Chapter 7.4, Problem 50E

To determine

To find: The reduced row-echelon form of the given matrix.

Answer to Problem 50E

The reduced row-echelon form of the given matrix is $\left[\begin{array}{cccc}1& 0& -2& 7\\ 0& 1& -3& 5\end{array}\right]$ .

Explanation of Solution

Given: The given matrix is $\left[\begin{array}{cccc}1& 0& -2& 7\\ 3& -4& 6& 1\end{array}\right]$ .

Concept Used: A matrix is said to be in reduced row echelon form if-

1. All rows consisting of only zeroes are at the bottom.
2. The leading coefficient of a non-zero row is always strictly to the right of the leading coefficient of the row above it.
3. The leading coefficient in each non-zero row is 1.
4. Each column containing a leading 1 has zeroes in all its other entries.

Reduced row echelon forms are obtained by elementary row operations or Gauss-Jordan elimination and are unique for every matrix.

Calculations: Let A = $\left[\begin{array}{cccc}1& 0& -2& 7\\ 3& -4& 6& 1\end{array}\right]$ .

Using elementary row operations, the above matrix transforms into-

A $=\left[\begin{array}{cccc}1& 0& -2& 7\\ 0& -4& 12& -20\end{array}\right]$   ( $R_2\to R_2-3R_1$ )

  $=\left[\begin{array}{cccc}1& 0& -2& 7\\ 0& 1& -3& 5\end{array}\right]$   ( $R_2\to \frac{-1}{4}{R}_2$ )

Thus, rref(A ) $=\left[\begin{array}{cccc}1& 0& -2& 7\\ 0& 1& -3& 5\end{array}\right]$