Question

With A and B as in Exercise 41, select a column v of A that was not used in the construction of B and determine if v is in the set spanned by the columns of B. (Describe your calculations.)

Short answer

Every linearly dependent column in A is in the set spanned by the columns of B because there are three pivot elements.

Step-by-step solution

Identify pivot position

To identify the pivot and the pivot position, observe the matrix’s leftmost column (nonzero column), that is, the pivot column. At the top of this column, 8 is the pivot.

Apply row operation

Consider matrix \(A = \left[ {\begin{array}{*{20}{c}}8&{ - 3}&0&{ - 7}&2\\{ - 9}&4&5&{11}&{ - 7}\\6&{ - 2}&2&{ - 4}&4\\5&{ - 1}&7&0&{10}\end{array}} \right]\).

Use code in the MATLAB to obtain the row-reduced echelon form as shown below:

\(\begin{array}{l} > > {\rm{ A }} = {\rm{ }}\left[ {{\rm{8 }} - 3{\rm{ }}0{\rm{ }} - 7{\rm{ 2}};{\rm{ }} - 9{\rm{ 4 5 }}11{\rm{ }} - {\rm{7}};{\rm{ 6 }} - {\rm{2 2 }} - {\rm{4 4}};{\rm{ 5 }} - 1{\rm{ 7 0 }}10} \right];\\ > > {\rm{ U}} = {\rm{rref}}\left( {\rm{A}} \right)\end{array}\)

\(\left[ {\begin{array}{*{20}{c}}8&{ - 3}&0&{ - 7}&2\\{ - 9}&4&5&{11}&{ - 7}\\6&{ - 2}&2&{ - 4}&4\\5&{ - 1}&7&0&{10}\end{array}} \right] \sim \left[ {\begin{array}{*{20}{c}}1&0&2&0&2\\0&1&{ - 1}&0&{ - 1}\\0&0&0&1&{ - 2}\\0&0&0&0&0\end{array}} \right]\)

Mark the pivot positions in the matrix

Mark the nonzero leading entries in columns 1, 2, and 4.

Columns 3 and 5 are the linear combinations of 1, 2, and 4 columns.

Now, mark the pivot columns of the given matrix as shown below:

Construct matrix B

Construct matrix B by using the 1, 2, and 4 pivot columns of the matrixas shown below:

\(B = \left[ {\begin{array}{*{20}{c}}8&{ - 3}&7\\{ - 9}&4&{11}\\6&{ - 2}&{ - 4}\\5&{ - 1}&0\end{array}} \right]\)

Choose one dependent vector and obtain the augmented matrix

Let one of the dependent vectors be\({\bf{v}} = \left[ {\begin{array}{*{20}{c}}{ - 4}\\{15}\\1\\{15}\end{array}} \right]\).

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

\(\left[ {\begin{array}{*{20}{c}}8&{ - 3}&7&{ - 4}\\{ - 9}&4&{11}&{15}\\6&{ - 2}&{ - 4}&1\\5&{ - 1}&0&{15}\end{array}} \right]\)

Use code in the MATLAB to obtain the row-reduced echelon form as shown below:

\(\begin{array}{l} > > {\rm{ A }} = {\rm{ }}\left[ {{\rm{8 }} - 3{\rm{ }} - 7{\rm{ - 4}};{\rm{ }} - 9{\rm{ 4 }}11{\rm{ 15}};{\rm{ 6 }} - {\rm{2 }} - {\rm{4 1}};{\rm{ 5 }} - 1{\rm{ 0 }}15} \right];\\ > > {\rm{ U}} = {\rm{rref}}\left( {\rm{A}} \right)\end{array}\)

\(\left[ {\begin{array}{*{20}{c}}8&{ - 3}&7&{ - 4}\\{ - 9}&4&{11}&{15}\\6&{ - 2}&{ - 4}&1\\5&{ - 1}&0&{15}\end{array}} \right] \sim \left[ {\begin{array}{*{20}{c}}1&0&0&2\\0&1&0&{ - 1}\\0&0&1&{ - 2}\\0&0&0&0\end{array}} \right]\)

Thus, every linearly dependent column in A is in the set spanned by the columns of B because there are three pivot elements.