Leading entry is to the right of the leading entry in the row above

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Introduction

In linear algebra, the reduced row-echelon form (RREF) is a method used to solve systems of linear equations. It is a way to transform a matrix into a simpler form, making it easier to find the solutions to the system. In this article, we will explore the concept of RREF and provide step-by-step solutions to two systems of linear equations.

What is Reduced Row-Echelon Form?

The reduced row-echelon form is a matrix that has been transformed into a specific form, where each row represents a linear equation. The RREF has the following properties:

Leading entries: Each row has a leading entry, which is the first non-zero entry in the row.
Leading entry is 1: The leading entry is always 1.
Leading entry is to the right of the leading entry in the row above: Each leading entry is to the right of the leading entry in the row above.
All entries below the leading entry are 0: All entries below the leading entry are 0.

Step-by-Step Guide to Finding RREF

To find the RREF of a matrix, follow these steps:

Write the matrix: Write the matrix in the form of a system of linear equations.
Perform row operations: Perform row operations to transform the matrix into RREF.
Check for leading entries: Check each row for leading entries.
Check for leading entry is 1: Check that the leading entry is 1.
Check for leading entry is to the right of the leading entry in the row above: Check that the leading entry is to the right of the leading entry in the row above.
Check for all entries below the leading entry are 0: Check that all entries below the leading entry are 0.

Solving the First System of Linear Equations

Let's solve the following system of linear equations:

{ \begin{align*} -4x + 2y &= -6 \\ 5x - 2y &= 8 \end{align*} \}

Step 1: Write the matrix

-4	2	-6
5	-2	8

Step 2: Perform row operations

To transform the matrix into RREF, we need to perform row operations. We can start by multiplying the first row by -1/4 to get a leading entry of 1.

1	-1/2	3/2
5	-2	8

Next, we can multiply the second row by 1/5 to get a leading entry of 1.

1	-1/2	3/2
1	-2/5	8/5

Now, we can subtract the second row from the first row to get a 0 in the second column.

1	-1/2	3/2
0	-1/5	-4/5

Step 3: Check for leading entries

We have a leading entry of 1 in the first row.

Step 4: Check for leading entry is 1

The leading entry is 1.

Step 5: Check for leading entry is to the right of the leading entry in the row above

The leading entry is to the right of the leading entry in the row above.

Step 6: Check for all entries below the leading entry are 0

All entries below the leading entry are 0.

Conclusion

The RREF of the first system of linear equations is:

1	-1/2	3/2
0	-1/5	-4/5

Solving the Second System of Linear Equations

Let's solve the following system of linear equations:

{ \begin{align*} -4x + 2y &= -6 \\ 5x - 2y &= 8 \end{align*} \}

We can use the same steps as before to find the RREF of this system.

Step 1: Write the matrix

-4	2	-6
5	-2	8

Step 2: Perform row operations

To transform the matrix into RREF, we need to perform row operations. We can start by multiplying the first row by -1/4 to get a leading entry of 1.

1	-1/2	3/2
5	-2	8

Next, we can multiply the second row by 1/5 to get a leading entry of 1.

1	-1/2	3/2
1	-2/5	8/5

Now, we can subtract the second row from the first row to get a 0 in the second column.

1	-1/2	3/2
0	-1/5	-4/5

Step 3: Check for leading entries

We have a leading entry of 1 in the first row.

Step 4: Check for leading entry is 1

The leading entry is 1.

Step 5: Check for leading entry is to the right of the leading entry in the row above

The leading entry is to the right of the leading entry in the row above.

Step 6: Check for all entries below the leading entry are 0

All entries below the leading entry are 0.

Conclusion

The RREF of the second system of linear equations is:

1	-1/2	3/2
0	-1/5	-4/5

Choosing the Correct Reduced Row-Echelon Form

When choosing the correct RREF, we need to make sure that the leading entry is 1, the leading entry is to the right of the leading entry in the row above, and all entries below the leading entry are 0.

Conclusion

In conclusion, the reduced row-echelon form is a method used to solve systems of linear equations. It is a way to transform a matrix into a simpler form, making it easier to find the solutions to the system. By following the steps outlined in this article, we can find the RREF of a system of linear equations and choose the correct RREF.

References

[1] Linear Algebra and Its Applications by Gilbert Strang
[2] Introduction to Linear Algebra by Gilbert Strang
[3] Linear Algebra by Jim Hefferon

Final Answer

Q&A: Reduced Row-Echelon Form

Q: What is the reduced row-echelon form?

A: The reduced row-echelon form is a method used to solve systems of linear equations. It is a way to transform a matrix into a simpler form, making it easier to find the solutions to the system.

Q: What are the properties of the reduced row-echelon form?

A: The reduced row-echelon form has the following properties:

Leading entries: Each row has a leading entry, which is the first non-zero entry in the row.
Leading entry is 1: The leading entry is always 1.
Leading entry is to the right of the leading entry in the row above: Each leading entry is to the right of the leading entry in the row above.
All entries below the leading entry are 0: All entries below the leading entry are 0.

Q: How do I find the reduced row-echelon form of a matrix?

A: To find the reduced row-echelon form of a matrix, follow these steps:

Write the matrix: Write the matrix in the form of a system of linear equations.
Perform row operations: Perform row operations to transform the matrix into RREF.
Check for leading entries: Check each row for leading entries.
Check for leading entry is 1: Check that the leading entry is 1.
Check for leading entry is to the right of the leading entry in the row above: Check that the leading entry is to the right of the leading entry in the row above.
Check for all entries below the leading entry are 0: Check that all entries below the leading entry are 0.

Q: What are some common row operations used to find the reduced row-echelon form?

A: Some common row operations used to find the reduced row-echelon form include:

Multiplying a row by a non-zero constant: Multiply a row by a non-zero constant to get a leading entry of 1.
Adding a multiple of one row to another row: Add a multiple of one row to another row to get a 0 in the second column.
Swapping two rows: Swap two rows to get a leading entry in the correct position.

Q: How do I choose the correct reduced row-echelon form?

A: When choosing the correct RREF, make sure that the leading entry is 1, the leading entry is to the right of the leading entry in the row above, and all entries below the leading entry are 0.

Q: What are some common mistakes to avoid when finding the reduced row-echelon form?

A: Some common mistakes to avoid when finding the reduced row-echelon form include:

Not checking for leading entries: Make sure to check each row for leading entries.
Not checking for leading entry is 1: Make sure that the leading entry is 1.
Not checking for leading entry is to the right of the leading entry in the row above: Make sure that the leading entry is to the right of the leading entry in the row above.
Not checking for all entries below the leading entry are 0: Make sure that all entries below the leading entry are 0.

Q: How do I use the reduced row-echelon form to solve systems of linear equations?

A: To use the reduced row-echelon form to solve systems of linear equations, follow these steps:

Write the matrix: Write the matrix in the form of a system of linear equations.
Find the reduced row-echelon form: Find the reduced row-echelon form of the matrix.
Read the solution: Read the solution from the reduced row-echelon form.

Conclusion

References

[1] Linear Algebra and Its Applications by Gilbert Strang
[2] Introduction to Linear Algebra by Gilbert Strang
[3] Linear Algebra by Jim Hefferon

Final Answer

The final answer is: $\boxed{\begin{pmatrix} 1 & -1/2 & 3/2 \\ 0 & -1/5 & -4/5 \end{pmatrix}}$