In Exercises 41 and 42, use as many columns of A as possible to construct a matrix B with the property that the equation \(B{\bf{x}} = 0\) has only the trivial solution. Solve \(B{\bf{x}} = 0\) to verify your work.

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

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.

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:

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

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

Mark the nonzero leading entries in columns 1, 2, and 5.

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

Construct matrix B by using the 1, 2, and 5 pivot columns of the matrixas shown below:

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

Matrix B can also be written using column 3 or column 4 of matrix A in the place of column 2 of matrix B as shown below:

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

Or,

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

Use matrix\(B = \left[ {\begin{array}{*{20}{c}}8&{ - 3}&2\\{ - 9}&4&{ - 7}\\6&{ - 2}&4\\5&{ - 1}&{10}\end{array}} \right]\)in the equation\(B{\bf{x}} = 0\)to show that the system has a trivial solution.

\(\left[ {\begin{array}{*{20}{c}}8&{ - 3}&2\\{ - 9}&4&{ - 7}\\6&{ - 2}&4\\5&{ - 1}&{10}\end{array}} \right]\left[ {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\\{{x_3}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}0\\0\\0\end{array}} \right]\)

The above matrix equation in the augmented matrix \[\left[ {\begin{array}{*{20}{c}}B&0\end{array}} \right]\] can be written as shown below:

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

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

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

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

In the equation, the matrix row can be written as shown below:

\(\begin{array}{l}{x_1} = 0\\{x_2} = 0\\{x_3} = 0\end{array}\)

Thus, \(B{\bf{x}} = 0\) has a trivial solution.