, \\ x + 2y + 5z = 10, \\ 2x + 3y + \lambda z = \mu \end{array}

has infinitely many solutions, then the value of λ+μ\lambda + \mu equals :

Options

A

1212

B

1616

C

2222

Correct
D

2828

Topics & Concepts

Step-by-Step Solution

To find the values of λ\lambda and μ\mu for which the given system of linear equations has infinitely many solutions, we observe the linear combinations of the equations.

The given system is: x+y+z=6— (1)x + y + z = 6 \quad \text{--- (1)} x+2y+5z=10— (2)x + 2y + 5z = 10 \quad \text{--- (2)} 2x+3y+λz=μ— (3)2x + 3y + \lambda z = \mu \quad \text{--- (3)}

Adding equation (1) and equation (2) gives: (x+y+z)+(x+2y+5z)=6+10(x + y + z) + (x + 2y + 5z) = 6 + 10 2x+3y+6z=16— (4)2x + 3y + 6z = 16 \quad \text{--- (4)}

For the system to have infinitely many solutions, equation (3) must be identical to equation (4). Comparing the coefficients and constant terms of equation (3) and equation (4): λ=6\lambda = 6 μ=16\mu = 16

Thus, the value of λ+μ\lambda + \mu is: λ+μ=6+16=22\lambda + \mu = 6 + 16 = 22

Hence, the correct option is C.

Value of Parameters for Infinitely Many Solutions in System of Equations | Mathematics PYQ Solution - JEE Challenger