Question: Find row echelon form and reduce row echelon form A = \(\left[\begin{array}{cc} 0&-1 &2&3&4\\2&1&3&4&1\\1&3&-1&2&3\\3&3&-6&4&7 \end{array}\right]\)
Solution:
Here given matrix is
A = \(\left[\begin{array}{cc} 0&-1 &2&3&4\\2&1&3&4&1\\1&3&-1&2&3\\3&3&-6&4&7 \end{array}\right]\)
Which is row echelon form of a matrix A. Now reduced echelon form,
\[\sim \left[\begin{array}{cc} 1&3 &-1&2&3\\0&1&-1&0&1\\0&0&1&0&\frac{-22}{25}\\0&0&0&1& \frac{49}{25} \end{array}\right] \ R_3 \Leftarrow R_3 - 3 R_4\] \[\sim \left[\begin{array}{cc} 1&3 &-1&2&3\\0&1&0&0&\frac{3}{25}\\0&0&1&0&\frac{-22}{25}\\0&0&0&1& \frac{49}{25} \end{array}\right] \ R_2 \Leftarrow R_2 + R_3\] \[\sim \left[\begin{array}{cc} 1&3 &-1&0&\frac{-23}{25}\\0&1&0&0&\frac{3}{25}\\0&0&1&0&\frac{-22}{25}\\0&0&0&1& \frac{49}{25} \end{array}\right] \ R_1 \Leftarrow R_1 - 2 R_4\] \[\sim \left[\begin{array}{cc} 1&3 &0&0&\frac{-45}{25}\\0&1&0&0&\frac{3}{25}\\0&0&1&0&\frac{-22}{25}\\0&0&0&1& \frac{49}{25} \end{array}\right] \ R_1 \Leftarrow R_1 + R_3\]\(\sim \left[\begin{array}{cc} 1&0 &0&0&\frac{-54}{25}\\0&1&0&0&\frac{3}{25}\\0&0&1&0&\frac{-22}{25}\\0&0&0&1& \frac{49}{25} \end{array}\right] \ R_1 \Leftarrow R_1 - 3R_3\)
Which is reduced row echelon form of A.
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