matrix doubt

newuser

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question is transform matrix A to echolen form by operations and find the range.
Pls see the attachment.
My answer is different: I got two rows same but answer is different.
Is it possible to get more than one answer?
 

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Row echelon forms are not unique, but reduced row echelon forms are unique

Your answer, however, is not in row echelon form. A matrix in row echelon form must be upper triangular, meaning zeros below the main diagonal.
 
Sure you can get more than one answer! A right one and any number of wrong ones!
:p

Here, you are row reducing the matrix
\(\displaystyle \begin{pmatrix}1 & 1 & -1 \\ 2 & 5 & 3 \\ 1 & 10 & -11\end{pmatrix}\)
and you start off by subtracting twice the first row from the secon row and subtracting the first row from the third:
\(\displaystyle \begin{pmatrix}1 & 1 & -1 \\ 0 & -3 & 5 \\ 0 & 6 & -10\end{pmatrix}\)
That is a good first step.

Now, your objective is to get a triangular matrix with "0"s below the main diagonal, right? So I don't understand why you would divide the third row by 2. It doesn't hurt, but it does't help! Instead, to get a "0" in the second column of the third row, you need to add two times the second row to the third: 6+ 2(-3)= 0 and -10+ 2(5)= 0. That is,
\(\displaystyle \begin{pmatrix}1 & 4 & -1 \\ 0 & -3 & 5 \\ 0 & 0 & 0\end{pmatrix}\).


As I said, while dividing the third row by 2, to get
\(\displaystyle \begin{pmatrix}1 & 4 & -1 \\ 0 & -3 & 5 \\ 0 & 3& -5\end{pmatrix}\)
doesn't help, it doesn't hurt- you just haven't finished. Now add the second row to the third row to get, again,
\(\displaystyle \begin{pmatrix}1 & 4 & -1 \\ 0 & -3 & 5 \\ 0 & 0 & 0\end{pmatrix}\).
 
Sure you can get more than one answer! A right one and any number of wrong ones!
:p

Here, you are row reducing the matrix
\(\displaystyle \begin{pmatrix}1 & 1 & -1 \\ 2 & 5 & 3 \\ 1 & 10 & -11\end{pmatrix}\)
and you start off by subtracting twice the first row from the secon row and subtracting the first row from the third:
\(\displaystyle \begin{pmatrix}1 & 1 & -1 \\ 0 & -3 & 5 \\ 0 & 6 & -10\end{pmatrix}\)
That is a good first step.

Now, your objective is to get a triangular matrix with "0"s below the main diagonal, right? So I don't understand why you would divide the third row by 2. It doesn't hurt, but it does't help! Instead, to get a "0" in the second column of the third row, you need to add two times the second row to the third: 6+ 2(-3)= 0 and -10+ 2(5)= 0. That is,
\(\displaystyle \begin{pmatrix}1 & 4 & -1 \\ 0 & -3 & 5 \\ 0 & 0 & 0\end{pmatrix}\).


As I said, while dividing the third row by 2, to get
\(\displaystyle \begin{pmatrix}1 & 4 & -1 \\ 0 & -3 & 5 \\ 0 & 3& -5\end{pmatrix}\)
doesn't help, it doesn't hurt- you just haven't finished. Now add the second row to the third row to get, again,
\(\displaystyle \begin{pmatrix}1 & 4 & -1 \\ 0 & -3 & 5 \\ 0 & 0 & 0\end{pmatrix}\).

Thanks for ur answer.
pls tell me our question was asking for echolen form that is why we made third row zero otherwise we can do like that what i did.
How can i know that my answer is wrong?
 
Row echelon forms are not unique, but reduced row echelon forms are unique

Your answer, however, is not in row echelon form. A matrix in row echelon form must be upper triangular, meaning zeros below the main diagonal.

Thanks for ur answer.
wat is the difference between row echelon form and reduced row echolen form?
 
wat (sic) is the difference between row echelon form and reduced row echolen form?

From this lesson page comes the following statement:

... the only real difference between row-echelon form and reduced row-echelon form is that a matrix in row-echelon form is only required to have zeroes below a leading 1 while a matrix in reduced row-echelon from must have zeroes both below and above a leading 1.
 
From this lesson page comes the following statement:

... the only real difference between row-echelon form and reduced row-echelon form is that a matrix in row-echelon form is only required to have zeroes below a leading 1 while a matrix in reduced row-echelon from must have zeroes both below and above a leading 1.
Thank u very much.
 
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