is it right?

5 votes

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

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

8 votes

Best answer

**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 $1$.