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13. For A as in exercise 11 i.e., \[A = \left[ {\begin{array}{*{20}{c}}{\bf{3}}&{\bf{2}}&{\bf{1}}&{ - {\bf{5}}}\\{ - {\bf{9}}}&{ - {\bf{4}}}&{\bf{1}}&{\bf{7}}\\{\bf{9}}&{\bf{2}}&{ - {\bf{5}}}&{\bf{1}}\end{array}} \right]\], find a nonzero vector in Nul A and a nonzero vector in Col A.

Short Answer

Expert verified

One of the nonzero vectors in Nul A is \[\left( {1,2, - 1,6} \right)\], and one of the nonzero vectors in Col A is \[\left( {3, - 9,9} \right)\].

Step by step solution

01

Use row reduction

First, solve the equation \[Ax = 0\].

Its augmented matrix is:

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

At row 2, multiply row 1 by 3 and add it to row 2, i.e., , and at row 3, multiply row 1 by 3 and subtract it from row 3, i.e., \[{R_3} \to {R_3} - 3{R_1}\]. Then,

\[ \sim \left[ {\begin{array}{*{20}{c}}3&2&1&{ - 5}&0\\0&2&4&{ - 8}&0\\0&{ - 4}&{ - 8}&{16}&0\end{array}} \right]\]

At row 3, multiply row 2 by 2 and add it to row 3, i.e., \[{R_3} \to {R_3} + 2{R_2}\].

\[ \sim \left[ {\begin{array}{*{20}{c}}3&2&1&{ - 5}&0\\0&2&4&{ - 8}&0\\0&0&0&0&0\end{array}} \right]\]

At row 1, subtract row 2 from row 1, i.e., \[{R_1} \to {R_1} - {R_2}\].

\[ \sim \left[ {\begin{array}{*{20}{c}}3&0&{ - 3}&3&0\\0&2&4&{ - 8}&0\\0&0&0&0&0\end{array}} \right]\]

At row 1, divide row 1 by 3, and at row 2, divide row 2 by 2.

\[\left[ {\begin{array}{*{20}{c}}A&0\end{array}} \right] \sim \left[ {\begin{array}{*{20}{c}}1&0&{ - 1}&1&0\\0&1&2&{ - 4}&0\\0&0&0&0&0\end{array}} \right]\]

This implies that

\[\begin{array}{c}{x_1} - {x_3} + {x_4} = 0\\{x_2} + 2{x_3} - 4{x_4} = 0\\0 = 0\end{array}\]

Thus, the system is consistent.

02

Find a nonzero vector in Nul A

The general solution is \[{x_1} = {x_3} - {x_4}\], and with \[{x_3},\] and are free. So, the nonzero vector in Nul A is given by \[{x_3},\] and .

Choose and then , and \[{x_2} = 6\] to obtain a nonzero vector \[\left( {1,2, - 1,6} \right)\] in Nul A.

03

Find a nonzero vector in Col A

By definition of column space, Col A is the span of \[\left\{ {\left( {3, - 9,9} \right),\left( {2, - 4,2} \right),\left( {1,1, - 5} \right),\left( { - 5,7,1} \right)} \right\}\].

So, every column of A belongs to Col A. Choose any nonzero column of A. So, is a nonzero vector in Col A.

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Most popular questions from this chapter

Let \(A = \left( {\begin{aligned}{*{20}{c}}{\bf{2}}&{\bf{5}}\\{ - {\bf{3}}}&{\bf{1}}\end{aligned}} \right)\) and \(B = \left( {\begin{aligned}{*{20}{c}}{\bf{4}}&{ - {\bf{5}}}\\{\bf{3}}&k\end{aligned}} \right)\). What value(s) of \(k\), if any will make \(AB = BA\)?

If a matrix \(A\) is \({\bf{5}} \times {\bf{3}}\) and the product \(AB\)is \({\bf{5}} \times {\bf{7}}\), what is the size of \(B\)?

Suppose the transfer function W(s) in Exercise 19 is invertible for some s. It can be showed that the inverse transfer function \(W{\left( s \right)^{ - {\bf{1}}}}\), which transforms outputs into inputs, is the Schur complement of \(A - BC - s{I_n}\) for the matrix below. Find the Sachur complement. See Exercise 15.

\(\left[ {\begin{array}{*{20}{c}}{A - BC - s{I_n}}&B\\{ - C}&{{I_m}}\end{array}} \right]\)

Exercises 15 and 16 concern arbitrary matrices A, B, and Cfor which the indicated sums and products are defined. Mark each statement True or False. Justify each answer.

16. a. If A and B are \({\bf{3}} \times {\bf{3}}\) and \(B = \left( {\begin{aligned}{*{20}{c}}{{{\bf{b}}_1}}&{{{\bf{b}}_2}}&{{{\bf{b}}_3}}\end{aligned}} \right)\), then \(AB = \left( {A{{\bf{b}}_1} + A{{\bf{b}}_2} + A{{\bf{b}}_3}} \right)\).

b. The second row of ABis the second row of Amultiplied on the right by B.

c. \(\left( {AB} \right)C = \left( {AC} \right)B\)

d. \({\left( {AB} \right)^T} = {A^T}{B^T}\)

e. The transpose of a sum of matrices equals the sum of their transposes.

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.

37. Construct a random \({\bf{4}} \times {\bf{4}}\) matrix Aand test whether \(\left( {A + I} \right)\left( {A - I} \right) = {A^2} - I\). The best way to do this is to compute \(\left( {A + I} \right)\left( {A - I} \right) - \left( {{A^2} - I} \right)\) and verify that this difference is the zero matrix. Do this for three random matrices. Then test \(\left( {A + B} \right)\left( {A - B} \right) = {A^2} - {B^{\bf{2}}}\) the same way for three pairs of random \({\bf{4}} \times {\bf{4}}\) matrices. Report your conclusions.

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