Combinatorics involved in matrix

Let matrix A = [ x y z 1 2 3 1 1 2 ] A= \begin{bmatrix} x & y & -z \\ 1 & 2 & 3 \\ 1 & 1 & 2 \end{bmatrix}

where x , y , z N x,y,z \in \mathbb{N} . If a d j ( a d j ( a d j ( a d j . A ) ) ) = 4 8 . 5 16 |adj(adj(adj(adj.A)))|=4^8.5^{16} , then find the no. of positive integral solutions.


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The answer is 36.

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3 solutions

Mvs Saketh
Feb 19, 2015

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

Nicely Done. Upvoted.

Keshav Tiwari - 6 years, 3 months ago
Aniket Verma
Mar 12, 2015

a d j ( a d j ( a d j ( a d j . A ) ) ) |adj(adj(adj(adj.A)))| = = A ( ( n 1 ) 2 ) 2 |A|^{((n-1)^2)^2} = A ( ( ( 3 1 ) 2 ) 2 ) = A 16 = |A|^{(((3-1)^2)^2)} = |A|^{16} , where n = o r d e r o f m a t r i x n= order~of~matrix

Therefore, A 16 = 4 8 × 5 16 |A|^{16} = 4^{8} \times 5^{16}

\rightarrow A = 10 |A| = 10

by simplifing we get A = x + y + z = 10 |A| = x+y+z = 10

so, number of positive integral solution = 9 C 2 = 36 = ^9C_2 = 36

Niranjan Sai
Mar 12, 2015

Did same as Aniket Verma.

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