JEE Main 2023 — Mathematics Algebra

Given,

The system of equations

$$x+2y+3z=3......(1)$$x+2y+3z=3......(1)

$$4x+3y-4z=4.........(2)$$4x+3y−4z=4.........(2)

$$8x+4y-\lambda z=9+\mu .....(3)$$8x+4y−λz=9+μ.....(3)

Now using the operation, equation $(1)\times 4-(2)$(1)×4−(2) we get,

$$5y+16z=8........(4)$$5y+16z=8........(4)

Now using the operation, equation $(2)\times 2-(3)$(2)×2−(3) we get,

$$2y+(\lambda -8)z=-1-\mu ........(5)$$2y+(λ−8)z=−1−μ........(5)

Now using the operation, equation $(4)\times 2-(3)\times 5$(4)×2−(3)×5 we get,

$$(72-5\lambda )z=21+5\mu$$(72−5λ)z=21+5μ

Now given system of equation has infinitely many solutions, so making coefficient of $z$z to be zero and equating $21+5\mu$21+5μ with zero we get,

$72-5\lambda =0\Rightarrow \lambda =\frac{72}{5}$72−5λ=0⇒λ=572​ and $21+5\mu =0\Rightarrow \mu =\frac{-21}{5}$21+5μ=0⇒μ=5−21​

Hence, the ordered pair $(\lambda ,\mu )$(λ,μ) is equal to $(\frac{72}{5},\frac{-21}{5})$(572​,5−21​).