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.

42. \(A = \left[ {\begin{array}{*{20}{c}}{12}&{10}&{ - 6}&{ - 3}&7&{10}\\{ - 7}&{ - 6}&4&7&{ - 9}&5\\9&9&{ - 9}&{ - 5}&5&{ - 1}\\{ - 4}&{ - 3}&1&6&{ - 8}&9\\8&7&{ - 5}&{ - 9}&{11}&{ - 8}\end{array}} \right]\)

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 \(A = \left[ {\begin{array}{*{20}{c}}{12}&{10}&{ - 6}&{ - 3}&7&{10}\\{ - 7}&{ - 6}&4&7&{ - 9}&5\\9&9&{ - 9}&{ - 5}&5&{ - 1}\\{ - 4}&{ - 3}&1&6&{ - 8}&9\\8&7&{ - 5}&{ - 9}&{11}&{ - 8}\end{array}} \right]\).

Apply the row operation by using MATLAB to obtain the row-reduced echelon form as shown below:

\(\left[ {\begin{array}{*{20}{c}}{12}&{10}&{ - 6}&{ - 3}&7&{10}\\{ - 7}&{ - 6}&4&7&{ - 9}&5\\9&9&{ - 9}&{ - 5}&5&{ - 1}\\{ - 4}&{ - 3}&1&6&{ - 8}&9\\8&7&{ - 5}&{ - 9}&{11}&{ - 8}\end{array}} \right] \sim \left[ {\begin{array}{*{20}{c}}1&0&2&0&2&0\\0&1&{ - 3}&0&{ - 2}&0\\0&0&0&1&{ - 1}&0\\0&0&0&0&0&1\\0&0&0&0&0&0\end{array}} \right]\)

Mark the nonzero leading entries in columns 1, 2, 4, and 6.

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

Construct matrix B by using 1, 2, 4, and 6 pivot columns of the matrixas shown below:

\(B = \left[ {\begin{array}{*{20}{c}}{12}&{10}&{ - 3}&{10}\\{ - 7}&{ - 6}&7&5\\9&9&{ - 5}&{ - 1}\\{ - 4}&{ - 3}&6&9\\8&7&{ - 9}&{ - 8}\end{array}} \right]\)

Matrix B can also be written using column 3 or column 5 of matrix A at the place of column 2 of matrix B or column 4 of matrix B,respectively,as shown below:

\(B = \left[ {\begin{array}{*{20}{c}}{12}&{ - 6}&{ - 3}&{10}\\{ - 7}&4&7&5\\9&{ - 9}&{ - 5}&{ - 1}\\{ - 4}&1&6&9\\8&{ - 5}&{ - 9}&{ - 8}\end{array}} \right]\)

Or,

\(B = \left[ {\begin{array}{*{20}{c}}{12}&{10}&{ - 3}&7\\{ - 7}&{ - 6}&7&{ - 9}\\9&9&{ - 5}&5\\{ - 4}&{ - 3}&6&{ - 8}\\8&7&{ - 9}&{11}\end{array}} \right]\)

Use matrix\(B = \left[ {\begin{array}{*{20}{c}}{12}&{10}&{ - 3}&{10}\\{ - 7}&{ - 6}&7&5\\9&9&{ - 5}&{ - 1}\\{ - 4}&{ - 3}&6&9\\8&7&{ - 9}&{ - 8}\end{array}} \right]\)in the equation\(B{\bf{x}} = 0\)to show that the system has a trivial solution.

\(\left[ {\begin{array}{*{20}{c}}{12}&{10}&{ - 3}&{10}\\{ - 7}&{ - 6}&7&5\\9&9&{ - 5}&{ - 1}\\{ - 4}&{ - 3}&6&9\\8&7&{ - 9}&{ - 8}\end{array}} \right]\left[ {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\\{{x_3}}\\{{x_4}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}0\\0\\0\\0\end{array}} \right]\)

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

\(\left[ {\begin{array}{*{20}{c}}{12}&{10}&{ - 3}&{10}&0\\{ - 7}&{ - 6}&7&5&0\\9&9&{ - 5}&{ - 1}&0\\{ - 4}&{ - 3}&6&9&0\\8&7&{ - 9}&{ - 8}&0\end{array}} \right]\)

Apply the row operation by using MATLAB to obtain the row-reduced echelon form as shown below:

\(\left[ {\begin{array}{*{20}{c}}{12}&{10}&{ - 3}&{10}&0\\{ - 7}&{ - 6}&7&5&0\\9&9&{ - 5}&{ - 1}&0\\{ - 4}&{ - 3}&6&9&0\\8&7&{ - 9}&{ - 8}&0\end{array}} \right] \sim \left[ {\begin{array}{*{20}{c}}1&0&0&0&0\\0&1&0&0&0\\0&0&1&0&0\\0&0&0&1&0\\0&0&0&0&0\end{array}} \right]\)

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

\(\begin{array}{l}{x_1} = 0\\{x_2} = 0\\{x_3} = 0\\{x_4} = 0\end{array}\)

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