Question

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

Short answer

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

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

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:

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

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

Mark the pivot positions in the matrix

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

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

Construct matrix B

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,

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

Show that \(B{\bf{x}} = 0\) has a trivial solution

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.