JEE Main 2025 — Mathematics Algebra

For infinitely many solution

$\begin{aligned}
& \Delta=0 \
& \left|\begin{array}{ccc}
1 & 5 & -1 \
4 & 3 & -3 \
24 & 1 & \lambda
\end{array}\right|=0 \
& \Rightarrow 1(3 \lambda+3)-5(4 \lambda+72)-1(4-72)=0 \
& \Rightarrow-17 \lambda+3-4 \times 72-4=0 \
& \Rightarrow 17 \lambda=-289
\end{aligned}$

$\begin{aligned} & \Delta 1=0 \\ & \Rightarrow\left|\begin{array}{ccc}1 & 5 & -1 \\ 7 & 3 & -3 \\ \mu & 1 & -17\end{array}\right|=0 \\ & \Rightarrow 1(-51+3)-5(-119+3 \mu)-1(7-3 \mu)=0 \\ & \Rightarrow-48+595-15 \mu-7+3 \mu=0 \\ & \Rightarrow 12 \mu=540\end{aligned}$​Δ1=0⇒​17μ​531​−1−3−17​​=0⇒1(−51+3)−5(−119+3μ)−1(7−3μ)=0⇒−48+595−15μ−7+3μ=0⇒12μ=540​

$$\begin{aligned} & x+5 y-z=1 \\ & 4 x+3 y-3 z=7 \\ & 24 x+y-17 z=45 \\ & \text { Let } z=1 \\ & x+5 y=1+\lambda] \times 4 \\ & 4 x+3 y=7+3 \lambda \\ & 4 x+20 y=4+4 \lambda \\ & \frac{-17 y=3-\lambda}{-17}\end{aligned}$$​x+5y−z=14x+3y−3z=724x+y−17z=45 Let z=1x+5y=1+λ]×44x+3y=7+3λ4x+20y=4+4λ−17−17y=3−λ​​

$$\begin{aligned} & \begin{aligned} \mathrm{y} & =\frac{\lambda-3}{17}, \mathrm{x}=1+\lambda-\frac{5 \lambda-15}{17} \\ & =\frac{32-12 \lambda}{17} \\ 7 & \leq \frac{\lambda-3}{17}+\frac{32+12 \lambda}{17}+\lambda \leq 77 \\ 7 & \leq \frac{30 \lambda+29}{17} \leq 77 \\ 3 & \leq \lambda \leq 42 \\ \lambda & =3,20,37\end{aligned}\end{aligned}$$​y773λ​=17λ−3​,x=1+λ−175λ−15​=1732−12λ​≤17λ−3​+1732+12λ​+λ≤77≤1730λ+29​≤77≤λ≤42=3,20,37​​