In Exercises 19–22, determine the value(s) of h such that the matrix is the augmented matrix of a consistent linear system.

22. \[\left[ {\begin{array}{*{20}{c}}2&{ - 3}&h\\{ - 6}&9&5\end{array}} \right]\]

The givenaugmented matrix of a consistent linear system is as follows:

\[\left[ {\begin{array}{*{20}{c}}2&{ - 3}&h\\{ - 6}&9&5\end{array}} \right]\]

A basic principle states that row operations do not affect the solution set of a linear system.

To eliminate the first and second terms of the second row,perform an elementary row operationon the matrix \[\left[ {\begin{array}{*{20}{c}}2&{ - 3}&h\\{ - 6}&9&5\end{array}} \right]\], as shown below.

Add three times of the first row to the second row;i.e., \({R_2} \to {R_2} + 3{R_1}\).

\[\left[ {\begin{array}{*{20}{c}}2&{ - 3}&h\\{ - 6 + 3\left( 2 \right)}&{9 + 3\left( { - 3} \right)}&{5 + 3\left( h \right)}\end{array}} \right]\]

After the row operation, the matrix becomes:

\[\left[ {\begin{array}{*{20}{c}}2&{ - 3}&h\\0&0&{5 + 3h}\end{array}} \right]\]

For the system to beconsistent, it should haveuniqueorinfinitely many solutions.

If \[5 + 3h = 0\], all the elements of the second row become zero, which means that the system has exactly one solution when

\[\begin{array}{c}\,\,5 + 3h = 0\\\,\,\,\,\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,h = - \frac{5}{3}.\end{array}\]

Thus, the system is consistent if and only if \[h = - \frac{5}{3}\].