We are given the following matrix equation: $ \begin{bmatrix} 4 & -2 \\ 3 & -6 \end{bmatrix} \begin{bmatrix} \frac{1}{3} & -6 & -15 \\ \frac{1}{3} & -3 & q \end{bmatrix} = \begin{bmatrix} 14 & -3 \\ 13 & -20 \end{bmatrix} $ We are asked to find the values of $p$ and $q$. Note, the $p$ variable in the first matrix is mistakenly written as a column. It should be in the first row. I assume that the first matrix is a $2 \times 2$ matrix, and the second matrix is a $2 \times 3$ matrix. Therefore, the product will be a $2 \times 3$ matrix. This differs from the given $2 \times 2$ matrix on the right-hand side. There must be a typo somewhere. Let's assume the second matrix is a $2 \times 2$ matrix, thus: $ \begin{bmatrix} 4 & -2 \\ 3 & -6 \end{bmatrix} \begin{bmatrix} -6 & -15 \\ -3 & q \end{bmatrix} = \begin{bmatrix} 14 & -3 \\ 13 & -20 \end{bmatrix} $ This changes the question to finding the values of $q$. This is still wrong because multiplying a $2 \times 2$ matrix with another $2 \times 2$ matrix will give a $2 \times 2$ matrix. Therefore, the actual question may have been $ \begin{bmatrix} p & -2 \\ 3 & -6 \end{bmatrix} \begin{bmatrix} \frac{1}{3} & -6 \\ \frac{1}{3} & q \end{bmatrix} = \begin{bmatrix} 14 & -3 \\ 13 & -20 \end{bmatrix} $ This means the question is to find $p$ and $q$.

1. Problem Description

We are given the following matrix equation:

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

We are asked to find the values of $p$ and $q$. Note, the $p$ variable in the first matrix is mistakenly written as a column. It should be in the first row. I assume that the first matrix is a $2 \times 2$ matrix, and the second matrix is a $2 \times 3$ matrix. Therefore, the product will be a $2 \times 3$ matrix. This differs from the given $2 \times 2$ matrix on the right-hand side. There must be a typo somewhere. Let's assume the second matrix is a $2 \times 2$ matrix, thus:

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

This changes the question to finding the values of $q$. This is still wrong because multiplying a $2 \times 2$ matrix with another $2 \times 2$ matrix will give a $2 \times 2$ matrix.

Therefore, the actual question may have been

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

This means the question is to find $p$ and $q$.

2. Solution Steps

Let the given matrix equation be:

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

The entry in the first row and first column of the product is $p(\frac{1}{3}) + (-2)(\frac{1}{3}) = \frac{p}{3} - \frac{2}{3} = \frac{p-2}{3}$. This must equal

1

4. $ \frac{p-2}{3} = 14 $

$p - 2 = 42$

$p = 44$

The entry in the first row and second column of the product is $p(-6) + (-2)(q) = -6p - 2q$. This must equal -

3. $ -6p - 2q = -3 $

Substituting $p = 44$, we get:

$-6(44) - 2q = -3$

$-264 - 2q = -3$

$-2q = 261$

$q = -130.5$

We can confirm using the second row:

Entry (2,1) is $3(\frac{1}{3}) + (-6)(\frac{1}{3}) = 1 - 2 = -1$. Given is 13, so there is a typo in the matrix entry.

Entry (2,2) is $3(-6) + (-6)(q) = -18 - 6q$. This must equal -

2

0. $ -18 - 6q = -20 $

$-6q = -2$

$q = \frac{1}{3}$

Assuming entry (2,1) is 13 is a typo: entry (2,1) is equal to

1.

3. Final Answer

Given the assumption that the entry (2,1) is incorrect in the final matrix, and instead equals -1:

$p = 44$

$q = -130.5$

If we assume the given matrix is correct as is and entry (2,1) is actually 13:

Given the assumption that the entry (1,1) is incorrect in the final matrix, and instead equals $14/3$:

$p = 44$

$q = \frac{1}{3}$

There seems to be a typo in the original question. There seems to be multiple versions to the correct answer given multiple typo assumptions.

Let us assume:

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

$\begin{bmatrix} 4/3 - 2/3 & -24 - 2q \\ 1-2 & -18-6q\end{bmatrix} = \begin{bmatrix} 14/3 & -3 \\ 1 & -20 \end{bmatrix}$

$2/3 = 14/3$ which is not correct

$-1 = 1$ which is not correct

There must be a typo.

Final Answer: There are typos. We cannot find unique solutions for $p$ and $q$.

Final Answer: The given problem has inconsistencies and typos, preventing a unique solution for $p$ and $q$.

FINAL ANSWER: There are typos, and cannot solve with a unique answer.