https://file.aiquickdraw.com/mathbot/file/1730607945184m4eisygd.jpg

Let's go through each row operation shown in the matrix, step-by-step, to complete the missing entries.

The matrix initially given is:

\[
\begin{bmatrix} 4 & -3 & 6 \\ -4 & 5 & -4 \end{bmatrix}
\]

### Step 1: Applying \(-4R_1\)

The operation \( -4R_1 \) means we multiply each element in the first row \( R_1 \) by \(-4\):

1. For the element in the first row and first column: \( 4 \times -4 = -16 \)
2. For the element in the first row and second column: \( -3 \times -4 = 12 \)
3. For the element in the first row and third column: \( 6 \times -4 = -24 \)

After performing this operation, the new matrix becomes:

\[
\begin{bmatrix} -16 & 12 & -24 \\ -4 & 5 & -4 \end{bmatrix}
\]

### Step 2: Applying \( R_2 + 3R_1 \)

For this step, we add \(3\) times the first row \( R_1 \) to the second row \( R_2 \):

1. For the element in the second row and first column: \( -4 + 3 \times 4 = -4 + 12 = 8 \)
2. For the element in the second row and second column: \( 5 + 3 \times -3 = 5 - 9 = -4 \)
3. For the element in the second row and third column: \( -4 + 3 \times 6 = -4 + 18 = 14 \)

After this operation, the matrix becomes:

\[
\begin{bmatrix} 4 & -3 & 6 \\ 8 & -4 & 14 \end{bmatrix}
\]

### Step 3: Applying \( R_1 - 4R_2 \)

In this operation, we subtract \(4\) times the second row \( R_2 \) from the first row \( R_1 \):

1. For the element in the first row and first column: \( 4 - 4 \times 8 = 4 - 32 = -28 \)
2. For the element in the first row and second column: \( -3 - 4 \times -4 = -3 + 16 = 13 \)
3. For the element in the first row and third column: \( 6 - 4 \times 14 = 6 - 56 = -50 \)

After this operation, the matrix becomes:

\[
\begin{bmatrix} -28 & 13 & -50 \\ 8 & -4 & 14 \end{bmatrix}
\]

### Summary of the Results

The resulting matrices after each row operation are:

1. After \( -4R_1 \):
   \[
   \begin{bmatrix} -16 & 12 & -24 \\ -4 & 5 & -4 \end{bmatrix}
   \]

2. After \( R_2 + 3R_1 \):
   \[
   \begin{bmatrix} 4 & -3 & 6 \\ 8 & -4 & 14 \end{bmatrix}
   \]

3. After \( R_1 - 4R_2 \):
   \[
   \begin{bmatrix} -28 & 13 & -50 \\ 8 & -4 & 14 \end{bmatrix}
   \]

Would you like further explanation on any of the steps?

### Relative Questions:
1. What is the purpose of row operations in matrix transformations?
2. How does multiplying a row by a scalar affect the determinant of the matrix?
3. What is the significance of the identity matrix in row operations?
4. Can matrix row operations change the rank of a matrix? Why or why not?
5. How do row operations help in solving systems of linear equations?

### Tip:
Row operations are fundamental for simplifying matrices in Gaussian elimination, which helps solve systems of linear equations efficiently.

Fill in the blanks of the resulting matrix after each given row operation is performed on the matrix: 

Initial matrix:
\[
\begin{bmatrix} 4 & -3 & 6 \\ -4 & 5 & -4 \end{bmatrix}
\]

1. Apply \(-4R_1\) to the first row.
2. Apply \(R_2 + 3R_1\) to the second row.
3. Apply \(R_1 - 4R_2\) to the first row.

Explore detailed steps to apply row operations on a 2x3 matrix. This example covers fundamental linear algebra concepts, including matrix transformations using row operations, making it ideal for high school students studying matrices in Grades 10-12.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation