Given, A=[aij]3×3 where aij=2j−i we get,
A=[121412121421]
A2=[121412121421][121412121421]
⇒A2=[3234363231263]=3A ⇒A3=3A2=9A
Now, A2+A3+A4+⋯A10
=3A+32A+33A+⋯39A
=3A[3−139−1]=A[2310−3]
Let A=[aij] be a square matrix of order 3 such that aij=2j−i, for all i,j=1,2,3. Then, the matrix A2+A3+…+A10 is equal to
Held on 29 Jun 2022 · Verified 6 Jul 2026.
(2310−1)A
(2310+1)A
(2310+3)A
(2310−3)A
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