Let the system of equations x + 5y - z = 1 4x + 3y - 3z = 7 24x + y + z = , R, have infini…

Question

Let the system of equations
x + 5y - z = 1
4x + 3y - 3z = 7
24x + y + \lambda z = \mu 
\lambda , \mu  \in  \mathbb{R}, have infinitely many solutions. Then the number of the solutions of this system, if x, y, z are integers and satisfy 7 \le  x + y + z \le  77, is :

Accepted Answer

$$\begin{aligned}
&\text { For infinitely many solution }\
&\begin{aligned}
& \Delta=0 \
& \left|\[\begin{array}{ccc} 1 & 5 & -1 \ 4 & 3 & -3 \ 24 & 1 & \lambda \end{array}\end{aligned}\end{aligned}$$\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 \
\& \Rightarrow \lambda=-17 \
\& \Delta 1=0

\]

$$\begin{aligned}
& \Rightarrow\left|\[\begin{array}{ccc} 1 & 5 & -1 \ 7 & 3 & -3 \ \mu & 1 & -17 \end{array}\end{aligned}$$\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 \
\& \mu=45 \
\& 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
\]

$\frac{\underset{-}{4 \mathrm{x}}+20 \mathrm{y}=\underset{-}{4+4 \lambda}}{-17 \mathrm{y}=3-\lambda}$

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

Solutions · 1