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

2. \(\left[ {\begin{array}{*{20}{c}}0\\0\\2\end{array}} \right]\), \(\left[ {\begin{array}{*{20}{c}}0\\5\\{ - 8}\end{array}} \right]\), \(\left[ {\begin{array}{*{20}{c}}{ - 3}\\4\\1\end{array}} \right]\)

The vectors are said to be linearly independent if the equation \({x_1}{{\bf{v}}_1} + {x_2}{{\bf{v}}_2} + {x_3}{{\bf{v}}_3} = 0\) has a trivial solution, where \({{\bf{v}}_1}\), \({{\bf{v}}_2}\), and \({{\bf{v}}_3}\) are vectors.

Consider the vectors \({{\bf{v}}_1} = \left[ {\begin{array}{*{20}{c}}0\\0\\2\end{array}} \right]\), \({{\bf{v}}_2} = \left[ {\begin{array}{*{20}{c}}0\\5\\{ - 8}\end{array}} \right]\), \({{\bf{v}}_3} = \left[ {\begin{array}{*{20}{c}}{ - 3}\\4\\1\end{array}} \right]\).

Substitute the above vectors in the equation \({x_1}{{\bf{v}}_1} + {x_2}{{\bf{v}}_2} + {x_3}{{\bf{v}}_3} = 0\), as shown below:

\(\begin{aligned}{c}{x_1}{{\bf{v}}_1} + {x_2}{{\bf{v}}_2} + {x_3}{{\bf{v}}_3} &= 0\\{x_1}\left[ {\begin{array}{*{20}{c}}0\\0\\2\end{array}} \right] + {x_2}\left[ {\begin{array}{*{20}{c}}0\\5\\{ - 8}\end{array}} \right] + {x_3}\left[ {\begin{array}{*{20}{c}}{ - 3}\\4\\1\end{array}} \right] &= \left[ {\begin{array}{*{20}{c}}0\\0\\0\end{array}} \right]\end{aligned}\)

Now, write the vector equation in the matrix equation as shown below:

\(\left[ {\begin{array}{*{20}{c}}0&0&{ - 3}\\0&5&4\\2&{ - 8}&1\end{array}} \right]\left[ {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\\{{x_3}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}0\\0\\0\end{array}} \right]\)

The matrix equation is in \(\left[ {\begin{array}{*{20}{c}}0&0&{ - 3}\\0&5&4\\2&{ - 8}&1\end{array}} \right]\left[ {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\\{{x_3}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}0\\0\\0\end{array}} \right]\), \(A{\bf{x}} = {\bf{0}}\) form.

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

\(\left[ {\begin{array}{*{20}{c}}0&0&{ - 3}&0\\0&5&4&0\\2&{ - 8}&1&0\end{array}} \right]\)

In the echelon form, at the top of the left-most column, the leading entry should be non-zero.

Interchange rows one and three.

\(\left[ {\begin{array}{*{20}{c}}0&0&{ - 3}&0\\0&5&4&0\\2&{ - 8}&1&0\end{array}} \right] \sim \left[ {\begin{array}{*{20}{c}}2&{ - 8}&1&0\\0&5&4&0\\0&0&{ - 3}&0\end{array}} \right]\)

Mark the non-zero leading entries in columns 1, 2 and 3.

Write the obtained matrix, ,in the equation notation.

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

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