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

19. \(\left[ {\begin{array}{*{20}{c}}1&h&4\\3&6&8\end{array}} \right]\)

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

Use the \({x_1}\) term in the first equation to eliminate the \(3{x_1}\) term from the second equation. Perform an elementaryrow operationon the matrix\(\left[ {\begin{array}{*{20}{c}}1&h&4\\3&6&8\end{array}} \right]\) as shown below.

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

\(\left[ {\begin{array}{*{20}{c}}1&h&4\\{3 - 3\left( 1 \right)}&{6 - 3\left( h \right)}&{8 - 3\left( 4 \right)}\end{array}} \right]\)

After performing the row operation, the matrix becomes

\(\left[ {\begin{array}{*{20}{c}}1&h&4\\0&{6 - 3h}&{ - 4}\end{array}} \right]\)

For the system of equations to be consistent, the solution must satisfy it.

Obtain the value of \(h\) for which \(6 - 3h = 0\).

\(\begin{array}{c}6 - 3h = 0\\3h = 6\\h = 2\end{array}\)

For \(h = 2\), the matrix becomes

\( \Rightarrow \left[ {\begin{array}{*{20}{c}}1&h&4\\0&{6 - 3\left( 2 \right)}&{ - 4}\end{array}} \right]\)

\( \Rightarrow \)\(\left[ {\begin{array}{*{20}{c}}1&2&4\\0&0&{ - 4}\end{array}} \right]\)

It can be observed that \(0 = - 4\). This can be re-written as \(0{x_1} + 0{x_2} + 0{x_3} = - 4\). No values of \({x_1}\), \({x_2}\), and \({x_3}\) can satisfy the equation \(0{x_1} + 0{x_2} + 0{x_3} = - 4\). This proves that the system is inconsistent when \(h = 2\).

Thus, for \(h \ne 2\), the system is consistent and has a solution.