To find the row echelon form of the given matrix:
[tex]$
\left[\begin{array}{ccc}
2 & 4 & 6 \\
-4 & 7 & 3 \\
4 & -1 & 2
\end{array}\right]
$[/tex]
we perform a series of row operations to transform the matrix into row echelon form, where the leading coefficient (pivot) of each row is 1 and appears to the right of the leading coefficient of the row above it. Additionally, all entries below each pivot must be zeros.
After performing the necessary row operations, the row echelon form of the given matrix is:
[tex]$
\left[\begin{array}{ccc}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right]
$[/tex]
Given the options:
A. [tex]$\left[\begin{array}{ccc}1 & 2 & 3 \\ 0 & 1 & 5 \\ 0 & 0 & 1\end{array}\right]$[/tex]
B. [tex]$\left[\begin{array}{ccc}1 & 2 & 3 \\ 0 & 15 & 15 \\ 4 & -1 & 2\end{array}\right]$[/tex]
C. [tex]$\left[\begin{array}{ccc}1 & 2 & 3 \\ 0 & 1 & 1 \\ 0 & 0 & 1\end{array}\right]$[/tex]
D. [tex]$\left[\begin{array}{ccc}1 & 2 & 3 \\ 0 & 1 & 1 \\ 0 & -9 & 10\end{array}\right]$[/tex]
None of these options match the row echelon form we calculated:
[tex]$
\left[\begin{array}{ccc}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right].
$[/tex]
Therefore, the correct row echelon form is not given in the options provided.
