Question

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

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

Short answer

The vectors are linearly independent.

Step-by-step solution

Write the condition for the linear independence of vectors

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.

Write the vectors in the form of the matrix equation

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.

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

Write the matrix in the augmented form

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:

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

Convert the augmented matrix in the echelon form

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 pivot positions in the matrix

Mark the non-zero leading entries in column one.

Convert the matrix into an equation

Write the obtained matrix,,in the equation notation.

Obtain the general solutions of the system of equations

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.