Q.

If the system of equations

(λ–1)x+(λ–4)y+λz=5

λx+(λ–1)y+(λ–4)z=7

(λ+1)x+(λ+2)y–(λ+2)z=9

has infinitely many solutions, then λ2+λ is equal to          [2025]

1 20  
2 12  
3 6  
4 10  

Ans.

(2)

For infinitely many solutions,

D=|(λ–1)(λ–4)λλ(λ–1)(λ–4)(λ+1)(λ+2)–(λ+2)|=0          ... (i)

On solving (i) along column (1), we get

2λ2–5λ–3=0

⇒ (2λ+1)(λ–3)=0

⇒ λ=3 or –12          ... (ii)

D2=0 ⇒|(λ–1)5λλ7(λ–4)(λ+1)9–(λ+2)|=0

⇒ 2λ2–13λ+21=0

⇒ (2λ–7)(λ–3)=0

⇒ λ=7/2 or λ=3          ... (iii)

Thus, λ=3          (From (ii) and (iii))

∴  λ2+λ=(3)2+3=9+3=12.