91Ó°ÊÓ

Vector \({{\bf{v}}_3}\) is said to be in the span \(\left\{ {{{\bf{v}}_1},{{\bf{v}}_2}} \right\}\) if the equation \({x_1}{{\bf{v}}_1} + {x_2}{{\bf{v}}_2} = {{\bf{v}}_3}\) has a solution, where \({{\bf{v}}_1}\), \({{\bf{v}}_2}\), and \({{\bf{v}}_3}\) are vectors.

Consider the vectors \({{\bf{v}}_1} = \left[ {\begin{array}{*{20}{c}}1\\{ - 5}\\{ - 3}\end{array}} \right]\), \({{\bf{v}}_2} = \left[ {\begin{array}{*{20}{c}}{ - 2}\\{10}\\6\end{array}} \right]\), \({{\bf{v}}_3} = \left[ {\begin{array}{*{20}{c}}2\\{ - 9}\\h\end{array}} \right]\).

Use the augmented matrix \(\left[ {\begin{array}{*{20}{c}}{{{\bf{v}}_1}}&{{{\bf{v}}_2}}&{{{\bf{v}}_3}}\end{array}} \right]\) as shown below:

\(\left[ {\begin{array}{*{20}{c}}1&{ - 2}&2\\{ - 5}&{10}&{ - 9}\\{ - 3}&6&h\end{array}} \right]\)

Use the \({x_1}\) term from the first equation to eliminate the \( - 5{x_1}\) term from the second equation. Add 5 times row one to row two.

\(\left[ {\begin{array}{*{20}{c}}1&{ - 2}&2\\0&0&1\\{ - 3}&6&h\end{array}} \right]\)

Use the \({x_1}\) term from the first equation to eliminate the \( - 3{x_1}\) term from the third equation. Add 3 times row one to row three.

\(\left[ {\begin{array}{*{20}{c}}1&{ - 2}&2\\0&0&1\\0&0&{h + 6}\end{array}} \right]\)

Write the obtained matrix, \(\left[ {\begin{array}{*{20}{c}}1&{ - 2}&2\\0&0&1\\0&0&{h + 6}\end{array}} \right]\),in the system of equations.

\(\begin{array}{c}{x_1} - 2{x_2} = 2\\0\left( {{x_1}} \right) + 0\left( {{x_2}} \right) = 1\\0\left( {{x_1}} \right) + 0\left( {{x_2}} \right) = h + 6\end{array}\)

Consider the equation \(0\left( {{x_1}} \right) + 0\left( {{x_2}} \right) = 1\). No values of \({x_1}\) and \({x_2}\) satisfy the equation \(0\left( {{x_1}} \right) + 0\left( {{x_2}} \right) = 1\).

Thus, there is no value of \(h\) for which \({{\bf{v}}_3}\) is in the span \(\left\{ {{{\bf{v}}_1},{{\bf{v}}_2}} \right\}\).

The vectors are said to be linearly independent if the equation \({x_1}{{\bf{v}}_1} + {x_2}{{\bf{v}}_2} + {x_3}{{\bf{v}}_3} = 0\) has a trivial solution, where \({{\bf{v}}_1}\), \({{\bf{v}}_2}\), and \({{\bf{v}}_3}\) are vectors.

Consider the vectors \({{\bf{v}}_1} = \left[ {\begin{array}{*{20}{c}}1\\{ - 5}\\{ - 3}\end{array}} \right]\), \({{\bf{v}}_2} = \left[ {\begin{array}{*{20}{c}}{ - 2}\\{10}\\6\end{array}} \right]\), \({{\bf{v}}_3} = \left[ {\begin{array}{*{20}{c}}2\\{ - 9}\\h\end{array}} \right]\).

Substitute these vectors in the equation \({x_1}{{\bf{v}}_1} + {x_2}{{\bf{v}}_2} + {x_3}{{\bf{v}}_3} = 0\) as shown below:

\(\begin{array}{c}{x_1}{{\bf{v}}_1} + {x_2}{{\bf{v}}_2} + {x_3}{{\bf{v}}_3} = 0\\{x_1}\left[ {\begin{array}{*{20}{c}}1\\{ - 5}\\{ - 3}\end{array}} \right] + {x_2}\left[ {\begin{array}{*{20}{c}}{ - 2}\\{10}\\6\end{array}} \right] + {x_3}\left[ {\begin{array}{*{20}{c}}2\\{ - 9}\\h\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}0\\0\\0\end{array}} \right]\end{array}\)

Now, write the vector equation into the matrix equation as shown below:

\(\left[ {\begin{array}{*{20}{c}}1&{ - 2}&2\\{ - 5}&{10}&{ - 9}\\{ - 3}&6&h\end{array}} \right]\left[ {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\\{{x_3}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}0\\0\\0\end{array}} \right]\)

The matrix equation is in \(\left[ {\begin{array}{*{20}{c}}1&{ - 2}&2\\{ - 5}&{10}&{ - 9}\\{ - 3}&6&h\end{array}} \right]\left[ {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\\{{x_3}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}0\\0\\0\end{array}} \right]\), \(A{\bf{x}} = {\bf{0}}\) form.

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

\(\left[ {\begin{array}{*{20}{c}}1&{ - 2}&2&0\\{ - 5}&{10}&{ - 9}&0\\{ - 3}&6&h&0\end{array}} \right]\)

Use the \({x_1}\) term from the first equation to eliminate the \( - 5{x_1}\) term from the second equation. Add 5 times row one to row two.

\(\left[ {\begin{array}{*{20}{c}}1&{ - 2}&2&0\\0&0&1&0\\{ - 3}&6&h&0\end{array}} \right]\)

Use the \({x_1}\) term from the first equation to eliminate the \( - 3{x_1}\) term from the third equation. Add 3 times row one to row three.

\(\left[ {\begin{array}{*{20}{c}}1&{ - 2}&2&0\\0&0&1&0\\0&0&{h + 6}&0\end{array}} \right]\)

Mark the non-zero leading entries in columns one and three.

According to the pivot positions in the obtained matrix, \({x_2}\) is a free variable.

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

So, for every value of \(h\), \(\left\{ {{{\bf{v}}_1},{{\bf{v}}_2},{{\bf{v}}_3}} \right\}\) is linearly dependent.