91Ó°ÊÓ

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 \( - 2{x_1}\) term from the second equation. Perform an elementary row operationon the matrix\(\left[ {\begin{array}{*{20}{c}}1&h&{ - 3}\\{ - 2}&4&6\end{array}} \right]\) as shown below.

Add 2 times the first row to the second row; i.e., \({R_2} \to {R_2} + 2{R_1}\).

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

After the row operation, the matrix becomes

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

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

Obtain the value of \(h\) for which \(4 + 2h = 0\).

\(\begin{array}{c}4 + 2h = 0\\2h = - 4\\h = - 2\end{array}\)

Now, for \(h = - 2\), the matrix becomes

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

It can be observed that \(0 = 0\). This can be re-written as \(0{x_1} + 0{x_2} + 0{x_3} = 0\). Infinite values of \({x_1}\), \({x_2}\), and \({x_3}\) satisfy the equation \(0{x_1} + 0{x_2} + 0{x_3} = 0\). It proves that the system is consistent for all values of \(h\).

Thus, for all values of \(h\), the system is consistent and has a solution.