Let
$e^T=\begin{bmatrix}1 & 1 & \cdots & 1\end{bmatrix}$.
Since every column sum of $A$ is equal to $1$, we have
$e^TA=e^T$.
Now,
$e^TA^2=(e^TA)A=e^TA=e^T$.
So every column of $A^2$ also sums to $1$, not $2$.
Hence statement I is false.
Similarly,
$e^TA^3=(e^TA)A^2=e^TA^2=e^T$.
So every column of $A^3$ also sums to $1$, not $3$.
Hence statement II is false.
Now since $A$ is invertible and $e^TA=e^T$, multiply both sides on the right by $A^{-1}$:
$e^TAA^{-1}=e^TA^{-1}$.
Thus,
$e^T=e^TA^{-1}$.
Therefore every column of $A^{-1}$ sums to $1$.
Hence statement III is true.
So only statement III is correct.
Therefore, the correct option is D.
$\underline{\textbf{Alternate Solution :}}$
Let $A=\begin{bmatrix} 1 &0 \\ 0 &1 \end{bmatrix}$
$A^{2} = A \cdot A = \begin{bmatrix} 1 &0 \\ 0 &1 \end{bmatrix}\cdot\begin{bmatrix} 1 &0 \\ 0 &1 \end{bmatrix} = \begin{bmatrix} 1 &0 \\ 0 &1 \end{bmatrix}$
- Every column in the matrix $A^{2}$ sums to $2.\implies \text{False}$
$A^{3} = A^{2} \cdot A = \begin{bmatrix} 1 &0 \\ 0 &1 \end{bmatrix}\cdot\begin{bmatrix} 1 &0 \\ 0 &1 \end{bmatrix} = \begin{bmatrix} 1 &0 \\ 0 &1 \end{bmatrix}$
- Every column in the matrix $A^{3}$ sums to $3.\implies \text{False}$
$A^{-1} = \begin{bmatrix} 1 &0 \\ 0 &1 \end{bmatrix}$
Every column in the matrix $A^{-1}$ sums to $1.\implies \text{True}$
Lets take another example for statement $3:$
$A = \begin{bmatrix} 3&-6 \\ -2 &7 \end{bmatrix}$
$A = \begin{bmatrix} a &b \\ c &d \end{bmatrix}^{-1} = \dfrac{1}{ad-bc}\begin{bmatrix} d &-b \\ -c &a \end{bmatrix}$
In other words: swap the positions of $a$ and $d,$ put negatives in front of $b$ and $c,$ and divide everything by the determinant $(ad-bc).$
$A^{-1} = \dfrac{1}{9}\begin{bmatrix}7 & 6\\ 2 &3 \end{bmatrix}$
- Every column in the matrix $A^{-1}$ sums to $1.\implies \text{True}$
So, the correct answer is D.
