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Linear Algebra and its Applications
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Linear Algebra and its Applications

Linear Algebra and its Applications

Book edition 5th
Author(s) David C. Lay, Steven R. Lay and Judi J. McDonald
Pages 483 pages
ISBN 978-03219822384

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Short Answer

The solutions \(\left( {x,y,z} \right)\) of a single linear equation \(ax + by + cz = d\)

form a plane in \({\mathbb{R}^3}\) when a, b, and c are not all zero. Construct sets of three linear equations whose graphs (a) intersect in a single line, (b) intersect in a single point, and (c) have no points in common. Typical graphs are illustrated in the figure.

Three planes intersecting in a line.

(a)

Three planes intersecting in a point.

(b)

Three planes with no intersection.

(c)

Three planes with no intersection.

(c’)

(a) The echelon form of the consistent linear system is , , or .

(b) The echelon form of the consistent linear system is the identity matrix of the order \(3 \times 3\).

(c) The inconsistent linear system of three variables and equations has no common point.

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Step by Step Solution

TABLE OF CONTENTS :
TABLE OF CONTENTS

(a) Step 1: Write the condition when the graphs intersect on a single line

Each point on the line is a solution to the given system of equations. And the solution set is infinite if the three planes cross at a single point. As a result, there must be two pivot components in the possible forms.

The echelon form of the consistent linear system is shown below:


Or,

Or,

Here, is the leading entry, and \(\left( * \right)\) can have any value, including 0.

(b) Step 2: Write the condition w\(3 \times 3\)hen the graphs intersect at a single point

The system of three equations is fulfilled if the three planes cross at a single location. As a consequence, the system is consistent, and it offers a unique solution. The echelon form that may be produced by solving this system of equations is an identity matrix of the order .

Thus, the echelon form of the consistent linear system is an identity matrix of the order \(3 \times 3\).

(c) Step 3: Write the condition when the graphs have no point in common

If there is no common point between the planes, their intersection is not a unique line or point. As a result, there is no way to solve it.

Thus, the inconsistent linear system of three variables and equations has no common point.

Most popular questions for Math Textbooks

Suppose A is an \(n \times n\) matrix with the property that the equation \(A{\mathop{\rm x}\nolimits} = 0\) has at least one solution for each b in \({\mathbb{R}^n}\). Without using Theorem 5 or 8, explain why each equation Ax = b has in fact exactly one solution.

Write the vector \(\left( {\begin{array}{*{20}{c}}5\\6\end{array}} \right)\) as the sum of two vectors, one on the line \(\left\{ {\left( {x,y} \right):y = {\bf{2}}x} \right\}\) and one on the line \(\left\{ {\left( {x,y} \right):y = x/{\bf{2}}} \right\}\).

Consider the problem of determining whether the following system of equations is consistent for all \({b_1},{b_2},{b_3}\):

\(\begin{aligned}{c}{\bf{2}}{x_1} - {\bf{4}}{x_2} - {\bf{2}}{x_3} = {b_1}\\ - {\bf{5}}{x_1} + {x_2} + {x_3} = {b_2}\\{\bf{7}}{x_1} - {\bf{5}}{x_2} - {\bf{3}}{x_3} = {b_3}\end{aligned}\)

  1. Define appropriate vectors, and restate the problem in terms of Span \(\left\{ {{{\bf{v}}_1},{{\bf{v}}_2},{{\bf{v}}_3}} \right\}\). Then solve that problem.

  1. Define an appropriate matrix, and restate the problem using the phrase “columns of A.”

  1. Define an appropriate linear transformation T using the matrix in (b), and restate the problem in terms of T.

In Exercises 10, write a vector equation that is equivalent to the given system of equations.

10. \(4{x_1} + {x_2} + 3{x_3} = 9\)

\(\begin{array}{c}{x_1} - 7{x_2} - 2{x_3} = 2\\8{x_1} + 6{x_2} - 5{x_3} = 15\end{array}\)

Let T be a linear transformation that maps \({\mathbb{R}^n}\) onto \({\mathbb{R}^n}\). Is \({T^{ - 1}}\) also one-to-one?

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