Question 21: Let \({v_1} = \left( {\begin{array}{*{20}{c}}1\\0\\{ - 1}\\0\end{array}} \right),{v_2} = \left( {\begin{array}{*{20}{c}}0\\{ - 1}\\0\\1\end{array}} \right),{v_3} = \left( {\begin{array}{*{20}{c}}1\\0\\0\\{ - 1}\end{array}} \right)\) .

First, reduce the row matrix\(\left\{ {{v_1},{v_2},{v_3}} \right\}\)to check whether it contains a pivot in each row.

So, the matrix \(\left\{ {{v_1},{v_2},{v_3}} \right\}\)can be written as:

\(\left\{ {{v_1},{v_2},{v_3}} \right\} = \left[ {\begin{array}{*{20}{c}}1&0&1\\0&{ - 1}&0\\{ - 1}&0&0\\0&1&{ - 1}\end{array}} \right]\)

Apply \({R_3} \to {R_3} + {R_1}\) in the given matrix.

\(\left\{ {{v_1},{v_2},{v_3}} \right\} = \left[ {\begin{array}{*{20}{c}}1&0&1\\0&{ - 1}&0\\0&0&1\\0&1&{ - 1}\end{array}} \right]\)

Now, apply row operation \({R_4} \to {R_4} + {R_2}\)again.

\(\left\{ {{v_1},{v_2},{v_3}} \right\} = \left[ {\begin{array}{*{20}{c}}1&0&1\\0&{ - 1}&0\\0&0&1\\0&0&{ - 1}\end{array}} \right]\)

Now, apply row operation\({R_2} \to - {R_2}\)and \({R_4} \to {R_4} + {R_3}\).

\(\left\{ {{v_1},{v_2},{v_3}} \right\} = \left[ {\begin{array}{*{20}{c}}1&0&1\\0&1&0\\0&0&1\\0&0&0\end{array}} \right]\)

From the above, it is concluded that the matrix \(\left\{ {{v_1},{v_2},{v_3}} \right\}\)does not contain a pivot in every row. As a result, span \({R^4}\) does not appear in the matrix column.

Hence, \(\left\{ {{v_1},{v_2},{v_3}} \right\}\)does not belong in span \({R^4}\).