In general, when finding the reduced row echolon for a matrix, is there only one solution, even if there is no solution?
$$\begin{pmatrix}2&3&5\\ \:-2&-3&-3\end{pmatrix}$$
I got the following answer by hand (no solution because $1$ does not $= -2$):
$$\begin{pmatrix}1&0&-2\\ 0&1&1\end{pmatrix}$$
But an online calculator gave this answer (no solution because $0$ does not $= 2$):
$$\begin{pmatrix}1&1.5&2.5\\ \:0&0&2\end{pmatrix}$$
Is mine wrong or are they both correct?
Thank you
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Following this step: $$\begin{pmatrix}1&1.5&2.5\\ \:0&0&2\end{pmatrix}$$
There are 2 steps. First, multiply the 2nd row by 0.5 to get: $$\begin{pmatrix}1&1.5&2.5\\ \:0&0&1\end{pmatrix}$$
Then, add -2.5 times to the 2nd row to the 1st row to get: $$\begin{pmatrix}1&1.5&0\\ \:0&0&1\end{pmatrix}$$
Your row echelon form is wrong. The second column is a scalar multiple of the first, which means you cannot have a leading one in the second column, because elementary row operation don't change linear relations among columns.
In your matrix, the first two columns form a linearly independent set: this rules out that this matrix can be obtained from the given one by elementary row operations.
A row echelon form from your matrix is obtained by the operations \begin{align} \begin{pmatrix} 2 & 3 & 5 \\ -2 & -3 & 2 \end{pmatrix} &\to \begin{pmatrix} 1 & 3/2 & 5/2 \\ -2 & -3 & 2 \end{pmatrix} && R_1\gets \tfrac{1}{2}R_1 \\&\to \begin{pmatrix} 1 & 3/2 & 5/2 \\ 0 & 0 & 7 \end{pmatrix} &&R_2\gets R_2+2R_1 \\&\to \begin{pmatrix} 1 & 3/2 & 5/2 \\ 0 & 0 & 1 \end{pmatrix} && R_2\gets\tfrac{1}{7}R_2 \\&\to \begin{pmatrix} 1 & 3/2 & 0 \\ 0 & 0 & 1 \end{pmatrix} && R_1\gets R_1-\tfrac{5}{2}R_2 \end{align} The last step produces the reduced row echelon form.