Find the value(s) of \(h\) for which the vectors are linearly dependent. Justify each answer.

\(\left[ {\begin{array}{*{20}{c}}1\\{ - 1}\\4\end{array}} \right],\left[ {\begin{array}{*{20}{c}}3\\{ - 5}\\7\end{array}} \right],\left[ {\begin{array}{*{20}{c}}{ - 1}\\5\\h\end{array}} \right]\)

When a set has more vectors than entries in each vector, it is said to be linearly dependent.

Let \({v_1},{v_2}\,\),and \({v_3}\) be the three vectors. The linear dependence of these three vectors in the form of an augmented matrix is \(\left[ {\begin{array}{*{20}{c}}{{v_1}}&{{v_2}}&{{v_3}}&0\end{array}} \right]\).

Hence, the augmented matrix is:

\(\left[ {\begin{array}{*{20}{c}}1&3&{ - 1}&0\\{ - 1}&{ - 5}&5&0\\4&7&h&0\end{array}} \right]\)

Apply row operation \({R_2} \to {R_1} + {R_2}\) to the augmented matrix above.

\(\left[ {\begin{array}{*{20}{c}}1&3&{ - 1}&0\\0&{ - 2}&4&0\\4&7&h&0\end{array}} \right]\)

Apply row operation \({R_3} \to {R_3} - 4{R_1}.\)

\(\left[ {\begin{array}{*{20}{c}}1&3&{ - 1}&0\\0&{ - 2}&4&0\\0&{ - 5}&{h + 4}&0\end{array}} \right]\)

Apply row operation \({R_3} \to {R_3} - \frac{5}{2}{R_2}.\)

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

The pivots in the echelon matrix are represented as:

\(S = \left\{ {{v_1},{v_2},.....{v_p}} \right\}\)

If the zero vector appears in a set in \({R^n}\) , the set is linearly dependent.

Thus, the equation can be written as \({x_1}{v_1} + {x_2}{v_2} + {x_3}{v_3} = 0\). The vector has a nontrivial solution if \(h - 6 = 0\).

Hence, the vectors are linearly dependent if \(h = 6\).