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Let $A$ be an invertible $10 \times 10$ matrix with real entries such that the sum of each row is $1$. Then

  1. The sum of the entries of each row of the inverse of $A$ is $1$
  2. The sum of the entries of each column of the inverse of $A$ is $1$
  3. The trace of the inverse of $A$ is non-zero
  4. None of the above
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Ans: 1. The sum of the entries of each row of the inverse of $A$ is $1$.


Explanation:

Let me first define a vector $e$, which is a vector of all $1$s. (Vector is nothing but a column matrix)

so,

$e = \begin{bmatrix} 1\\ 1\\ ..\\ 1 \end{bmatrix}_{10 \times 1}$

Now, what will be $Ae =\ ?$

Think about it.

It turns out, that since the sum of each row of $A$ is $1$, $Ae$ will be equal to $e$. This is an important point.

so,

$Ae = e$

Multiply both sides by $A^{-1}$

$\implies A^{-1}Ae = A^{-1}e$

$\implies e = A^{-1}e$

$\implies A^{-1}e = e$

The above equation means that when we multiply $A^{-1}$ with $e$, we get $e$. What this implies?

This means that the sum of each row in $A^{-1}$ is $1$.

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One of the easiest method to solve this question is that we can assume A to be an identity matrix . As we know in identity matrix  sum of each row ia “1” also inverse of identity matrix is also identity matrix of same order , hence answere will be A)
Answer:

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