Q.

Let A be a 3×3 matrix such that XTAX='O' for all nonzero 3×1 matrices X=[xyz]. If A[111]=[14–5], A[121]=[04–8], and det(adj(2(A+I)))=2α3β5γ, α, β, γ∈ℕ, then α2+β2+γ2 is __________.          [2025]


Ans.

(44)

Given, XTAX=O

∴  [XYZ][a1a2a3b1b2b3c1c2c3][XYZ]=[000], where A =[a1a2a3b1b2b3c1c2c3] 

⇒ X(a1X+a2Y+a3Z)+Y(b1X+b2Y+b3Z)+Z(c1X+c2Y+c3Z)=0

On comparing cofficients, we get

⇒ a1=0, b2=0, c3=0 and a2+b1=0, a3+c1=0, b3+c2=0

∴  A=[0a2a3–a20b3–a3–b30]=[0xy–x0z–y–z0] (Let)

∴  A=[0xy–x0z–y–z0], which is skew-symmetric matrix

Given, A[111]=[14–5] ⇒ A=[0xy–x0z–y–z0][111]=[14–5]

x + y = 1, – x + z = 4, y + z = – 5

[0xy–z0z–y–x0][121]=[04–8]

2x + y = 0, – x + z = 4, – y – 2z = – 8

⇒ x=–1, y=2, z=3

∴  A=[0–12103–2–30]

∴  2(A+I)=[2–24226–4-62]

∴  det(adj(2(A+I)))=|2(A+I)|2=(120)2

2α3β5γ=26×32×52

∴  α=6, β=2, γ=2

∴  α2+β2+γ2=36+4+4=44.