Q.

If the system of equations

2x+λy+3z=5

3x+2y–z=7

4x+5y+μz=9

has infinitely many solutions, then (λ2+μ2) is equal to:          [2025]

1 22  
2 18  
3 26  
4 30  

Ans.

(3)

For infinitely man solutions, we have △=△1=△2=△3=0

Now, △=|2λ332–145μ|=0

⇒ 2(2μ+5)–λ(3μ+4)+3(15–8)=0

⇒ 4μ+10–3λμ–4λ+21=0

⇒ 4μ–4λ–3λμ+31=0           ... (i)

Now, △2=|25337–149μ|=0

⇒ 2(7μ+9)–5(3μ+4)+3(27–28)

⇒ 14μ+18–15μ–20–3=0

⇒ –μ+18–23=0 ⇒ μ=–5

Substituting the value of μ in (i), we get

   –20–4λ+15λ+31=0

⇒ 11λ+11=0 ⇒ λ=–1

∴  λ2+μ2=(–1)2+(–5)2=26.