In Exercises 1–4, determine if the vectors are linearly independent. Justify each answer.

4. \(\left[ {\begin{array}{*{20}{c}}{ - 1}\\4\end{array}} \right]\), \(\left[ {\begin{array}{*{20}{c}}{ - 2}\\{ - 8}\end{array}} \right]\)

The vectors are said to be linearly independent if the equation \({x_1}{{\bf{v}}_1} + {x_2}{{\bf{v}}_2} = 0\) has a trivial solution, where \({{\bf{v}}_1}\), and \({{\bf{v}}_3}\) are vectors.

Consider the vectors \({{\bf{v}}_1} = \left[ {\begin{array}{*{20}{c}}{ - 1}\\4\end{array}} \right]\), \({{\bf{v}}_2} = \left[ {\begin{array}{*{20}{c}}{ - 2}\\{ - 8}\end{array}} \right]\).

Substitute these vectors in the equation \({x_1}{{\bf{v}}_1} + {x_2}{{\bf{v}}_2} = 0\) as shown below:

\(\begin{aligned}{c}{x_1}{{\bf{v}}_1} + {x_2}{{\bf{v}}_2} &= 0\\{x_1}\left[ {\begin{array}{*{20}{c}}{ - 1}\\4\end{array}} \right] + {x_2}\left[ {\begin{array}{*{20}{c}}{ - 2}\\{ - 8}\end{array}} \right] &= 0\end{aligned}\)

Now, write the vector equation in the matrix form.

\(\left[ {\begin{array}{*{20}{c}}{ - 1}&{ - 2}\\4&{ - 8}\end{array}} \right]\left[ {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}0\\0\end{array}} \right]\)

The matrix equation is in \(\left[ {\begin{array}{*{20}{c}}{ - 1}&{ - 2}\\4&{ - 8}\end{array}} \right]\left[ {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}0\\0\end{array}} \right]\), \(A{\bf{x}} = {\bf{0}}\) form.

Write the augmented matrix \(\left[ {\begin{array}{*{20}{c}}A&{\bf{0}}\end{array}} \right]\) as shown below:

\(\left[ {\begin{array}{*{20}{c}}{ - 1}&{ - 2}&0\\4&{ - 8}&0\end{array}} \right]\)

To obtain \({x_1}\) as a term in the first equation, multiply the first equation by \( - 1\).

\(\left[ {\begin{array}{*{20}{c}}1&2&0\\4&{ - 8}&0\end{array}} \right]\)

Use the \({x_1}\) term in the first equation to eliminate the \(4{x_1}\) term from the second equation. Add \( - 4\) times row one to row two.

\(\left[ {\begin{array}{*{20}{c}}1&2&0\\4&{ - 8}&0\end{array}} \right] \sim \left[ {\begin{array}{*{20}{c}}1&2&0\\0&{ - 16}&0\end{array}} \right]\)

Mark the non-zero leading entries in column one.

Write the obtained matrix,,in the equation notation.

According to the pivot positions in the obtained matrix, there are no free variables.

Thus, the homogeneous equation has a trivial solution, which means the vectors are linearly independent.