Question

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

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

Short answer

The set of vectors is linearly independent.

Step-by-step solution

State the condition for linear independency of the set of vectors

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

Write the vectors in the matrix form

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

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

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

Check the linear independency of the set of vectors

Obtain the determinant of matrix A, as shown below:

\(\begin{aligned}{c}\det \left( A \right) = \left| {\begin{aligned}{*{20}{c}}4&{ - 7}&{ - 3}\\6&0&{ - 5}\\2&7&{ - 2}\end{aligned}} \right|\\ = 4 \cdot \left| {\begin{aligned}{*{20}{c}}0&{ - 5}\\7&{ - 2}\end{aligned}} \right| + 7 \cdot \left| {\begin{aligned}{*{20}{c}}6&{ - 5}\\2&{ - 2}\end{aligned}} \right| - 3 \cdot \left| {\begin{aligned}{*{20}{c}}6&0\\2&7\end{aligned}} \right|\\ = 4\left( {0\left( { - 2} \right) + 5\left( 7 \right)} \right) + 7\left( {6\left( { - 2} \right) - 2\left( { - 5} \right)} \right) - 3\left( {6\left( 7 \right) - 0\left( 2 \right)} \right)\\ = 140 - 14 - 126\\ = 0\end{aligned}\)

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