Q.

Let A be a 3×3 real matrix such that A2(A–2I)–4(A–I)=O, where I and O are the identity and null matrices, respectively. If A5=αA2+βA+γI, where α, β and γ are real constants, then α+β+γ is equal to :          [2025]

1 20  
2 76  
3 12  
4 4  

Ans.

(3)

We have, A2(A–2I)–4(A–I)=O

⇒ A3–2A2–4A+4I=O

⇒ A3=2A2+4A–4I          ... (i)

Now, A4=A·A3=A(2A2+4A–4I)

      =2A3+4A2–4A=2(2A2+4A–4I)+4A2–4A          [From (i)]

      =4A2+8A–8I+4A2–4A=8A2+ 4A–8I

Similarly, A5=A·A4=A(8A2+4A–8I)

      =8A3+4A2–8A=8(2A2+4A–4I)+4A2–8A          [From (i)]

      =20A2+24A–32I

⇒ α=20, β=24 and γ=–32

∴   α+β+γ=20+24–32=12.