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If the system of equations

x−3y+5z=3x - 3y + 5z = 3

x−2y+4z=4x - 2y + 4z = 4

2x−7y+λz=52x - 7y + \lambda z = 5

has infinite number of solutions, then the value of λ\lambda is:

Solution

✅ Correct Option: 4

For a system of equations to have infinite solutions, the determinant of the coefficient matrix must be zero.

The coefficient matrix is:

A=∣1−351−242−7λ∣A = \begin{vmatrix} 1 & -3 & 5 \\ 1 & -2 & 4 \\ 2 & -7 & \lambda \end{vmatrix}


Expanding the determinant along Row 1:

det⁡(A)=1[(−2)(λ)−(4)(−7)]−(−3)[(1)(λ)−(4)(2)]+5[(1)(−7)−(−2)(2)]\det(A) = 1[(-2)(\lambda) - (4)(-7)] - (-3)[(1)(\lambda) - (4)(2)] + 5[(1)(-7) - (-2)(2)]

=1[−2λ+28]+3[λ−8]+5[−7+4]= 1[-2\lambda + 28] + 3[\lambda - 8] + 5[-7 + 4]

=−2λ+28+3λ−24−15= -2\lambda + 28 + 3\lambda - 24 - 15

=λ+28−24−15= \lambda + 28 - 24 - 15

=λ−11= \lambda - 11


Setting the determinant equal to zero:

λ−11=0\lambda - 11 = 0

λ=11\lambda = 11

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