# Factoring Matrices with Elementary Row Operations

• I
• cbarker1
In summary: As mentioned before, the sequence of elementary matrices is not unique, as long as the end result is the same. It is up to personal preference which steps to use. In summary, the conversation is about finding a sequence of elementary matrices for a given matrix A. The speaker used a specific sequence, while the book's answer key used a different sequence. Both sequences result in the same final matrix A, as confirmed by a calculator. The difference lies in the specific steps used in each sequence, which can vary as the sequence of elementary matrices is not unique.
cbarker1
Gold Member
MHB
TL;DR Summary
I am working on reviewing some Linear Algebra for a Graduate course in the Spring. I thought I did it correctly when I finished. But I looked in the book a different answer. I used my calculator to check the book answer and gives the correct matrix.
Dear Everybody,

I have some trouble with this problem: Finding a sequence of elementary matrix for this matrix A.

Let ##A=\begin{bmatrix} 4 & -1 \\ 3& -1\end{bmatrix}##. I first used the ##\frac{1}{4}R1##-> ##R1##. So the ##E_1=\begin{bmatrix} \frac{1}{4} & 0 \\ 0& 1\end{bmatrix}##. So the matrix ##A= \begin{bmatrix}1 & \frac{-1}{4} \\ 3& -1\end{bmatrix}## we can use ##-3R1+R2->R2##. ##A''= \begin{bmatrix}1 & \frac{-1}{4} \\ 0& \frac{-1}{4}\end{bmatrix}## and ##E_2=\begin{bmatrix} 1 & 0 \\ -3& 1\end{bmatrix}##. We multiply ##4R2->R2##,##A'''= \begin{bmatrix} 1 & \frac{-1}{4} \\ 0& 1\end{bmatrix}## and ##E_3=\begin{bmatrix} 1 & 0 \\ 0& 4\end{bmatrix}##. Then we multiply 1/4 to row 2 and add to row 1,##A''''= \begin{bmatrix}1 & 0 \\ 0& 1\end{bmatrix}## and ##E_4=\begin{bmatrix} 1 & \frac{1}{4} \\ 0& 1\end{bmatrix}##. So ##A={E_1}^{-1}{E_2}^{-1}{E_3}^{-1}{E_4}^{-1}##. But in the book's answer key, it said that ##A={E_2}^{-1}{E_3}^{-1}{E_4}^{-1}##.

I am confused as to why the book's answer is different from mine. I understand that the sequence is not unique. Here is the study guide's answer as well.

#### Attachments

• CalcChat _ Elementary Linear Algebra 7e Chapter 2 section 4 with problem 29.pdf
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The pdf shows that row 1 doesn’t have a 4. Are you looking at the right solution?

When we did row reduction in Linear Algebra, we were taught to avoid adding/ subtracting fractions if at all possible so the notion of dividing by 4 to get a 1 in that row would not be considered. Instead we would add / subtract the rows to get a 1 meaning we’d go for the -1 column.

cbarker1
jedishrfu said:
The pdf shows that row 1 doesn’t have a 4. Are you looking at the right solution?
The PDF shows the resulting matrix after the row operation has been performed. The steps shown in the PDF are correct.

cbarker1 said:
So ##A={E_1}^{-1}{E_2}^{-1}{E_3}^{-1}{E_4}^{-1}##. But in the book's answer key, it said that ##A={E_2}^{-1}{E_3}^{-1}{E_4}^{-1}##.
I haven't taken the time to calculate all of the above inverses. Does the product you show come out to A? If so, then your work is correct, albeit slightly longer than what is shown in the PDF.
cbarker1 said:
I am confused as to why the book's answer is different from mine. I understand that the sequence is not unique. Here is the study guide's answer as well.
They used some different steps. You could have shortened your work a bit in step 3 by -1/4R2 + R1 --> R1, instead of what you did.

sysprog and cbarker1
Mark44 said:
The PDF shows the resulting matrix after the row operation has been performed. The steps shown in the PDF are correct.I haven't taken the time to calculate all of the above inverses. Does the product you show come out to A? If so, then your work is correct, albeit slightly longer than what is shown in the PDF.

They used some different steps. You could have shortened your work a bit in step 3 by -1/4R2 + R1 --> R1, instead of what you did.
I checked with my calculator. My sequence is the same as the matrix given.

cbarker1 said:
I checked with my calculator. My sequence is the same as the matrix given.
Then the difference is just that you used some different steps.

jedishrfu and cbarker1

## 1. What are elementary row operations in matrix factoring?

Elementary row operations are a set of three operations that can be performed on a matrix to simplify or transform it. These operations include multiplying a row by a constant, adding one row to another, and interchanging two rows.

## 2. How do elementary row operations help in factoring matrices?

Elementary row operations help in factoring matrices by simplifying the matrix into a form that is easier to factor. By using these operations, we can transform the matrix into a triangular form, making it easier to identify the factors.

## 3. Can elementary row operations change the solution of a matrix?

No, elementary row operations do not change the solution of a matrix. These operations only change the way the matrix is written, but the solution remains the same.

## 4. What is the difference between elementary row operations and elementary column operations?

Elementary row operations are performed on rows of a matrix, while elementary column operations are performed on columns. Both operations are used to simplify the matrix, but they have different effects on the structure of the matrix.

## 5. Are there any limitations to using elementary row operations in factoring matrices?

Yes, there are some limitations to using elementary row operations in factoring matrices. These operations can only be used on square matrices, and they may not always result in a triangular form that is easy to factor. In some cases, other methods may be more efficient in factoring matrices.

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