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Let A be a $5 × 5$ invertible matrix with row sums $1$. That is $\sum_{j=1}^{5} a_{ij} = 1$ for $1 \leq i\leq 5$. Then, what is the sum of all entries of $A^{-1}$.
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Check the question again... I found Some contradiction
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@Shaik Masthan Can you elaborate the contradictory part here?

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@pankaj_vir, sorry it's my mistake

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@Shaik Masthan not an issue.

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very well explained by Rishabh Gupta 2

A similar question was asked in TIFR 2015. But it asked for the row sum of $A^{-1}$.

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Consider any matrix which is invertible say 2*2 in which sum of rows is 1, what we get is sum of all elements of A-1 as 2.

similarly u can check it for a 3*3 matrix it comes out to be 3.

and for 4*4 it comes out to be 4

and so on for 5*5 matrix.

and we can conclude that its equal to n for n*n matrix.

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can you explain it with an example.
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for 2*2 matrix say A

 2 -1 0 1

consider the matrix it is invertible as |A|=2.

sum of each row value is 1.

now inverse is : adj(A)/|A| :

 1/2 0 1/2 1

so sum of entire element is 2.(as its a 2*2 matrix)

similarly u can take any value where matrix is invertible and row sum is 1.

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@ashwashil lanjewar : no its not (-1/2) its 1/2. you have missed on finding adjoint. go through the concept once.
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yep i got that

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