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Artist 52 [7]
3 years ago
10

For the following linear system, put the augmented coefficient matrix into reduced row-echelon form.

Mathematics
1 answer:
zhenek [66]3 years ago
7 0

The augmented matrix is

\left[\begin{array}{ccc|c}2&3&-1&14\\1&2&1&4\\5&9&2&7\end{array}\right]

Swap the first rows:

\left[\begin{array}{ccc|c}1&2&1&4\\2&3&-1&14\\5&9&2&7\end{array}\right]

Add -2(row 1) to row 2, and add -5(row 1) to row 3:

\left[\begin{array}{ccc|c}1&2&1&4\\0&-1&-3&6\\0&-1&-3&-13\end{array}\right]

Multiply through row 2 by -1:

\left[\begin{array}{ccc|c}1&2&1&4\\0&1&3&-6\\0&-1&-3&-13\end{array}\right]

Add row 2 to row 3:

\left[\begin{array}{ccc|c}1&2&1&4\\0&1&3&-6\\0&0&0&-19\end{array}\right]

Multiply through row 3 by -1/19:

\left[\begin{array}{ccc|c}1&2&1&4\\0&1&3&-6\\0&0&0&1\end{array}\right]

Add -4(row 3) to row 1, and add 6(row 3) to row 2:

\left[\begin{array}{ccc|c}1&2&1&0\\0&1&3&0\\0&0&0&1\end{array}\right]

Add -2(row 2) to row 1:

\left[\begin{array}{ccc|c}1&0&-5&0\\0&1&3&0\\0&0&0&1\end{array}\right]

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