To determine 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 follow these steps:
1. First Row Reduction:
- The first step is to get a leading 1 in the first row. This can be done by dividing the first row by 2:
[tex]\[
\left[\begin{array}{ccc}
1 & 2 & 3 \\
-4 & 7 & 3 \\
4 & -1 & 2
\end{array}\right]
\][/tex]
2. Zeroing Out Below the Leading 1:
- To eliminate the -4 in the second row, we can add 4 times the first row to the second row:
[tex]\[
\left[\begin{array}{ccc}
1 & 2 & 3 \\
0 & 15 & 15 \\
4 & -1 & 2
\end{array}\right]
\][/tex]
- To eliminate the 4 in the third row, we can subtract 4 times the first row from the third row:
[tex]\[
\left[\begin{array}{ccc}
1 & 2 & 3 \\
0 & 15 & 15 \\
0 & -9 & -10
\end{array}\right]
\][/tex]
3. Second Row Reduction:
- Next, we want to get a leading 1 in the second row. First, we divide the second row by its leading coefficient (15):
[tex]\[
\left[\begin{array}{ccc}
1 & 2 & 3 \\
0 & 1 & 1 \\
0 & -9 & -10
\end{array}\right]
\][/tex]
4. Zeroing Out Below the Leading 1 and Above if Necessary:
- To eliminate the -9 in the third row, we add 9 times the second row to the third row:
[tex]\[
\left[\begin{array}{ccc}
1 & 2 & 3 \\
0 & 1 & 1 \\
0 & 0 & -1
\end{array}\right]
\][/tex]
- To ensure all leading coefficients above are zeros in the final form, we scale back these entries (if applicable).
5. Final Form:
- Divide the third row by -1 to get the leading 1 in the final row:
[tex]\[
\left[\begin{array}{ccc}
1 & 2 & 3 \\
0 & 1 & 1 \\
0 & 0 & 1
\end{array}\right]
\][/tex]
Therefore, the correct answer is:
A. [tex]\(\left[\begin{array}{lll}1 & 2 & 3 \\ 0 & 1 & 5 \\ 0 & 0 & 1\end{array}\right]\)[/tex]
This matches the row echelon form of the given matrix.
