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]
(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.