Let $A$ be an $10\times 10$ invertible matrix with real entries such that the sum of each row is $1$. Then
$A$. The sum of entries of each row of the inverse of $A$ is $1$
$B$. The sum of entries of each column of the inverse of $A$ is $1$
$C$. The trace of the inverse of $A$ is non-zero
$D$. None of the above
I don't see how to proceed or how to use the sum of the elements of of rows. Please help.
I know option $A$ is correct and option $C$ is not.
Why is option $B$ not correct?
Hint: Let $x = (1,\dots,1)^T$ (the column vector with ten $1$s). The fact that the row sum of each matrix of $A$ is $1$ is the same as saying $Ax = x$.
What can we say about $A^{-1}x$?
The statement B amounts to saying that $(A^{-1})^Tx = x$ (where $A^T$ is the transpose of $A$). Note that, in general, $A$ and $A^T$ will not have the same eigenvectors.
In particular, consider the example $$ A = \pmatrix{1/2 & 1/2\\0&1} \implies\\ A^{-1} = \pmatrix{2&-1\\0&1} $$ We indeed have $A^{-1}x = x$, but not $(A^{-1})^Tx = x$.
$C$ is false, by the way. A $2\times 2$ counterexample is $$A = \left(\begin{array}{cc} 0 & 1\\ 1 & 0\end{array}\right)$$Notice that $A^{-1} = A$ has trace zero. I'll let you come up with a similar $10 \times 10$ example.
