- #1
kostoglotov
- 234
- 6
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 = [itex]
\begin{bmatrix}
1 & 2\\
3 & 8
\end{bmatrix}
[/itex]
[tex]\\ \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}
[/tex]
Now I would have thought that this last matrix, A = [itex]
\begin{bmatrix}
1 & 2\\
0 & 1
\end{bmatrix}
[/itex] would be the ref(A).
But my TI-89 gives ref(A) = [itex]
\begin{bmatrix}
1 & \frac{8}{3}\\
0 & 1
\end{bmatrix}
[/itex] and this is not the rref(A), the rref(A) is just the 2x2 identity matrix.
Compute ref(A) where A = [itex]
\begin{bmatrix}
1 & 2\\
3 & 8
\end{bmatrix}
[/itex]
[tex]\\ \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}
[/tex]
Now I would have thought that this last matrix, A = [itex]
\begin{bmatrix}
1 & 2\\
0 & 1
\end{bmatrix}
[/itex] would be the ref(A).
But my TI-89 gives ref(A) = [itex]
\begin{bmatrix}
1 & \frac{8}{3}\\
0 & 1
\end{bmatrix}
[/itex] and this is not the rref(A), the rref(A) is just the 2x2 identity matrix.