Let matrix A = ⎣ ⎡ x 1 1 y 2 1 − z 3 2 ⎦ ⎤
where x , y , z ∈ N . If ∣ a d j ( a d j ( a d j ( a d j . A ) ) ) ∣ = 4 8 . 5 1 6 , then find the no. of positive integral solutions.
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Nicely Done. Upvoted.
∣ a d j ( a d j ( a d j ( a d j . A ) ) ) ∣ = ∣ A ∣ ( ( n − 1 ) 2 ) 2 = ∣ A ∣ ( ( ( 3 − 1 ) 2 ) 2 ) = ∣ A ∣ 1 6 , where n = o r d e r o f m a t r i x
Therefore, ∣ A ∣ 1 6 = 4 8 × 5 1 6
→ ∣ A ∣ = 1 0
by simplifing we get ∣ A ∣ = x + y + z = 1 0
so, number of positive integral solution = 9 C 2 = 3 6
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det(adj A)= (det A )^(2)
repeating this rule, we see that det(A) = 10
now simply expand the determinant, ( no need of any advanced tricks or row transformation )
to get x+y+z=10
then use formula for positive integral solution (10-1)C(3-1) = 9C2=36
, and yes these are often the sort of problems in JEE, moulding a problem so weirdly so that people will be scared to even think about it, but once thought, they often become simple