Question

In Exercises 24–26, use determinants to decide if the set of vectors is linearly independent.

26. \(\left[ {\begin{aligned}{*{20}{c}}3\\5\\{ - 6}\\4\end{aligned}} \right]\), \(\left[ {\begin{aligned}{*{20}{c}}2\\{ - 6}\\0\\7\end{aligned}} \right]\), \(\left[ {\begin{aligned}{*{20}{c}}{ - 2}\\{ - 1}\\3\\0\end{aligned}} \right]\), \(\left[ {\begin{aligned}{*{20}{c}}0\\0\\0\\{ - 2}\end{aligned}} \right]\)

Short answer

The set of vectors is linearly independent.

Step-by-step solution

State the condition for linear independence of the set of vectors

The set of vectors\({{\bf{b}}_1}\),\({{\bf{b}}_2}\),\({{\bf{b}}_3}\),\({{\bf{b}}_4}\)are said to belinearly independent if thedeterminant of the matrix \(\left[ {\begin{aligned}{*{20}{c}}{{{\bf{b}}_1}}&{{{\bf{b}}_2}}&{{{\bf{b}}_3}}&{{{\bf{b}}_4}}\end{aligned}} \right]\) is 0 (\(\left| {\begin{aligned}{*{20}{c}}{{{\bf{b}}_1}}&{{{\bf{b}}_2}}&{{{\bf{b}}_3}}&{{{\bf{b}}_4}}\end{aligned}} \right| = 0\)).

Write the vectors in the matrix form

Consider the vectors\({{\bf{b}}_1} = \left[ {\begin{aligned}{*{20}{c}}3\\5\\{ - 6}\\4\end{aligned}} \right]\), \({{\bf{b}}_2} = \left[ {\begin{aligned}{*{20}{c}}2\\{ - 6}\\0\\7\end{aligned}} \right]\), \({{\bf{b}}_3} = \left[ {\begin{aligned}{*{20}{c}}{ - 2}\\{ - 1}\\3\\0\end{aligned}} \right]\),and\({{\bf{b}}_4} = \left[ {\begin{aligned}{*{20}{c}}0\\0\\0\\{ - 2}\end{aligned}} \right]\).

Constructmatrix\(A = \left[ {\begin{aligned}{*{20}{c}}{{{\bf{b}}_1}}&{{{\bf{b}}_2}}&{{{\bf{b}}_3}}&{{{\bf{b}}_4}}\end{aligned}} \right]\)by using the vectors\({{\bf{b}}_1}\),\({{\bf{b}}_2}\),\({{\bf{b}}_3}\),\({{\bf{b}}_4}\), as shown below:

\(A = \left[ {\begin{aligned}{*{20}{c}}3&2&{ - 2}&0\\5&{ - 6}&{ - 1}&0\\{ - 6}&0&3&0\\4&7&0&{ - 2}\end{aligned}} \right]\)

Check the linear independence of the set of vectors

Expand along the fourth column to obtain the determinant of matrix A, as shown below:

\(\begin{aligned}{c}\det \left( A \right) = \left| {\begin{aligned}{*{20}{c}}3&2&{ - 2}&0\\5&{ - 6}&{ - 1}&0\\{ - 6}&0&3&0\\4&7&0&{ - 2}\end{aligned}} \right|\\ = {\left( { - 1} \right)^{1 + 1}} \cdot \left( 0 \right)\left| {\begin{aligned}{*{20}{c}}5&{ - 6}&{ - 1}\\{ - 6}&0&3\\4&7&0\end{aligned}} \right| + {\left( { - 1} \right)^{1 + 2}} \cdot \left( 0 \right)\left| {\begin{aligned}{*{20}{c}}3&2&{ - 2}\\{ - 6}&0&3\\4&7&0\end{aligned}} \right|\\ + {\left( { - 1} \right)^{1 + 3}} \cdot \left( 0 \right)\left| {\begin{aligned}{*{20}{c}}3&2&{ - 2}\\5&{ - 6}&{ - 1}\\4&7&0\end{aligned}} \right| + {\left( { - 1} \right)^{1 + 4}} \cdot \left( { - 2} \right)\left| {\begin{aligned}{*{20}{c}}3&2&{ - 2}\\5&{ - 6}&{ - 1}\\{ - 6}&0&3\end{aligned}} \right|\\ = 0 + 0 + 0 - \left( { - 2} \right)\left| {\begin{aligned}{*{20}{c}}3&2&{ - 2}\\5&{ - 6}&{ - 1}\\{ - 6}&0&3\end{aligned}} \right|\\ = 2\left| {\begin{aligned}{*{20}{c}}3&2&{ - 2}\\5&{ - 6}&{ - 1}\\{ - 6}&0&3\end{aligned}} \right|\end{aligned}\)

Compute the determinant\(\left| {\begin{aligned}{*{20}{c}}3&2&{ - 2}\\5&{ - 6}&{ - 1}\\{ - 6}&0&3\end{aligned}} \right|\).

\(\begin{aligned}{c}\left| {\begin{aligned}{*{20}{c}}3&2&{ - 2}\\5&{ - 6}&{ - 1}\\{ - 6}&0&3\end{aligned}} \right| = 3\left( { - 6\left( 3 \right) - 0\left( { - 1} \right)} \right) - 2\left( {5\left( 3 \right) - \left( { - 6} \right)\left( { - 1} \right)} \right) - 2\left( {5\left( 0 \right) + 6\left( { - 6} \right)} \right)\\ = - 54 - 18 + 72\\ = 0\end{aligned}\)

Obtain the determinant of matrix A.

\(\begin{aligned}{c}\det \left( A \right) = 2\left( 0 \right)\\ = 0\end{aligned}\)

Since\(\det \left( A \right) = 0\), the vectors are linearly independent.

Thus, the set of vectors is linearly independent.