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

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1. Sep 29, 2015

### 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 = $\begin{bmatrix} 1 & 2\\ 3 & 8 \end{bmatrix}$

$$\\ \begin{bmatrix}1 & 2\\ 3 & 8\end{bmatrix} \ r_2 \rightarrow r_2 - 3 \times r_1 \\ \\ \begin{bmatrix}1 & 2\\ 0 & 2 \end{bmatrix} \ r_2 \rightarrow \frac{1}{2} \times r_2 \\ \\ \begin{bmatrix}1 & 2\\ 0 & 1 \end{bmatrix}$$

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

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

2. Sep 29, 2015

### Staff: Mentor

Yes, I agree.
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.