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 $(D).$