Is this the correct way to compute the row echelon form?

[kostoglotov]

This is actually a pretty simple thing, but the ref(A) that I compute on paper is different from the ref(A) that my TI-89 gives me.

Compute ref(A) where A = <br /> \begin{bmatrix}<br /> 1 &amp; 2\\ <br /> 3 &amp; 8<br /> \end{bmatrix}<br />

\\ \begin{bmatrix}1 &amp; 2\\ 3 &amp; 8\end{bmatrix} \ r_2 \rightarrow r_2 - 3 \times r_1 \\ \\ \begin{bmatrix}1 &amp; 2\\ 0 &amp; 2 \end{bmatrix} \ r_2 \rightarrow \frac{1}{2} \times r_2 \\ \\ \begin{bmatrix}1 &amp; 2\\ 0 &amp; 1 \end{bmatrix}<br />

Now I would have thought that this last matrix, A = <br /> \begin{bmatrix}<br /> 1 &amp; 2\\ <br /> 0 &amp; 1<br /> \end{bmatrix}<br /> would be the ref(A).

But my TI-89 gives ref(A) = <br /> \begin{bmatrix}<br /> 1 &amp; \frac{8}{3}\\ <br /> 0 &amp; 1<br /> \end{bmatrix}<br /> and this is not the rref(A), the rref(A) is just the 2x2 identity matrix.

 

[Mark44]

This is actually a pretty simple thing, but the ref(A) that I compute on paper is different from the ref(A) that my TI-89 gives me.

Compute ref(A) where A = <br /> \begin{bmatrix}<br /> 1 &amp; 2\\<br /> 3 &amp; 8<br /> \end{bmatrix}<br />

\\ \begin{bmatrix}1 &amp; 2\\ 3 &amp; 8\end{bmatrix} \ r_2 \rightarrow r_2 - 3 \times r_1 \\ \\ \begin{bmatrix}1 &amp; 2\\ 0 &amp; 2 \end{bmatrix} \ r_2 \rightarrow \frac{1}{2} \times r_2 \\ \\ \begin{bmatrix}1 &amp; 2\\ 0 &amp; 1 \end{bmatrix}<br />

Now I would have thought that this last matrix, A = <br /> \begin{bmatrix}<br /> 1 &amp; 2\\<br /> 0 &amp; 1<br /> \end{bmatrix}<br /> would be the ref(A).

Yes, I agree.

But my TI-89 gives ref(A) = <br /> \begin{bmatrix}<br /> 1 &amp; \frac{8}{3}\\<br /> 0 &amp; 1<br /> \end{bmatrix}<br /> and this is not the rref(A), the rref(A) is just the 2x2 identity matrix.

I'm guessing that your calculator switched the two rows, and then did row reduction. If you start with this matrix --
\begin{bmatrix}
3 & 8\\
1 & 2 \end{bmatrix}
-- row reduction gives you the matrix the calculator shows.