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Perform the indicated operation.

[tex]\[ \left[\begin{array}{ccc|c}
4 & 0 & -4 & -8 \\
0 & 1 & 0 & -1 \\
0 & 0 & 1 & 3
\end{array}\right] \xrightarrow{4R_3 + R_1 \rightarrow R_1} \left[\begin{array}{ccc|c}
4 & 0 & 0 & 4 \\
0 & 1 & 0 & -1 \\
0 & 0 & 1 & 3
\end{array}\right] \][/tex]

Multiply row 1 by the reciprocal of 4:

[tex]\[ \left[\begin{array}{ccc|c}
4 & 0 & 0 & 4 \\
0 & 1 & 0 & -1 \\
0 & 0 & 1 & 3
\end{array}\right] \xrightarrow{\frac{1}{4} R_1 \rightarrow R_1} \left[\begin{array}{ccc|c}
1 & 0 & 0 & 1 \\
0 & 1 & 0 & -1 \\
0 & 0 & 1 & 3
\end{array}\right] \][/tex]

Sagot :

GinnyAnswer
Let's go through this problem step-by-step.

First, we need to perform the indicated row operation on the given matrix:
[tex]\[ \left[\begin{array}{ccc|c} 4 & 0 & -4 & -8 \\ 0 & 1 & 0 & -1 \\ 0 & 0 & 1 & 3 \end{array}\right] \xrightarrow{4 R_3 + R_1 \rightarrow R_1} \][/tex]

### Step 1: Applying the Row Operation

We are instructed to replace row 1 ([tex]\( R_1 \)[/tex]) with [tex]\( 4R_3 + R_1 \)[/tex]:
- The new first row's first element: [tex]\( 4 \times 0 + 4 = 4 \)[/tex]
- The new first row's second element: [tex]\( 4 \times 0 + 0 = 0 \)[/tex]
- The new first row's third element: [tex]\( 4 \times 1 + (-4) = 4 - 4 = 0 \)[/tex]
- The new first row's augmented part: [tex]\( 4 \times 3 + (-8) = 12 - 8 = 4 \)[/tex]

Thus, the matrix after performing this row operation becomes:
[tex]\[ \left[\begin{array}{ccc|c} 4 & 0 & 0 & 4 \\ 0 & 1 & 0 & -1 \\ 0 & 0 & 1 & 3 \end{array}\right] \][/tex]

### Step 2: Multiplying the First Row by the Reciprocal of 4

Next, we need to multiply the entire first row ([tex]\( R_1 \)[/tex]) by [tex]\(\frac{1}{4}\)[/tex]:
- The new first row's first element: [tex]\( \frac{4}{4} = 1 \)[/tex]
- The new first row's second element: [tex]\( \frac{0}{4} = 0 \)[/tex]
- The new first row's third element: [tex]\( \frac{0}{4} = 0 \)[/tex]
- The new first row's augmented part: [tex]\( \frac{4}{4} = 1 \)[/tex]

So, the final matrix after this operation becomes:
[tex]\[ \left[\begin{array}{ccc|c} 1 & 0 & 0 & 1 \\ 0 & 1 & 0 & -1 \\ 0 & 0 & 1 & 3 \end{array}\right] \][/tex]

Thus, the complete solution results in:
[tex]\[ \left[\begin{array}{ccc|c} 1 & 0 & 0 & 1 \\ 0 & 1 & 0 & -1 \\ 0 & 0 & 1 & 3 \end{array}\right] \][/tex]
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