To solve the problem of performing the elementary row operation [tex]\( 5R_2 + R_1 \rightarrow R_1 \)[/tex] on the matrix
[tex]\[
\left[\begin{array}{ccc}
0 & -6 & 2 \\
0 & -5 & 8
\end{array}\right]
\][/tex]
we will follow a step-by-step approach.
### Step 1: Identify the Multiples
First, we need to determine the multiples of the elements of the second row [tex]\( R_2 \)[/tex].
[tex]\[
5 \times R_2 = 5 \times \left[\begin{array}{ccc}
0 & -5 & 8
\end{array}\right] = \left[\begin{array}{ccc}
0 & -25 & 40
\end{array}\right]
\][/tex]
### Step 2: Add the Result to the First Row
Next, we add the result from step 1 to the elements of the first row [tex]\( R_1 \)[/tex]:
[tex]\[
R_1 + 5R_2 = \left[\begin{array}{ccc}
0 & -6 & 2
\end{array}\right] + \left[\begin{array}{ccc}
0 & -25 & 40
\end{array}\right] = \left[\begin{array}{ccc}
0 & -31 & 42
\end{array}\right]
\][/tex]
### Step 3: Replace the First Row with the Result
Finally, we replace the first row [tex]\( R_1 \)[/tex] with the result from step 2:
[tex]\[
\left[\begin{array}{ccc}
0 & -31 & 42 \\
0 & -5 & 8
\end{array}\right]
\][/tex]
### Final Matrix Configuration
The matrix after performing the elementary row operation [tex]\( 5R_2 + R_1 \rightarrow R_1 \)[/tex] is:
[tex]\[
\left[\begin{array}{ccc}
0 & -31 & 42 \\
0 & -5 & 8
\end{array}\right]
\][/tex]
Thus, the resulting rows are:
- [tex]\(R_1 = [0, -31, 42] \)[/tex]
- [tex]\(R_2 = [0, -5, 8]\)[/tex]
This is the final answer.
