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Multiply row 2 by the reciprocal of -40:

[tex]\[
\left[\begin{array}{ccc|c}
4 & -1 & -4 & -7 \\
0 & -40 & 0 & 40 \\
0 & 0 & 1 & 3
\end{array}\right] \xrightarrow{-\frac{1}{40} R_2 \rightarrow R_2} \left[\begin{array}{ccc|c}
4 & -1 & -4 & -7 \\
0 & 1 & 0 & -1 \\
0 & 0 & 1 & 3
\end{array}\right]
\][/tex]

Sagot :

GinnyAnsweridn
Sure! Let's solve the problem step-by-step using the given matrix:

Original Matrix:
[tex]\[ \left[\begin{array}{ccc|c} 4 & -1 & -4 & -7 \\ 0 & -40 & 0 & 40 \\ 0 & 0 & 1 & 3 \end{array}\right] \][/tex]

We need to multiply the second row by the reciprocal of -40. The reciprocal of -40 is [tex]\(-\frac{1}{40}\)[/tex].

By performing this operation, we will update each element in the second row as follows:

1. Element in column 1: [tex]\(0 \times -\frac{1}{40} = 0\)[/tex]

2. Element in column 2: [tex]\(-40 \times -\frac{1}{40} = 1\)[/tex]

3. Element in column 3: [tex]\(0 \times -\frac{1}{40} = 0\)[/tex]

4. Element in column 4: [tex]\(40 \times -\frac{1}{40} = -1\)[/tex]

Updating the second row with these results, the matrix becomes:
[tex]\[ \left[\begin{array}{ccc|c} 4 & -1 & -4 & -7 \\ 0 & 1 & 0 & -1 \\ 0 & 0 & 1 & 3 \end{array}\right] \][/tex]

So, the final matrix after multiplying the second row by the reciprocal of -40 is:
[tex]\[ \left[\begin{array}{ccc|c} 4 & -1 & -4 & -7 \\ 0 & 1 & 0 & -1 \\ 0 & 0 & 1 & 3 \end{array}\right] \][/tex]
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