The problem gives a matrix equation and asks us to find the values of $p$ and $q$. The equation is: $4 \begin{bmatrix} p & -2 \\ 3 & -6 \end{bmatrix} - \frac{1}{3} \begin{bmatrix} -6 & -15 \\ -3 & q \end{bmatrix} = \begin{bmatrix} 14 & -3 \\ 13 & -20 \end{bmatrix}$

1. Problem Description

The problem gives a matrix equation and asks us to find the values of $p$ and $q$. The equation is:

$4 \begin{bmatrix} p & -2 \\ 3 & -6 \end{bmatrix} - \frac{1}{3} \begin{bmatrix} -6 & -15 \\ -3 & q \end{bmatrix} = \begin{bmatrix} 14 & -3 \\ 13 & -20 \end{bmatrix}$

2. Solution Steps

First, we distribute the scalars to the matrices:

$\begin{bmatrix} 4p & -8 \\ 12 & -24 \end{bmatrix} - \begin{bmatrix} -2 & -5 \\ -1 & \frac{q}{3} \end{bmatrix} = \begin{bmatrix} 14 & -3 \\ 13 & -20 \end{bmatrix}$

Next, we subtract the two matrices on the left-hand side:

$\begin{bmatrix} 4p - (-2) & -8 - (-5) \\ 12 - (-1) & -24 - \frac{q}{3} \end{bmatrix} = \begin{bmatrix} 14 & -3 \\ 13 & -20 \end{bmatrix}$

$\begin{bmatrix} 4p + 2 & -3 \\ 13 & -24 - \frac{q}{3} \end{bmatrix} = \begin{bmatrix} 14 & -3 \\ 13 & -20 \end{bmatrix}$

Now, we can equate the corresponding entries of the two matrices:

$4p + 2 = 14$

$-24 - \frac{q}{3} = -20$

From the first equation, we solve for $p$:

$4p = 14 - 2$

$4p = 12$

$p = \frac{12}{4}$

$p = 3$

From the second equation, we solve for $q$:

$-\frac{q}{3} = -20 + 24$

$-\frac{q}{3} = 4$

$q = -12$

3. Final Answer

$p = 3$

$q = -12$