Perform the indicated row operation:

[tex]\[
\left[\begin{array}{ccc|c}
2 & -1 & -4 & 7 \\
-1 & 2 & 1 & -1 \\
3 & 4 & 5 & 5
\end{array}\right] \xrightarrow{R_3 + 3R_2 \rightarrow R_3} \left[\begin{array}{ccc|c}
2 & -1 & -4 & 7 \\
-1 & 2 & 1 & -1 \\
0 & 10 & 8 & 2
\end{array}\right]
\][/tex]



Answer :

Let's perform the row operation [tex]\( R3 + 3R2 \rightarrow R3 \)[/tex] on the given matrix step-by-step.

We start with the initial matrix:
[tex]\[ \left[\begin{array}{ccc|c} 2 & -1 & -4 & 7 \\ -1 & 2 & 1 & -1 \\ 3 & 4 & 5 & 5 \end{array}\right] \][/tex]

Step 1: Identify the third row [tex]\( R3 \)[/tex] and the second row [tex]\( R2 \)[/tex].

[tex]\[ R3 = [3, 4, 5, 5] \][/tex]
[tex]\[ R2 = [-1, 2, 1, -1] \][/tex]

Step 2: Compute [tex]\( 3R2 \)[/tex].

[tex]\[ 3R2 = 3 \times [-1, 2, 1, -1] = [-3, 6, 3, -3] \][/tex]

Step 3: Add [tex]\( 3R2 \)[/tex] to [tex]\( R3 \)[/tex] to obtain the new [tex]\( R3 \)[/tex], i.e., [tex]\( R3 + 3R2 \rightarrow R3 \)[/tex].

[tex]\[ R3 + 3R2 = [3, 4, 5, 5] + [-3, 6, 3, -3] = [0, 10, 8, 2] \][/tex]

Replacing the old [tex]\( R3 \)[/tex] with the new [tex]\( R3 \)[/tex] in the matrix, we have:

[tex]\[ \left[\begin{array}{ccc|c} 2 & -1 & -4 & 7 \\ -1 & 2 & 1 & -1 \\ 0 & 10 & 8 & 2 \end{array}\right] \][/tex]

Step 4: Extract only the first two rows to form the final matrix as required:

[tex]\[ \left[\begin{array}{ccc|c} 2 & -1 & -4 & 7 \\ -1 & 2 & 1 & -1 \end{array}\right] \][/tex]

Therefore, the final matrix after performing the row operation [tex]\( R3 + 3R2 \rightarrow R3 \)[/tex] and extracting the first two rows is:

[tex]\[ \left[\begin{array}{ccc|c} 2 & -1 & -4 & 7 \\ -1 & 2 & 1 & -1 \end{array}\right] \][/tex]