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]
(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.