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Chapter 2: Q2.8-25E (page 93)

Exercises 23-26 display a matrix A and echelon form of A. Find a basis for Col A and a basis for Nul A.

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

Short Answer

Expert verified

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

Step by step solution

01

Identify the pivot column using the echelon form

From the echelon form, it can be observed that columns 1, 2, and 4 are the pivot columns

02

Find the basis of Col A

The basis of Col A is

\(\left[ {\begin{array}{*{20}{c}}1\\{ - 1}\\{ - 2}\\3\end{array}} \right]\), \(\left[ {\begin{array}{*{20}{c}}4\\2\\2\\6\end{array}} \right]\), \(\left[ {\begin{array}{*{20}{c}}{ - 3}\\3\\5\\{ - 5}\end{array}} \right]\).

03

Find the basis of Nul A

Nul A is given by the equation \(Ax = 0\). Then,

\[\left[ {\begin{array}{*{20}{c}}1&4&8&0&5\\0&2&5&0&{ - 1}\\0&0&0&1&4\\0&0&0&0&0\end{array}} \right]\left[ {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\\{{x_3}}\\{{x_4}}\\{{x_5}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}0\\0\\0\\0\end{array}} \right]\].

So, the system of equations is

\({x_4} + 4{x_5} = 0\),

\(2{x_2} + 5{x_3} - {x_5} = 0\),

\({x_1} + 4{x_2} + 8{x_3} + 5{x_5} = 0\).

04

Simplify the equation Nul A

For the equations, \({x_3}\) and \({x_4}\) are the free variables.

\(\begin{array}{c}{x_1} = - 4{x_2} - 8{x_3} - 5{x_5}\\ = - 4\left( { - \frac{5}{2}{x_3} + \frac{1}{2}{x_5}} \right) - 8{x_3} - 5{x_5}\\ = 2{x_3} - 7{x_5}\end{array}\)

The solution set is obtained as shown below:

\(\begin{array}{c}\left[ {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\\{{x_3}}\\{{x_4}}\\{{x_5}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{2{x_3} - 7{x_5}}\\{ - \frac{5}{2}{x_3} + \frac{1}{2}{x_5}}\\{{x_3}}\\{ - 4{x_5}}\\{{x_5}}\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}2\\{ - \frac{5}{2}}\\1\\0\\0\end{array}} \right]{x_3} + \left[ {\begin{array}{*{20}{c}}{ - 7}\\{\frac{1}{2}}\\0\\{ - 4}\\1\end{array}} \right]{x_5}\end{array}\)

So, the basis of Col A is \(\left\{ {\left[ {\begin{array}{*{20}{c}}1\\{ - 1}\\{ - 2}\\3\end{array}} \right],\left[ {\begin{array}{*{20}{c}}4\\2\\2\\6\end{array}} \right],\left[ {\begin{array}{*{20}{c}}{ - 3}\\3\\5\\{ - 5}\end{array}} \right]} \right\}\), and the basis of Nul A is \(\left\{ {\left[ {\begin{array}{*{20}{c}}2\\{ - \frac{5}{2}}\\1\\0\\0\end{array}} \right],\left[ {\begin{array}{*{20}{c}}{ - 7}\\{\frac{1}{2}}\\0\\{ - 4}\\1\end{array}} \right]} \right\}\).

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Most popular questions from this chapter

Suppose Ais a \(3 \times n\) matrix whose columns span \({\mathbb{R}^3}\). Explain how to construct an \(n \times 3\) matrix Dsuch that \(AD = {I_3}\).

A useful way to test new ideas in matrix algebra, or to make conjectures, is to make calculations with matrices selected at random. Checking a property for a few matrices does not prove that the property holds in general, but it makes the property more believable. Also, if the property is actually false, you may discover this when you make a few calculations.

36. Write the command(s) that will create a \(6 \times 4\) matrix with random entries. In what range of numbers do the entries lie? Tell how to create a \(3 \times 3\) matrix with random integer entries between \( - {\bf{9}}\) and 9. (Hint:If xis a random number such that 0 < x < 1, then \( - 9.5 < 19\left( {x - .5} \right) < 9.5\).

(M) Read the documentation for your matrix program, and write the commands that will produce the following matrices (without keying in each entry of the matrix).

  1. A \({\bf{5}} \times {\bf{6}}\) matrix of zeros
  2. A \({\bf{3}} \times {\bf{5}}\) matrix of ones
  3. The \({\bf{6}} \times {\bf{6}}\) identity matrix
  4. A \({\bf{5}} \times {\bf{5}}\) diagonal matrix, with diagonal entries 3, 5, 7, 2, 4

Use the inverse found in Exercise 3 to solve the system

\(\begin{aligned}{l}{\bf{8}}{{\bf{x}}_{\bf{1}}} + {\bf{5}}{{\bf{x}}_{\bf{2}}} = - {\bf{9}}\\ - {\bf{7}}{{\bf{x}}_{\bf{1}}} - {\bf{5}}{{\bf{x}}_{\bf{2}}} = {\bf{11}}\end{aligned}\)

In Exercise 10 mark each statement True or False. Justify each answer.

10. a. A product of invertible \(n \times n\) matrices is invertible, and the inverse of the product of their inverses in the same order.

b. If A is invertible, then the inverse of \({A^{ - {\bf{1}}}}\) is A itself.

c. If \(A = \left( {\begin{aligned}{*{20}{c}}a&b\\c&d\end{aligned}} \right)\) and \(ad = bc\), then A is not invertible.

d. If A can be row reduced to the identity matrix, then A must be invertible.

e. If A is invertible, then elementary row operations that reduce A to the identity \({I_n}\) also reduce \({A^{ - {\bf{1}}}}\) to \({I_n}\).
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