Factoring Matrices with Elementary Row Operations

[cbarker1]

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.

 

[jedishrfu]

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.

 

[Mark44]

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.

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.

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.

 

[cbarker1]

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.

 

[Mark44]

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.