Question

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

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

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 \({x_1}{{\bf{v}}_1} + {x_2}{{\bf{v}}_2} + {x_3}{{\bf{v}}_3} = 0\) has a trivial solution, where \({{\bf{v}}_1}\), \({{\bf{v}}_2}\), and \({{\bf{v}}_3}\) are vectors.

Write the vectors in the form of the matrix equation

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

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:

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

Write the matrix in the augmented form

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:

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

Convert the augmented matrix in the echelon form

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 pivot positions in the matrix

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

Convert the matrix into the 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.

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