Q.

Let the system of equations

x+5y–z=1

4x+3y–3z=7

24x+y+λz=μ

λ,μ∈R, have infinitely many solutions. Then the number of the solutions of this system, if x, y, z are integers and satisfy 7≤x+y+z≤77, is :          [2025]

1 3  
2 5  
3 6  
4 4  

Ans.

(1)

For infinitely many solutions, we have D = 0

⇒ |15–143–3241λ|=0

⇒ 1(3λ+3)–5(4λ+72)–1(4–72)=0

⇒ 17λ=–289 ⇒  λ=–17          ... (i)

Now, D1=0

⇒ |15–173–3μ1–17|=0

⇒ 1(–51+3)–5(–119+3μ)–1(7–3μ)=0

⇒ –48+595–15μ–7+3μ=0

⇒ 12μ=540

⇒  μ=45          ... (ii)

Using (i) and (ii), we get

x + 5y – z = 1, 4x + 3y – 3z =7, 24x + y –17z = 45

⇒ z=x+5y–1

∴  4x+3y–3x–15y+3=7

⇒ x–12y=4

⇒ x=4+12y and z=4+12y+5y–1=3+17y

∴  (x,y,z)=(4+12k,k,3+17k)          (∵ Assume y = k)

Also, 7≤7+30k≤77

⇒ 0≤30k<70

⇒ 0≤k≤2.3

⇒ k = 0, 1, 2

Thus, there are three possible solutions.