Chapter 8, Problem 150RE

To determine

Give the explanation for the given matrices.

Explanation of Solution

Given information:

The given reduced matrices are$\left[\begin{array}{cccc}1& 2& :& 3\\ 0& 1& :& 1\end{array}\right]$

  $\left[\begin{array}{cccc}1& 0& :& 1\\ 0& 1& :& 1\end{array}\right]$

  $\left[\begin{array}{cccc}1& 2& :& 3\\ 0& 0& :& 0\end{array}\right]$

Calculation:

There are three types of solutions.

For unique solution, the row-reduced matrix has the same number of non-zero rows as variables

Example $\left[\begin{array}{ccccc}1& 0& 0& :& 5\\ 0& 1& 0& :& 1\\ 0& 0& 1& :& 3\end{array}\right]$

  $x=5,y=1,z=3$

For many solutions, the row-reduced matrix has more variables than non-zero rows.

Example $\left[\begin{array}{ccccc}1& 0& 4& :& 5\\ 0& 1& -3& :& 4\\ 0& 0& 0& :& 0\end{array}\right]$

In the augmented matrix

  $z$ column is not cleared.

If $z=a$

Then,$\begin{array}{l}x=5-4a\\ y=4+3a\end{array}$

For no solution, the row-reduced matrix has a row of zeros on the left side, but the right hand side is not zero.

Example $\left[\begin{array}{ccccc}1& 0& 4& :& 5\\ 0& 1& -3& :& 4\\ 0& 0& 0& :& 2\end{array}\right]$

From the above condition, the first and second matrix has a unique solution and the third matrix has many solutions.

Hence , they cannot be all correct.