Question

Exercise 15 and 16 use the notation of Example 1 for matrices in echelon form. Suppose each matrix represents the augmented matrix for a system of linear equations. In each case, determine if the system is consistent. If the system is consistent, determine if the solution is unique.

15.

a.

\(\left( {\begin{array}{*{20}{c}} \square & * & * & * \\ 0&\square & * & * \\ 0&0&\square &0\end{array}} \right)\)

b. \(\left( {\begin{array}{*{20}{c}}0&\square & * & * & * \\ 0&0&\square & * & * \\ 0&0&0&0&\square \end{array}} \right)\)

Short answer

(a)Thus, the system of linear equations of the matrix is consistent and has a unique solution

(b)Thus, the system of linear equations of the matrix is inconsistent and does not have a unique solution.

Step-by-step solution

Use the notation of example 1 for matrices in the echelon form

In example 1, the following matrices are in the echelon form. The leading entries \(\left( \square \right)\) may have any nonzero value; the starred entries \(\left( * \right)\) may have any value (including zero).

\(\left( {\begin{array}{*{20}{c}} \square & * & * & * \\ 0&\square & * & * \\ 0&0&0&0 \\ 0&0&0&0 \end{array}} \right),\left( {\begin{array}{*{20}{c}} 0&\square & * & * & * & * & * & * & * & * \\ 0&0&0&\square & * & * & * & * & * & * \\ 0&0&0&0&\square & * & * & * & * & * \\ 0&0&0&0&0&\square & * & * & * & * \\ 0&0&0&0&0&0&0&0&\square & * \end{array}} \right)\)

Thus, the leading entries \(\left( \square \right)\) represent any nonzero values, and the starred entries \(\left( * \right)\) represent any values (including zero).

(a)

\(\left( {\begin{array}{*{20}{c}} \square & * & * & * \\ 0&\square & * & * \\ 0&0&\square &0\end{array}} \right)\)

(b)

\(\left( {\begin{array}{*{20}{c}}0&\square & * & * & * \\ 0&0&\square & * & * \\ 0&0&0&0&\square \end{array}} \right)\)

Write the matrices in the reduced echelon form

In example 1,the matrices are in the reduced echelon form because the leading entries are 1’s and there are 0’s in column above each leading entry 1.

\(\left[ {\begin{array}{*{20}{c}}1&0& * & * \\0&1& * & * \\0&0&0&0\\0&0&0&0\end{array}} \right],\left[ {\begin{array}{*{20}{c}}0&1& * &0&0&0& * & * &0& * \\0&0&0&1&0&0& * & * &0& * \\0&0&0&0&1&0& * & * &0& * \\0&0&0&0&0&1& * & * &0& * \\0&0&0&0&0&0&0&0&1& * \end{array}} \right]\)

Write the given matrices in the reduced echelon form.

(a)\(\left[ {\begin{array}{*{20}{c}}1&0&0& * \\0&1&0& * \\0&0&1&0\end{array}} \right]\)

(b) \(\left[ {\begin{array}{*{20}{c}}0&1&0& * & * \\0&0&1& * & * \\0&0&0&0&1\end{array}} \right]\)

Determine if the system of linear equations is consistent

The system is inconsistentif the reduced echelon form has a row with a nonzero number in the last column and zeros in all other columns.

The system isconsistent and has a unique solution if pivots are in columns 1 and 2 but not in the last column.

1. There is no pivot in the last column,so the system is consistent.

\(\left[ {\begin{array}{*{20}{c}}1&0&0& * \\0&1&0& * \\0&0&1&0\end{array}} \right]\)

b. There is a pivot in the last column,so the system is inconsistent.

\(\left[ {\begin{array}{*{20}{c}}0&1&0& * & * \\0&0&1& * & * \\0&0&0&0&1\end{array}} \right]\)

Determine if the solution is unique

A pivot positionin matrix \(A\) is a location that corresponds to a leading 1 in the reduced echelon form of\(A\). A pivot column is a column that contains a pivot position.

(a)There are pivots in every other column, and there are no free variables, so the solution is unique.

\(\left[ {\begin{array}{*{20}{c}}1&0&0& * \\0&1&0& * \\0&0&1&0\end{array}} \right]\)

(b)The first column has no pivot, and the first and fourth variablesare free. Thus, the system does not have a unique solution.

\(\left[ {\begin{array}{*{20}{c}}0&1&0& * & * \\0&0&1& * & * \\0&0&0&0&1\end{array}} \right]\)

(a)Thus, the matrix is consistent and has a unique solution.

(b)Thus, the matrix is inconsistent and does not have a unique solution.