In Exercises 5–8, determine if the columns of the matrix form a

linearly independent set. Justify each answer.

6. \(\left[ {\begin{array}{*{20}{c}}{ - 4}&{ - 3}&0\\0&{ - 1}&4\\1&0&3\\5&4&6\end{array}} \right]\)

The vectors are said to be linearly independent if the equation \(A{\bf{x}} = 0\) has a trivial solution, where A is the matrix and xis the vector.

Consider the matrix \(\left[ {\begin{array}{*{20}{c}}{ - 4}&{ - 3}&0\\0&{ - 1}&4\\1&0&3\\5&4&6\end{array}} \right]\).Asthe matrix has three columns, there should be three entries in the vector.

Thus, the matrix equation is \(\left[ {\begin{array}{*{20}{c}}{ - 4}&{ - 3}&0\\0&{ - 1}&4\\1&0&3\\5&4&6\end{array}} \right]\left[ {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\\{{x_3}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}0\\0\\0\\0\end{array}} \right]\), and it is in \(A{\bf{x}} = {\bf{0}}\) form.

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

\(\left[ {\begin{array}{*{20}{c}}{ - 4}&{ - 3}&0&0\\0&{ - 1}&4&0\\1&0&3&0\\5&4&6&0\end{array}} \right]\)

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

Interchange rows one and three.

\(\left[ {\begin{array}{*{20}{c}}{ - 4}&{ - 3}&0&0\\0&{ - 1}&4&0\\1&0&3&0\\5&4&6&0\end{array}} \right] \sim \left[ {\begin{array}{*{20}{c}}1&0&3&0\\0&{ - 1}&4&0\\{ - 4}&{ - 3}&0&0\\5&4&6&0\end{array}} \right]\)

Add 4 times row one to row three to eliminate the \( - 4{x_1}\) term from the third equation. Add \( - 5\) times row one to row four to eliminate the \(5{x_1}\) term from the fourth equation.

\(\left[ {\begin{array}{*{20}{c}}1&0&3&0\\0&{ - 1}&4&0\\{ - 4}&{ - 3}&0&0\\5&4&6&0\end{array}} \right] \sim \left[ {\begin{array}{*{20}{c}}1&0&3&0\\0&1&{ - 4}&0\\0&{ - 3}&{12}&0\\0&4&{ - 9}&0\end{array}} \right]\)

Add 3 times row two to row three to eliminate the \( - 3{x_2}\) term from the third equation. Add \( - 4\) times row two to row four to eliminate the \(4{x_2}\) term from the fourth equation. Interchange rows three and four.

\(\left[ {\begin{array}{*{20}{c}}1&0&3&0\\0&{ - 1}&4&0\\0&{ - 3}&{12}&0\\0&4&{ - 9}&0\end{array}} \right] \sim \left[ {\begin{array}{*{20}{c}}1&0&3&0\\0&1&{ - 4}&0\\0&0&1&0\\0&0&0&0\end{array}} \right]\)

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

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

Thus, the homogeneous equation has a trivial solution (only one solution of the equation), which means the vectors are linearly independent.