Question

Question: Find a basis for the set of all vectors of the form

\(\left[ {\begin{array}{*{20}{c}}{a - {\bf{2}}b + {\bf{5}}c}\\{{\bf{2}}a + {\bf{5}}b - {\bf{8}}c}\\{ - a - {\bf{4}}b + {\bf{7}}c}\\{{\bf{3}}a + b + c}\end{array}} \right]\).

Short answer

The basis of the matrix is \(\left\{ {\left[ {\begin{array}{*{20}{c}}1\\2\\{ - 1}\\3\end{array}} \right],\,\,\left[ {\begin{array}{*{20}{c}}{ - 2}\\5\\{ - 4}\\1\end{array}} \right]} \right\}\).

Step-by-step solution

Step 1:Form the matrix using the vectors

The matrix using the set of vectors can be written in the following manner:

\(\left[ {\begin{array}{*{20}{c}}1&{ - 2}&5\\2&5&{ - 8}\\{ - 1}&{ - 4}&7\\3&1&1\end{array}} \right]\)

Row reduce the matrix

Use the row operationson the above matrix as shown below:

\(\begin{array}{c}\left[ {\begin{array}{*{20}{c}}1&{ - 2}&5\\2&5&{ - 8}\\{ - 1}&{ - 4}&7\\3&1&1\end{array}} \right] = \,\,\left[ {\begin{array}{*{20}{c}}1&{ - 2}&5\\0&9&{ - 18}\\0&{ - 6}&{12}\\0&7&{ - 14}\end{array}} \right]\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left\{ \begin{array}{l}{R_2} \to {R_2} - 2{R_1}\\{R_3} \to {R_3} + {R_1}\\{R_4} \to {R_4} - 3{R_1}\end{array} \right\}\\ = \left[ {\begin{array}{*{20}{c}}1&{ - 2}&5\\0&1&{ - 2}\\0&{ - 6}&{12}\\0&7&{ - 14}\end{array}} \right]\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left\{ {{R_2} \to \frac{1}{9}{R_2}} \right\}\\ = \left[ {\begin{array}{*{20}{c}}1&0&1\\0&1&{ - 2}\\0&0&0\\0&0&0\end{array}} \right]\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left\{ \begin{array}{l}{R_1} \to {R_1} + 2{R_2}\\{R_3} \to {R_3} + 6{R_2}\end{array} \right\}\end{array}\)

From the row reduced form, it can be observed that column 1 and column 2 are the pivot columns.

So, the basis of the matrix is \(\left\{ {\left[ {\begin{array}{*{20}{c}}1\\2\\{ - 1}\\3\end{array}} \right],\,\,\left[ {\begin{array}{*{20}{c}}{ - 2}\\5\\{ - 4}\\1\end{array}} \right]} \right\}\).