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Q6E

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Linear Algebra With Applications
Found in: Page 18
Linear Algebra With Applications

Linear Algebra With Applications

Book edition 5th
Author(s) Otto Bretscher
Pages 442 pages
ISBN 9780321796974

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Short Answer

In Exercises 1 through 12, find all solutions of the equations with paper and pencil using Gauss–Jordan elimination. Show all your work.

|x1-7x2+x5=3x3-2x5=2x4+x5=1|

For the given system of equation x5,x5.are free variables and

x3=2+2x5,x4=1-x5,=3+7x2-x5.

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :
TABLE OF CONTENTS

Step 1: Write the augmented matrix

First of all we will make the augmented matrix to the corresponding system of equations.

x1-7x2+x5=3x3-2x5=2x4+x5=1

The corresponding augmented matrix is written as:

1-70010010-200011321

Step 2: Solving the Augmented matrix

The above augmented matrix obtained above is already in reduced form, then the

System of equation to the corresponding above matrix will be

x1=3+7x2-x5x3=2+2x5x4=1-x5

Step 3: Finding free variables

According to system x1=3+7x2-x5x3=2+2x5x4=1-x5, the variables x3,x4,x1 depend on x2,x5. So, x2,x5.are free variables.

Step 4: Final answer

Thus, for the given system of equation x2,x5 are free variables and x3=2+2x5,x4 =1-x5x1 =3+7x2-x5.

Most popular questions for Math Textbooks

in exercises 1 through 10, find all solutions of the linear systems using elimination.Then check your solutions.

6. x+2y+3z=8x+3y+3z=10x+2y+4z=9

Find the polynomial f(t) of degree 3 such that f(1)=(1),f(2)=5,f'(1)=2,and f'(2)=9 , where f'(t) is the derivative of f(t) . Graph this polynomial.

Find the least square solutions of the system Ax⇀=b⇀ where

A=[1110-100010-10] and b⇀=[110-1010-10].

In exercises, 1 through 10, find all solutions of the linear systems using elimination.Then check your solutions.

8.x+2y+3z=04x+5y+6z=07x+8y+10z=0

Consider a solutionx1→of the linear systemAx→=b→. Justify the facts stated in parts (a) and (b):

a. Ifx→his a solution of the systemAx→=0→, thenx1→+xh→ is a solution of the system A=x→=b→.

b. Ifx2→is another solution of the systemAx→=b→, thenx1→+xh→is a solution of the system Ax→+0→.

c. Now suppose A is a2×2matrix. A solution vectorx1→of the systemAx→+b→is shown in the accompanying figure. We are told that the solutions of the systemAx→=0→form the line shown in the sketch. Draw the line consisting of all solutions of the systemAx→=b→.

If you are puzzled by the generality of this problem, think about an example first:

A=(1    23    6),b→=[39]andx1→=[11]
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