Math Problem Statement
What is the rank of the given matrix?
Solution
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:
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.Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Linear Algebra
Matrix Rank
Formulas
Row Echelon Form (REF)
Reduced Row Echelon Form (RREF)
Theorems
Rank-Nullity Theorem
Suitable Grade Level
Undergraduate Level