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The image shows a handwritten matrix and a question that appears to be asking for the rank of the matrix. Here's the matrix provided:

\[
\begin{pmatrix}
6 & 1 & 2 & 8 \\
4 & 7 & 8 & -1 \\
16 & 0 & 12 & 15
\end{pmatrix}
\]

### To find the rank of the matrix:
1. **Step 1: Convert the matrix to Row Echelon Form (REF) or Reduced Row Echelon Form (RREF)**.
   
   The goal is to make as many rows of zeros as possible while keeping the matrix's structure.

2. **Step 2: Identify the number of non-zero rows**. The rank of the matrix is the number of non-zero rows in the row echelon form.

Let's proceed with this.

### Step 1: Row Operations to get REF:
We start with the given matrix:

\[
\begin{pmatrix}
6 & 1 & 2 & 8 \\
4 & 7 & 8 & -1 \\
16 & 0 & 12 & 15
\end{pmatrix}
\]

- **First, make the first entry of the first row (the pivot) 1** by dividing the first row by 6:

\[
\begin{pmatrix}
1 & \frac{1}{6} & \frac{1}{3} & \frac{4}{3} \\
4 & 7 & 8 & -1 \\
16 & 0 & 12 & 15
\end{pmatrix}
\]

- **Eliminate the entries below the pivot** in the first column by subtracting appropriate multiples of the first row from the other rows:

   - Row 2 = Row 2 - 4 * Row 1
   - Row 3 = Row 3 - 16 * Row 1

This will give:

\[
\begin{pmatrix}
1 & \frac{1}{6} & \frac{1}{3} & \frac{4}{3} \\
0 & \frac{38}{6} & \frac{20}{3} & -\frac{19}{3} \\
0 & -\frac{8}{3} & \frac{4}{3} & -\frac{37}{3}
\end{pmatrix}
\]

- **Simplify Row 2** by multiplying by 6/38:

\[
\begin{pmatrix}
1 & \frac{1}{6} & \frac{1}{3} & \frac{4}{3} \\
0 & 1 & \frac{10}{19} & -\frac{38}{19} \\
0 & -\frac{8}{3} & \frac{4}{3} & -\frac{37}{3}
\end{pmatrix}
\]

- **Eliminate the entry below the pivot in the second column** by adding (8/3) times Row 2 to Row 3:

\[
\begin{pmatrix}
1 & \frac{1}{6} & \frac{1}{3} & \frac{4}{3} \\
0 & 1 & \frac{10}{19} & -\frac{38}{19} \\
0 & 0 & \frac{152}{57} & -\frac{152}{57}
\end{pmatrix}
\]

### Step 2: Count the non-zero rows
The final matrix has three non-zero rows:

\[
\begin{pmatrix}
1 & \frac{1}{6} & \frac{1}{3} & \frac{4}{3} \\
0 & 1 & \frac{10}{19} & -\frac{38}{19} \\
0 & 0 & 1 & -1
\end{pmatrix}
\]

Thus, **the rank of the matrix is 3**.

---

Would you like any further details on this problem, or have any questions?

Here are five related questions you could explore:

1. What is the difference between Row Echelon Form (REF) and Reduced Row Echelon Form (RREF)?
2. How can the rank of a matrix help in determining the solutions of a system of linear equations?
3. What is the significance of the rank of a matrix in linear algebra?
4. How would the process change if you were finding the rank of a larger matrix?
5. How can you determine if a matrix is invertible based on its rank?

**Tip:** When working with matrices, mastering row operations will greatly simplify tasks like finding the rank, solving systems of equations, and determining invertibility.

Determine the rank of a 3x4 matrix using Row Echelon Form (REF) and Reduced Row Echelon Form (RREF). This problem is suitable for undergraduate students studying linear algebra and focuses on the concepts of matrix rank and the Rank-Nullity Theorem.

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