Question 42: Find a column of matrix in \(\left[ {\begin{array}{*{20}{c}}8&{11}&{ - 6}&{ - 7}&{13}\\{ - 7}&{ - 8}&5&6&{ - 9}\\{11}&7&{ - 7}&{ - 9}&{ - 6}\\{ - 3}&4&1&8&7\end{array}} \right]\) that can be deleted and yet have the remaining matrix columns still span\({R^4}\). Can you delete more than one column?

Let the given matrix be\(A = \left[ {\begin{array}{*{20}{c}}8&{11}&{ - 6}&{ - 7}&{13}\\{ - 7}&{ - 8}&5&6&{ - 9}\\{11}&7&{ - 7}&{ - 9}&{ - 6}\\{ - 3}&4&1&8&7\end{array}} \right]\)and after deleting column 2 of the given matrix A,you get\(A' = \left[ {\begin{array}{*{20}{c}}8&{ - 6}&{ - 7}&{13}\\{ - 7}&5&6&{ - 9}\\{11}&{ - 7}&{ - 9}&{ - 6}\\{ - 3}&1&8&7\end{array}} \right]\).

Apply the Gaussian elimination method in matrix A to find the pivot columns in the matrix.

Apply row operation\({R_2} \to {R_2} + \frac{7}{8}{R_1}\).

\(\left[ {\begin{array}{*{20}{c}}8&{11}&{ - 6}&{ - 7}&{13}\\0&{\frac{{13}}{8}}&{ - \frac{1}{4}}&{ - \frac{1}{8}}&{\frac{{19}}{8}}\\{11}&7&{ - 7}&{ - 9}&{ - 6}\\{ - 3}&4&1&8&7\end{array}} \right]\)

Now, apply row operation\({R_3} \to {R_3} - \frac{{11}}{8}{R_1}\)again in the above matrix.

\(\left[ {\begin{array}{*{20}{c}}8&{11}&{ - 6}&{ - 7}&{13}\\0&{\frac{{13}}{8}}&{ - \frac{1}{4}}&{ - \frac{1}{8}}&{\frac{{19}}{8}}\\0&{ - \frac{{65}}{8}}&{\frac{5}{4}}&{\frac{5}{8}}&{ - \frac{{191}}{8}}\\{ - 3}&4&1&8&7\end{array}} \right]\)

Apply row operation\({R_4} \to {R_4} + \frac{3}{8}{R_1}\)in the above matrix.

\(\left[ {\begin{array}{*{20}{c}}8&{11}&{ - 6}&{ - 7}&{13}\\0&{\frac{{13}}{8}}&{ - \frac{1}{4}}&{ - \frac{1}{8}}&{\frac{{19}}{8}}\\0&{ - \frac{{65}}{8}}&{\frac{5}{4}}&{\frac{5}{8}}&{ - \frac{{191}}{8}}\\0&{\frac{{65}}{8}}&{ - \frac{5}{4}}&{\frac{{43}}{8}}&{\frac{{95}}{8}}\end{array}} \right]\)

Apply row operation\({R_3} \to {R_3} + 5{R_2}\)in the above matrix.

\(\left[ {\begin{array}{*{20}{c}}8&{11}&{ - 6}&{ - 7}&{13}\\0&{\frac{{13}}{8}}&{ - \frac{1}{4}}&{ - \frac{1}{8}}&{\frac{{19}}{8}}\\0&0&0&0&{ - 12}\\0&{\frac{{65}}{8}}&{ - \frac{5}{4}}&{\frac{{43}}{8}}&{\frac{{95}}{8}}\end{array}} \right]\)

Apply row operation\({R_4} \to {R_4} - 5{R_2}\)in the above matrix to get the pivot of a matrix.

\(\left[ {\begin{array}{*{20}{c}}8&{11}&{ - 6}&{ - 7}&{13}\\0&{\frac{{13}}{8}}&{ - \frac{1}{4}}&{ - \frac{1}{8}}&{\frac{{19}}{8}}\\0&0&0&0&{ - 12}\\0&0&0&6&0\end{array}} \right]\)

Now, after interchanging the rows\({R_3} \leftrightarrow {R_4}\)in the above matrix, you get:

\(\left[ {\begin{array}{*{20}{c}}8&{11}&{ - 6}&{ - 7}&{13}\\0&{\frac{{13}}{8}}&{ - \frac{1}{4}}&{ - \frac{1}{8}}&{\frac{{19}}{8}}\\0&0&0&6&0\\0&0&0&0&{ - 12}\end{array}} \right]\)

The pivots in the matrix columns are represented as:

\(\left[ {\begin{array}{*{20}{c}} {\boxed8}&{11}&{ - 6}&{ - 7}&{13} \\ 0&{\boxed{\frac{{13}}{8}}}&{ - \frac{1}{4}}&{ - \frac{1}{8}}&{\frac{{19}}{8}} \\ 0&0&0&{\boxed6}&0 \\ 0&0&0&0&{\boxed{ - 12}} \end{array}} \right]\)

The matrix has a pivot in every row.

Let this echelon matrix be \(B = \left[ {\begin{array}{*{20}{c}}8&{11}&{ - 6}&{ - 7}&{13}\\0&{\frac{{13}}{8}}&{ - \frac{1}{4}}&{ - \frac{1}{8}}&{\frac{{19}}{8}}\\0&0&0&6&0\\0&0&0&0&{ - 12}\end{array}} \right]\)and after deleting column 4,you get matrix \(B' = \left[ {\begin{array}{*{20}{c}}8&{11}&{ - 6}&{ - 7}\\0&{\frac{{13}}{8}}&{ - \frac{1}{4}}&{\frac{1}{8}}\\0&0&0&6\\0&0&0&0\end{array}} \right]\).

\(A'\)to\(B'\)is reduced by the same sequence of row operations that reduces A to B. Since\(B'\)is in the echelon form, each row provides a pivot position for\(A'\).

Hence,\(A'\)columns of the matrix are in span\({R^4}\).

Instead of column 3, it is feasible to delete column 2 of A. Only one column, however, can be eliminated. If two or more columns in column A are removed, the resulting matrix will have fewer than four columns and thus fewer than four pivot places. In this situation, not every row could have a pivot position, and the matrix's columns wouldnot cover \({R^4}\).