The problem requires us to solve for $p$, $q$, and $r$ given the matrix equation: $\begin{bmatrix} p & 6 \\ 2 & -4 \end{bmatrix} \begin{bmatrix} p & 6 \\ 2 & -4 \end{bmatrix} = \begin{bmatrix} r & -12 \\ -4 & 7q \end{bmatrix}$

1. Problem Description

The problem requires us to solve for $p$, $q$, and $r$ given the matrix equation:

$\begin{bmatrix} p & 6 \\ 2 & -4 \end{bmatrix} \begin{bmatrix} p & 6 \\ 2 & -4 \end{bmatrix} = \begin{bmatrix} r & -12 \\ -4 & 7q \end{bmatrix}$

2. Solution Steps

First, perform the matrix multiplication:

$\begin{bmatrix} p & 6 \\ 2 & -4 \end{bmatrix} \begin{bmatrix} p & 6 \\ 2 & -4 \end{bmatrix} = \begin{bmatrix} p^2 + 6(2) & 6p + 6(-4) \\ 2p - 4(2) & 2(6) + (-4)(-4) \end{bmatrix} = \begin{bmatrix} p^2 + 12 & 6p - 24 \\ 2p - 8 & 12 + 16 \end{bmatrix} = \begin{bmatrix} p^2 + 12 & 6p - 24 \\ 2p - 8 & 28 \end{bmatrix}$

Now, equate the result to the right-hand side matrix:

$\begin{bmatrix} p^2 + 12 & 6p - 24 \\ 2p - 8 & 28 \end{bmatrix} = \begin{bmatrix} r & -12 \\ -4 & 7q \end{bmatrix}$

This gives us the following equations:

$p^2 + 12 = r$

$6p - 24 = -12$

$2p - 8 = -4$

$28 = 7q$

From $6p - 24 = -12$, we have:

$6p = 12$

$p = 2$

From $2p - 8 = -4$, we have:

$2p = 4$

$p = 2$

Since both equations give $p=2$, it is consistent.

Now substitute $p = 2$ into $p^2 + 12 = r$:

$r = (2)^2 + 12 = 4 + 12 = 16$

From $28 = 7q$, we have:

$q = \frac{28}{7} = 4$

Therefore, $p = 2$, $q = 4$, and $r = 16$.

3. Final Answer

$p = 2$

$q = 4$

$r = 16$