Question

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

25. \(\left( {\begin{aligned}{*{20}{c}}7\\{ - 4}\\{ - 6}\end{aligned}} \right)\), \(\left( {\begin{aligned}{*{20}{c}}{ - {\bf{8}}}\\{\bf{5}}\\7\end{aligned}} \right)\), \(\left( {\begin{aligned}{*{20}{c}}{\bf{7}}\\{\bf{0}}\\{ - {\bf{5}}}\end{aligned}} \right)\)

Short answer

The set of vectors is not linearly independent.

Step-by-step solution

State the condition for linear independence of the set of vectors

The set of vectors\({{\bf{b}}_1}\),\({{\bf{b}}_2}\),\({{\bf{b}}_3}\)is 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\({{\bf{b}}_1} = \left( {\begin{aligned}{*{20}{c}}7\\{ - 4}\\{ - 6}\end{aligned}} \right)\), \({{\bf{b}}_2} = \left( {\begin{aligned}{*{20}{c}}{ - 8}\\5\\7\end{aligned}} \right)\),and\({{\bf{b}}_3} = \left( {\begin{aligned}{*{20}{c}}7\\0\\{ - 5}\end{aligned}} \right)\).

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

\(A = \left( {\begin{aligned}{*{20}{c}}7&{ - 8}&7\\{ - 4}&5&0\\{ - 6}&7&{ - 5}\end{aligned}} \right)\)

Check the linear independence of the set of vectors

Obtain the determinant of matrix A, as shown below:

\(\begin{aligned}{c}\det \left( A \right) = \left| {\begin{aligned}{*{20}{c}}7&{ - 8}&7\\{ - 4}&5&0\\{ - 6}&7&{ - 5}\end{aligned}} \right|\\ = 7 \cdot \left| {\begin{aligned}{*{20}{c}}5&0\\7&{ - 5}\end{aligned}} \right| + 8 \cdot \left| {\begin{aligned}{*{20}{c}}{ - 4}&0\\{ - 6}&{ - 5}\end{aligned}} \right| + 7 \cdot \left| {\begin{aligned}{*{20}{c}}{ - 4}&5\\{ - 6}&7\end{aligned}} \right|\\ = 7\left( {5\left( { - 5} \right) - 0\left( 7 \right)} \right) + 8\left( { - 4\left( { - 5} \right) - 0\left( { - 6} \right)} \right) + 7\left( { - 4\left( 7 \right) - 5\left( { - 6} \right)} \right)\\ = - 175 + 160 + 14\\ = - 1\end{aligned}\)

Since \(\det \left( A \right) \ne 0\), the vectors are not linearly independent.