91Ó°ÊÓ

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

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

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

Add 3 times row one to row two to eliminate the \( - 3{x_1}\) term from the second equation.

\(\left[ {\begin{array}{*{20}{c}}1&{ - 3}&3&{ - 2}&0\\0&{ - 2}&8&{ - 4}&0\\0&1&{ - 4}&3&0\end{array}} \right]\)

Add \(\frac{1}{2}\) times row two to row three to eliminate the \({x_2}\) term from the third equation.

\(\left[ {\begin{array}{*{20}{c}}1&{ - 3}&3&{ - 2}&0\\0&{ - 2}&8&{ - 4}&0\\0&0&0&1&0\end{array}} \right]\)

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.