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Repeat Exercise 43 with the matrices A and B from Exercise 42. Then give an explanation for what you discover, assuming that B was constructed as specified.

Short Answer

Expert verified

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

Step by step solution

01

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, 12 is the pivot.

02

Apply row operation

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

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

\(\begin{array}{l} > > {\rm{ A }} = {\rm{ }}\left[ \begin{array}{l}12{\rm{ }}10{\rm{ }} - 6{\rm{ }} - 3{\rm{ }}7{\rm{ }}10;{\rm{ }} - 7 - 6{\rm{ }}4{\rm{ }}7{\rm{ }} - 9{\rm{ }}5;{\rm{ }}9{\rm{ }}9{\rm{ }} - 9{\rm{ }} - 5{\rm{ }}5{\rm{ }} - 1;{\rm{ }}\\ - 4{\rm{ }} - 3{\rm{ }}1{\rm{ }}6{\rm{ }} - 8{\rm{ }}9;{\rm{ }}8{\rm{ }}7{\rm{ }} - 5{\rm{ }} - 9{\rm{ }}11{\rm{ }} - 8\end{array} \right];\\ > > {\rm{ U}} = {\rm{rref}}\left( {\rm{A}} \right)\end{array}\)

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

03

Mark the pivot positions in the matrix

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

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

04

Construct matrix B

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

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

05

Choose one dependent vector and obtain the augmented matrix

Let one of the dependent vectors be\({\bf{v}} = \left[ {\begin{array}{*{20}{c}}{ - 14}\\9\\{ - 18}\\1\\{ - 11}\end{array}} \right]\).

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

\(\left[ {\begin{array}{*{20}{c}}{12}&{10}&{ - 3}&{10}&{ - 14}\\{ - 7}&{ - 6}&7&5&9\\9&9&{ - 5}&{ - 1}&{ - 18}\\{ - 4}&{ - 3}&6&9&1\\8&7&{ - 9}&{ - 8}&{ - 11}\end{array}} \right]\)

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

\(\begin{array}{l} > > {\rm{ A }} = {\rm{ }}\left[ \begin{array}{l}12{\rm{ }}10{\rm{ }} - 3{\rm{ 10 }} - {\rm{14}};{\rm{ }} - 7 - 6{\rm{ 7 5 9}};{\rm{ }}9{\rm{ }}9{\rm{ }} - 5{\rm{ - 1 }} - {\rm{18}};{\rm{ }}\\ - 4{\rm{ }} - 3{\rm{ 6 9 1}};{\rm{ }}8{\rm{ }}7{\rm{ }} - 9{\rm{ }} - {\rm{8 }} - 11\end{array} \right];\\ > > {\rm{ U}} = {\rm{rref}}\left( {\rm{A}} \right)\end{array}\)

\(\left[ {\begin{array}{*{20}{c}}{12}&{10}&{ - 3}&{10}&{ - 14}\\{ - 7}&{ - 6}&7&5&9\\9&9&{ - 5}&{ - 1}&{ - 18}\\{ - 4}&{ - 3}&6&9&1\\8&7&{ - 9}&{ - 8}&{ - 11}\end{array}} \right] \sim \left[ {\begin{array}{*{20}{c}}1&0&0&0&1\\0&1&0&0&{ - 2}\\0&0&1&0&1\\0&0&0&1&0\\0&0&0&0&0\end{array}} \right]\)

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

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Most popular questions from this chapter

Suppose a linear transformation \(T:{\mathbb{R}^n} \to {\mathbb{R}^n}\) has the property that \(T\left( {\mathop{\rm u}\nolimits} \right) = T\left( {\mathop{\rm v}\nolimits} \right)\) for some pair of distinct vectors u and v in \({\mathbb{R}^n}\). Can Tmap \({\mathbb{R}^n}\) onto \({\mathbb{R}^n}\)? Why or why not?

Determine the values(s) of \(h\) such that matrix is the augmented matrix of a consistent linear system.

17. \(\left[ {\begin{array}{*{20}{c}}2&3&h\\4&6&7\end{array}} \right]\)

In Exercises 33 and 34, Tis a linear transformation from \({\mathbb{R}^2}\) into \({\mathbb{R}^2}\). Show that T is invertible and find a formula for \({T^{ - 1}}\).

34. \(T\left( {{x_1},{x_2}} \right) = \left( {6{x_1} - 8{x_2}, - 5{x_1} + 7{x_2}} \right)\)

Let \({{\mathop{\rm v}\nolimits} _1} = \left[ {\begin{array}{*{20}{c}}1\\0\\{ - 2}\end{array}} \right],{v_2} = \left[ {\begin{array}{*{20}{c}}{ - 3}\\1\\8\end{array}} \right],\) and \({\rm{y = }}\left[ {\begin{array}{*{20}{c}}h\\{ - 5}\\{ - 3}\end{array}} \right]\). For what values(s) of \(h\) is \(y\) in the plane generated by \({{\mathop{\rm v}\nolimits} _1}\) and \({{\mathop{\rm v}\nolimits} _2}\)

An important concern in the study of heat transfer is to determine the steady-state temperature distribution of a thin plate when the temperature around the boundary is known. Assume the plate shown in the figure represents a cross section of a metal beam, with negligible heat flow in the direction perpendicular to the plate. Let \({T_1},...,{T_4}\) denote the temperatures at the four interior nodes of the mesh in the figure. The temperature at a node is approximately equal to the average of the four nearest nodes—to the left, above, to the right, and below. For instance,

\({T_1} = \left( {10 + 20 + {T_2} + {T_4}} \right)/4\), or \(4{T_1} - {T_2} - {T_4} = 30\)

33. Write a system of four equations whose solution gives estimates

for the temperatures \({T_1},...,{T_4}\).

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