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Chapter 2: Q2.8-23E (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.

\[A = \left[ {\begin{array}{*{20}{c}}{\bf{4}}&{\bf{5}}&{\bf{9}}&{ - {\bf{2}}}\\{\bf{6}}&{\bf{5}}&{\bf{1}}&{{\bf{12}}}\\{\bf{3}}&{\bf{4}}&{\bf{8}}&{ - {\bf{3}}}\end{array}} \right] \sim \left[ {\begin{array}{*{20}{c}}{\bf{1}}&{\bf{2}}&{\bf{6}}&{ - {\bf{5}}}\\{\bf{0}}&{\bf{1}}&{\bf{5}}&{ - {\bf{6}}}\\{\bf{0}}&{\bf{0}}&{\bf{0}}&{\bf{0}}\end{array}} \right]\]

Short Answer

Expert verified

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

Step by step solution

01

Identify the pivot column using the echelon form

From the echelon form, it can be observed columns 1 and 2 are thepivot columns.

02

Find the basis of Col A

The basis of Col A is

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

03

Find the basis of Nul A

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

\[\left[ {\begin{array}{*{20}{c}}1&2&6&{ - 5}\\0&1&5&{ - 6}\\0&0&0&0\end{array}} \right]\left[ {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\\{{x_3}}\\{{x_4}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}0\\0\\0\\0\end{array}} \right]\].

So, the equations are

\({x_2} + 5{x_3} - 6{x_4} = 0\),

\({x_1} + 2{x_2} + 6{x_3} - 5{x_4} = 0\).

04

Simplify the equation Nul A

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

\({x_2} = - 5{x_3} + 6{x_4}\)

And

\(\begin{array}{c}{x_1} + 2\left( { - 5{x_3} + 6{x_4}} \right) + 6{x_3} - 5{x_4} = 0\\{x_1} = 4{x_3} - 7{x_4}\end{array}\)

The solution set is obtained as shown below:

\(\begin{array}{c}\left[ {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\\{{x_3}}\\{{x_4}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{4{x_3} - 7{x_4}}\\{ - 5{x_3} + 6{x_4}}\\{{x_3}}\\{{x_4}}\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}{4{x_3}}\\{ - 5{x_3}}\\{{x_3}}\\{{x_4}}\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}{ - 7{x_4}}\\{6{x_4}}\\{{x_3}}\\{{x_4}}\end{array}} \right]\\ = {x_3}\left[ {\begin{array}{*{20}{c}}4\\{ - 5}\\1\\0\end{array}} \right] + {x_4}\left[ {\begin{array}{*{20}{c}}{ - 7}\\6\\0\\1\end{array}} \right]\end{array}\)

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

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Most popular questions from this chapter

In the rest of this exercise set and in those to follow, you should assume that each matrix expression is defined. That is, the sizes of the matrices (and vectors) involved match appropriately.

Compute \(A - {\bf{5}}{I_{\bf{3}}}\) and \(\left( {{\bf{5}}{I_{\bf{3}}}} \right)A\)

\(A = \left( {\begin{aligned}{*{20}{c}}{\bf{9}}&{ - {\bf{1}}}&{\bf{3}}\\{ - {\bf{8}}}&{\bf{7}}&{ - {\bf{6}}}\\{ - {\bf{4}}}&{\bf{1}}&{\bf{8}}\end{aligned}} \right)\)

3. Find the inverse of the matrix \(\left( {\begin{aligned}{*{20}{c}}{\bf{8}}&{\bf{5}}\\{ - {\bf{7}}}&{ - {\bf{5}}}\end{aligned}} \right)\).

Let \({{\bf{r}}_1} \ldots ,{{\bf{r}}_p}\) be vectors in \({\mathbb{R}^{\bf{n}}}\), and let Qbe an\(m \times n\)matrix. Write the matrix\(\left( {\begin{aligned}{*{20}{c}}{Q{{\bf{r}}_1}}& \cdots &{Q{{\bf{r}}_p}}\end{aligned}} \right)\)as a productof two matrices (neither of which is an identity matrix).

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.

38. Use at least three pairs of random \(4 \times 4\) matrices Aand Bto test the equalities \({\left( {A + B} \right)^T} = {A^T} + {B^T}\) and \({\left( {AB} \right)^T} = {A^T}{B^T}\). (See Exercise 37.) Report your conclusions. (Note:Most matrix programs use \(A'\) for \({A^{\bf{T}}}\).

Let \(X\) be \(m \times n\) data matrix such that \({X^T}X\) is invertible., and let \(M = {I_m} - X{\left( {{X^T}X} \right)^{ - {\bf{1}}}}{X^T}\). Add a column \({x_{\bf{0}}}\) to the data and form

\(W = \left[ {\begin{array}{*{20}{c}}X&{{x_{\bf{0}}}}\end{array}} \right]\)

Compute \({W^T}W\). The \(\left( {{\bf{1}},{\bf{1}}} \right)\) entry is \({X^T}X\). Show that the Schur complement (Exercise 15) of \({X^T}X\) can be written in the form \({\bf{x}}_{\bf{0}}^TM{{\bf{x}}_{\bf{0}}}\). It can be shown that the quantity \({\left( {{\bf{x}}_{\bf{0}}^TM{{\bf{x}}_{\bf{0}}}} \right)^{ - {\bf{1}}}}\) is the \(\left( {{\bf{2}},{\bf{2}}} \right)\)-entry in \({\left( {{W^T}W} \right)^{ - {\bf{1}}}}\). This entry has a useful statistical interpretation, under appropriate hypotheses.

In the study of engineering control of physical systems, a standard set of differential equations is transformed by Laplace transforms into the following system of linear equations:

\(\left[ {\begin{array}{*{20}{c}}{A - s{I_n}}&B\\C&{{I_m}}\end{array}} \right]\left[ {\begin{array}{*{20}{c}}{\bf{x}}\\{\bf{u}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{\bf{0}}\\{\bf{y}}\end{array}} \right]\)

Where \(A\) is \(n \times n\), \(B\) is \(n \times m\), \(C\) is \(m \times n\), and \(s\) is a variable. The vector \({\bf{u}}\) in \({\mathbb{R}^m}\) is the 鈥渋nput鈥� to the system, \({\bf{y}}\) in \({\mathbb{R}^m}\) is the 鈥渙utput鈥� and \({\bf{x}}\) in \({\mathbb{R}^n}\) is the 鈥渟tate鈥� vector. (Actually, the vectors \({\bf{x}}\), \({\bf{u}}\) and \({\bf{v}}\) are functions of \(s\), but we suppress this fact because it does not affect the algebraic calculations in Exercises 19 and 20.)
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