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

linearly independent set. Justify each answer.

7. \(\left[ {\begin{array}{*{20}{c}}1&4&{ - 3}&0\\{ - 2}&{ - 7}&5&1\\{ - 4}&{ - 5}&7&5\end{array}} \right]\)

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 \(\left[ {\begin{array}{*{20}{c}}1&4&{ - 3}&0\\{ - 2}&{ - 7}&5&1\\{ - 4}&{ - 5}&7&5\end{array}} \right]\).As the matrix has four columns, there should be four entries in the vector.

Thus, the matrix equation is \(\left[ {\begin{array}{*{20}{c}}1&4&{ - 3}&0\\{ - 2}&{ - 7}&5&1\\{ - 4}&{ - 5}&7&5\end{array}} \right]\left[ {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\\{{x_3}}\\{{x_4}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}0\\0\\0\end{array}} \right]\), 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:

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

Add 4 times row one to row three to eliminate the \( - 4{x_1}\) term from the third equation. Add 2 times row one to row two to eliminate the \( - 2{x_1}\) term from the second equation.

\(\left[ {\begin{array}{*{20}{c}}1&4&{ - 3}&0&0\\0&1&{ - 1}&1&0\\0&{11}&{ - 5}&5&0\end{array}} \right]\)

Add \( - 11\) times row one to row three to eliminate the \(11{x_2}\) term from the third equation.

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

To obtain 1 as the coefficient of \(6{x_3}\), multiply row three by \(\frac{1}{6}\).

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

Add rows two and three to get row two. Then, add \( - 4\) times row two to row one to get row one.

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

Add 3 times row three to row one to eliminate the \( - 3{x_3}\) from the first equation.

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

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

In the obtained matrix, there are three pivot positions and four variables.

Thus, the homogeneous equation has a non-trivial solution, which means the vectors are linearly dependent.