Chapter 8.1, Problem 94E

To determine

Comparing solutions of two systems.

Answer to Problem 94E

There is No Same Solution.

Explanation of Solution

Given information: determine whether the two systems of linear equations yield the same solution. If so, find the solution using matrices..

a) $\{\begin{array}{ccccc}x& +3y& -z& =& 19\\ & y& +6z& =& -18\\ & & z& =& -4\end{array}$

b) $\{\begin{array}{ccccc}x& -y& +3z& =& -15\\ & y& -2z& =& 14\\ & & z& =& -4\end{array}$

Calculation:

The system of equations having same solution may be have different row-echelon form , may have same reduced row-echelon form.

Properties of row-echelon form:

1. Any row have all elements as zero in the bottom of the matrix
2. Each row that does not contain entirely of zeros has the first nonzero entry as one.
3. For successive nonzero rows the leading one in the higher row is farther than the leading in the lower row.
A matrix is in reduced row-echelon form if every column that has a leading one has zeroes in every position above and below of its leading one.

a) The given system of equations is,

  $\{\begin{array}{ccccc}x& +3y& -z& =& 19\\ & y& +6z& =& -18\\ & & z& =& -4\end{array}$

The augmented matrix of above system of equations is,

  $\left[\begin{array}{ccccc}1& 3& -1& =& 19\\ 0& 1& 6& =& -18\\ 0& 0& 1& =& -4\end{array}\right]$

Now apply row-echelon form.

  $-6R_3+R_2\to R_2$ $\left[\begin{array}{ccccc}1& 3& -1& =& 19\\ 0& 1& 0& =& 6\\ 0& 0& 1& =& -4\end{array}\right]$

  $-3R_2+R_1\to R_1$ $\left[\begin{array}{ccccc}1& 0& -1& =& 1\\ 0& 1& 0& =& 6\\ 0& 0& 1& =& -4\end{array}\right]$

  $R_3+R_1\to R_1$ $\left[\begin{array}{ccccc}1& 0& 0& =& -3\\ 0& 1& 0& =& 6\\ 0& 0& 1& =& -4\end{array}\right]$

b) The given system of equations is.

  $\{\begin{array}{ccccc}x& -y& +3z& =& -15\\ & y& -2z& =& 14\\ & & z& =& -4\end{array}$

The augmented matrix of above system of equations is,

  $\left[\begin{array}{ccccc}1& -1& 3& =& -15\\ 0& 1& -2& =& 14\\ 0& 0& 1& =& -4\end{array}\right]$

Now apply row-echelon form.

  $2R_3+R_2\to R_2$ $\left[\begin{array}{ccccc}1& -1& 3& =& -15\\ 0& 1& 0& =& 6\\ 0& 0& 1& =& -4\end{array}\right]$

  $R_2+R_1\to R_1$ $\left[\begin{array}{ccccc}1& 0& 3& =& -9\\ 0& 1& 0& =& 6\\ 0& 0& 1& =& -4\end{array}\right]$

  $-3R_3+R_1\to R_1$ $\left[\begin{array}{ccccc}1& 0& 0& =& 3\\ 0& 1& 0& =& 6\\ 0& 0& 1& =& -4\end{array}\right]$

The reduced matrices are not same, so that the two system of linear equation is not yield the same solution

Hence there is No Same Solution.