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QJEEM24Ma9b80: JEE Main 2024 Mathematics - Class 12 JEEM24Ma9b80 Solution: If the system of equations has infinitely many solutions, then…

If the system of equations

amp;2x+7y+λz=3amp;3x+2y+5z=4amp;x+μy+32z=−1\begin{aligned}& 2 x+7 y+\lambda z=3 \\& 3 x+2 y+5 z=4 \\& x+\mu y+32 z=-1\end{aligned}

has infinitely many solutions, then (λ−μ)(\lambda-\mu) is equal to ______ :

Step-by-Step Verified Solution

To determine if the system of equations:

2x+7y+λz=33x+2y+5z=4x+μy+32z=−1\begin{aligned} 2x + 7y + \lambda z = 3 \\ 3x + 2y + 5z = 4 \\ x + \mu y + 32z = -1 \end{aligned}

has infinitely many solutions, we must use Cramer's rule.

The determinants are calculated as follows:

Δamp;=−2λ+3λμ−10μ−509Δ1amp;=2λ+3λμ−15μ−739Δ2amp;=−7λ−7Δ3amp;=μ+39\begin{aligned} \Delta &= -2\lambda + 3\lambda\mu - 10\mu - 509 \\ \Delta_1 &= 2\lambda + 3\lambda\mu - 15\mu - 739 \\ \Delta_2 &= -7\lambda - 7 \\ \Delta_3 &= \mu + 39 \end{aligned}

To have infinitely many solutions, the determinants must satisfy:

Δ=Δ1=Δ2=Δ3=0\begin{aligned} \Delta = \Delta_1 = \Delta_2 = \Delta_3 = 0 \end{aligned}

Solving these equations, we find:

λ=−1,μ=−39\lambda = -1, \mu = -39

Thus, the value of λ−μ\lambda - \mu is:

λ−μ=38\lambda - \mu = 38

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