Question

Short answer

The columns of the matrix are not in span \({R^4}\) .

Step-by-step solution

Solve matrix

First, resolve the matrix into a pivot matrix.

Apply row operation \({R_2} \to {R_2} + \frac{5}{7}{R_1}\).

\(\left[ {\begin{array}{*{20}{c}}7&2&{ - 5}&8\\0&{ - \frac{{11}}{7}}&{\frac{3}{7}}&{ - \frac{{23}}{7}}\\6&{10}&{ - 2}&7\\{ - 7}&9&2&{15}\end{array}} \right]\)

Now, apply row operation \({R_3} \to {R_3} - \frac{6}{7}{R_1}\) again in the above matrix.

\(\left[ {\begin{array}{*{20}{c}}7&2&{ - 5}&8\\0&{ - \frac{{11}}{7}}&{\frac{3}{7}}&{ - \frac{{23}}{7}}\\0&{\frac{{58}}{7}}&{\frac{{16}}{7}}&{\frac{1}{7}}\\{ - 7}&9&2&{15}\end{array}} \right]\)

Row operation in the matrix

Apply row operation \({R_4} \to {R_4} + {R_1}\) in the above matrix.

\(\left[ {\begin{array}{*{20}{c}}7&2&{ - 5}&8\\0&{ - \frac{{11}}{7}}&{\frac{3}{7}}&{ - \frac{{23}}{7}}\\0&{\frac{{58}}{7}}&{\frac{{16}}{7}}&{\frac{1}{7}}\\0&{11}&{ - 3}&{23}\end{array}} \right]\)

Apply row operation \({R_3} \to {R_3} + \frac{{58}}{{11}}{R_2}\) in the above matrix.

\(\left[ {\begin{array}{*{20}{c}}7&2&{ - 5}&8\\0&{ - \frac{{11}}{7}}&{\frac{3}{7}}&{ - \frac{{23}}{7}}\\0&0&{\frac{{50}}{{11}}}&{\frac{{ - 189}}{{11}}}\\0&{11}&{ - 3}&{23}\end{array}} \right]\)

Pivot of a matrix

Apply row operation \({R_4} \to {R_4} + 7{R_2}\) in the above matrix to get the pivot of a matrix.

\(\left[ {\begin{array}{*{20}{c}}7&2&{ - 5}&8\\0&{ - \frac{{11}}{7}}&{\frac{3}{7}}&{ - \frac{{23}}{7}}\\0&0&{\frac{{50}}{{11}}}&{\frac{{ - 189}}{{11}}}\\0&0&0&0\end{array}} \right]\)

Determine the span

The above matrix in terms of pivot columns is represented as:

The matrix does not have a pivot in every row.

Hence, the columns of the matrix are not in span \({R^4}\).