The problem asks us to find $A^2$ where $A$ is the matrix $A = \begin{bmatrix} 0 & 3 & 1 \\ 5 & 4 & 2 \\ 0 & 3 & 1 \end{bmatrix}$. $A^2$ means $A \times A$.

1. Problem Description

The problem asks us to find $A^2$ where $A$ is the matrix

$A = \begin{bmatrix} 0 & 3 & 1 \\ 5 & 4 & 2 \\ 0 & 3 & 1 \end{bmatrix}$.

$A^2$ means $A \times A$.

2. Solution Steps

To find $A^2$, we multiply the matrix $A$ by itself.

$A^2 = A \times A = \begin{bmatrix} 0 & 3 & 1 \\ 5 & 4 & 2 \\ 0 & 3 & 1 \end{bmatrix} \begin{bmatrix} 0 & 3 & 1 \\ 5 & 4 & 2 \\ 0 & 3 & 1 \end{bmatrix}$

The resulting matrix will be a 3x3 matrix. The entry in the $i$-th row and $j$-th column of $A^2$ is the dot product of the $i$-th row of the first $A$ and the $j$-th column of the second $A$.

Let $A^2 = \begin{bmatrix} a & b & c \\ d & e & f \\ g & h & i \end{bmatrix}$.

$a = (0)(0) + (3)(5) + (1)(0) = 0 + 15 + 0 = 15$

$b = (0)(3) + (3)(4) + (1)(3) = 0 + 12 + 3 = 15$

$c = (0)(1) + (3)(2) + (1)(1) = 0 + 6 + 1 = 7$

$d = (5)(0) + (4)(5) + (2)(0) = 0 + 20 + 0 = 20$

$e = (5)(3) + (4)(4) + (2)(3) = 15 + 16 + 6 = 37$

$f = (5)(1) + (4)(2) + (2)(1) = 5 + 8 + 2 = 15$

$g = (0)(0) + (3)(5) + (1)(0) = 0 + 15 + 0 = 15$

$h = (0)(3) + (3)(4) + (1)(3) = 0 + 12 + 3 = 15$

$i = (0)(1) + (3)(2) + (1)(1) = 0 + 6 + 1 = 7$

Therefore,

$A^2 = \begin{bmatrix} 15 & 15 & 7 \\ 20 & 37 & 15 \\ 15 & 15 & 7 \end{bmatrix}$

3. Final Answer

$\begin{bmatrix} 15 & 15 & 7 \\ 20 & 37 & 15 \\ 15 & 15 & 7 \end{bmatrix}$