Chapter 10, Problem 5T

(a)

To determine

To check: Whether the given matrix is in reduced row-echelon form, row echelon form or neither.

Answer to Problem 5T

The matrix $\left[\begin{array}{cccc}1& 2& 4& -6\\ 0& 1& -3& 0\end{array}\right]$ is in row-echelon form.

Explanation of Solution

Given:

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

A matrix is in row-echelon form it satisfies the following conditions.

(1)The first non zero number in row is 1.

(2)The leading entry in each row is to the right of the leading entry in the row immediately above.

(3)All rows consisting entirely of zeros are at the bottom of the matrix.

A matrix is said to be in reduced row-echelon form if it is in row echelon form and every number above and below of each leading entry is 0.

Use elementary row operation to convert the matrix in row echelon form.

$\begin{array}{c}\left[\begin{array}{cccc}1& 2& 4& -6\\ 0& 1& -3& 0\end{array}\right]=\left[\begin{array}{cccc}1& 2-2& 4+6& -6\\ 0& 1& -3& 0\end{array}\right]\left[\because R_1=R_1-2R_2\right]\\ =\left[\begin{array}{cccc}1& 0& 10& -6\\ 0& 1& -3& 0\end{array}\right]\end{array}$

The first non zero number of the row is 1.

Thus, matrix can be reduced in row-echelon form.

(b)

To determine

To check: Whether the given matrix is in reduced row-echelon form, row echelon form or neither.

Answer to Problem 5T

The matrix $\left[\begin{array}{ccccc}1& 0& -1& 0& 0\\ 0& 1& 3& 0& 0\\ 0& 0& 0& 1& 0\\ 0& 0& 0& 0& 1\end{array}\right]$ is in reduced row-echelon form.

Explanation of Solution

Given:

The given matrix is $\left[\begin{array}{ccccc}1& 0& -1& 0& 0\\ 0& 1& 3& 0& 0\\ 0& 0& 0& 1& 0\\ 0& 0& 0& 0& 1\end{array}\right]$.

Use elementary row operation to convert the matrix in row echelon form.

$\begin{array}{c}\left[\begin{array}{ccccc}1& 0& -1& 0& 0\\ 0& 1& 3& 0& 0\\ 0& 0& 0& 1& 0\\ 0& 0& 0& 0& 1\end{array}\right]=\left[\begin{array}{ccccc}1& 0& 0& 0& 0\\ 0& 1& 3& 0& 0\\ 0& 0& 0& 1& 0\\ 0& 0& 0& 0& 1\end{array}\right]\left[\because C_3-C_1\right]\\ =\left[\begin{array}{ccccc}1& 0& 0& 0& 0\\ 0& 1& 0& 0& 0\\ 0& 0& 0& 1& 0\\ 0& 0& 0& 0& 1\end{array}\right]\left[\because C_3-3C_2\right]\end{array}$

The first non zero number of the row is 1 and every number above and below of each leading entry is 0.

Therefore, the matrix $\left[\begin{array}{ccccc}1& 0& -1& 0& 0\\ 0& 1& 3& 0& 0\\ 0& 0& 0& 1& 0\\ 0& 0& 0& 0& 1\end{array}\right]$ is in reduced row-echelon form.

(c)

To determine

To check: Whether the given matrix is in reduced row-echelon form, row echelon form or neither.

Answer to Problem 5T

The matrix $\left[\begin{array}{ccc}1& 1& 0\\ 0& 0& 1\\ 0& 1& 3\end{array}\right]$ is neither row-echelon nor reduced row-echelon.

Explanation of Solution

Given:

The given matrix is $\left[\begin{array}{ccc}1& 1& 0\\ 0& 0& 1\\ 0& 1& 3\end{array}\right]$.

Use elementary row operation to convert the matrix in row echelon form.

$\begin{array}{c}\left[\begin{array}{ccc}1& 1& 0\\ 0& 0& 1\\ 0& 1& 3\end{array}\right]=\left[\begin{array}{ccc}1& 0& -3\\ 0& 0& 1\\ 0& 1& 3\end{array}\right]\left[\because R_1-R_3\right]\\ =\left[\begin{array}{ccc}1& 0& -3\\ 0& 1& 4\\ 0& 1& 3\end{array}\right]\left[\because R_2+R_3\right]\\ =\left[\begin{array}{ccc}1& 0& -3\\ 0& 1& 4\\ 0& 0& -1\end{array}\right]\left[\because R_3-R_2\right]\end{array}$

The matrix $\left[\begin{array}{ccc}1& 1& 0\\ 0& 0& 1\\ 0& 1& 3\end{array}\right]$ cannot be converted in to row-echelon form.

Therefore, the matrix $\left[\begin{array}{ccc}1& 1& 0\\ 0& 0& 1\\ 0& 1& 3\end{array}\right]$ is neither row-echelon nor reduced row-echelon.