Certainly! Let's work through the row operation step-by-step.
We are given the initial matrix:
[tex]\[
\left[\begin{array}{ccc|c}
2 & -1 & -4 & 7 \\
-1 & 2 & 1 & -1 \\
0 & 10 & 8 & 2
\end{array}\right]
\][/tex]
We need to perform the row operation [tex]\(2R_2 + R_1 \rightarrow R_2\)[/tex], which means we will replace the second row [tex]\(R_2\)[/tex] with [tex]\(2 \times R_2 + R_1\)[/tex].
Let's break down the operation for each element of the second row:
1. First element of [tex]\(R_2\)[/tex]:
[tex]\[
2 \times (-1) + 2 = -2 + 2 = 0
\][/tex]
2. Second element of [tex]\(R_2\)[/tex]:
[tex]\[
2 \times 2 + (-1) = 4 - 1 = 3
\][/tex]
3. Third element of [tex]\(R_2\)[/tex]:
[tex]\[
2 \times 1 + (-4) = 2 - 4 = -2
\][/tex]
4. Fourth element of [tex]\(R_2\)[/tex]:
[tex]\[
2 \times (-1) + 7 = -2 + 7 = 5
\][/tex]
So, the new second row after the operation [tex]\(2R_2 + R_1\)[/tex] becomes:
[tex]\[
[0, 3, -2, 5]
\][/tex]
Now, we substitute this new row back into the original matrix, leaving the first and third rows unchanged:
[tex]\[
\left[\begin{array}{ccc|c}
2 & -1 & -4 & 7 \\
0 & 3 & -2 & 5 \\
0 & 10 & 8 & 2
\end{array}\right]
\][/tex]
Therefore, the resulting matrix after the row operation [tex]\(2R_2 + R_1 \rightarrow R_2\)[/tex] is:
[tex]\[
\left[\begin{array}{ccc|c}
2 & -1 & -4 & 7 \\
0 & 3 & -2 & 5 \\
0 & 10 & 8 & 2
\end{array}\right]
\][/tex]
