a. Answer [ 4 8 0 ⋮ − 8 − 3 − 5 − 1 ⋮ − 4 3 6 1 ⋮ − 2 ] [ 1 2 0 ⋮ − 2 − 3 − 5 − 1 ⋮ − 4 3 6 1 ⋮ − 2 ] [ 1 2 0 ⋮ − 2 0 1 0 ⋮ − 6 3 6 1 ⋮ − 2 ] [ 1 2 0 ⋮ − 2 0 1 0 ⋮ − 6 0 0 1 ⋮ 4 ] [ 1 0 0 ⋮ 10 0 1 0 ⋮ − 6 0 0 1 ⋮ 4 ] Explanation Given information: Perform the row operations to solve the augmented matrix. [ 7 13 1 ⋮ − 4 − 3 − 5 − 1 ⋮ − 4 3 6 1 ⋮ − 2 ] Add R 2 to R 1 . Multiply R 1 by 1 / 4 . Add R 3 to R 2 . Add − 3 times R 1 to R 3 . Add − 2 times R 2 to R 1 . Calculation: Add R 2 to R 1 . [ 7 − 3 13 − 5 1 − 1 ⋮ − 4 − 4 − 3 − 5 − 1 ⋮ − 4 3 6 1 ⋮ − 2 ] ⇒ [ 4 8 0 ⋮ − 8 − 3 − 5 − 1 ⋮ − 4 3 6 1 ⋮ − 2 ] Hence the resultant matrix is [ 4 8 0 ⋮ − 8 − 3 − 5 − 1 ⋮ − 4 3 6 1 ⋮ − 2 ] Multiply R 1 by 1 / 4 . [ 4 / 4 8 / 4 0 / 4 ⋮ − 8 / 4 − 3 − 5 − 1 ⋮ − 4 3 6 1 ⋮ − 2 ] ⇒ [ 1 2 0 ⋮ − 2 − 3 − 5 − 1 ⋮ − 4 3 6 1 ⋮ − 2 ] Hence, the resultant matrix is [ 1 2 0 ⋮ − 2 − 3 − 5 − 1 ⋮ − 4 3 6 1 ⋮ − 2 ] Add R 3 to R 2 . [ 1 2 0 ⋮ − 2 − 3 + 3 − 5 + 6 − 1 + 1 ⋮ − 4 − 2 3 6 1 ⋮ − 2 ] ⇒ [ 1 2 0 ⋮ − 2 0 1 0 ⋮ − 6 3 6 1 ⋮ − 2 ] Hence, the resultant matrix is [ 1 2 0 ⋮ − 2 0 1 0 ⋮ − 6 3 6 1 ⋮ − 2 ] Add − 3 times R 1 to R 3 . [ 1 2 0 ⋮ − 2 0 1 0 ⋮ − 6 3 − 3 6 − 6 1 − 0 ⋮ − 2 + 6 ] ⇒ [ 1 2 0 ⋮ − 2 0 1 0 ⋮ − 6 0 0 1 ⋮ 4 ] Hence, the resultant matrix is [ 1 2 0 ⋮ − 2 0 1 0 ⋮ − 6 0 0 1 ⋮ 4 ] Add − 2 times R 2 to R 1 . [ 1 − 0 2 − 2 0 − 0 ⋮ − 2 + 12 0 1 0 ⋮ − 6 0 0 1 ⋮ 4 ] ⇒ [ 1 0 0 ⋮ 10 0 1 0 ⋮ − 6 0 0 1 ⋮ 4 ] Hence, the resultant matrix is [ 1 0 0 ⋮ 10 0 1 0 ⋮ − 6 0 0 1 ⋮ 4 ] b. To determine Write and solve the system of linear equations. Answer x = 10 , y = − 6 , z = 4 Explanation Given information: Write and solve the system of linear equations (in variables x , y and z , if applicable) represented by the augmented matrix. [ 7 13 1 ⋮ − 4 − 3 − 5 − 1 ⋮ − 4 3 6 1 ⋮ − 2 ] Calculation: Solve the equations, 7 x + 13 y + z = − 4..... ( 1 ) − 3 x − 5 y − z = − 4...... ( 2 ) 3 x + 6 y + z = − 2..... ( 3 ) Eliminate x by adding equation ( 2 ) and ( 3 ) , − 3 x − 5 y − z = − 4 3 x + 6 y + z = − 2 y = − 6 Now eliminate z . 7 x + 13 y z = − 4 − 3 x − 5 y − z = − 4 4 x + 8 y = − 8 Substitute y = − 6 in the equation, 4 x + 8 y = − 8 4 x + 8 ( − 6 ) = − 8 4 x − 48 = − 8 4 x = 40 x = 10 Now put y = − 6 and x = 10 in equation ( 3 ) we get, 3 x + 6 y + z = − 2 3 ( 10 ) + 6 ( − 6 ) + z = − 2 30 − 36 + z = − 2 z = − 2 + 6 z = 4 Hence, the solution is x = 10 , y = − 6 , z = 4 . c. To determine Identify the best method. Answer Row opration method Explanation Given information: Compare the two solution methods. Which do you prefer? [ 7 13 1 ⋮ − 4 − 3 − 5 − 1 ⋮ − 4 3 6 1 ⋮ − 2 ] Calculation: Both the methods are best applicable, but row opration is the easy metod, Hence we prefer Row opration method .

4th Edition

ISBN: 9781337516853

Author: Larson

Publisher: CENGAGE CO

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Question

Chapter 8.1, Problem 40E

To determine

Perform the row operations to solve the augmented matrix.

Expert Solution

Answer to Problem 40E

[480⋮−8−3−5−1⋮−4361⋮−2]

[120⋮−2−3−5−1⋮−4361⋮−2]

[120⋮−2010⋮−6361⋮−2]

[120⋮−2010⋮−6001⋮4]

[100⋮10010⋮−6001⋮4]

Explanation of Solution

Given information:

Perform the row operations to solve the augmented matrix.

[7131⋮−4−3−5−1⋮−4361⋮−2]

Add R2 to R1 .

Multiply R1 by 1/4 .

Add R3 to R2 .

Add −3 times R1 to R3 .

Add −2 times R2 to R1 .

Calculation:

Add R2 to R1 .

[7−313−51−1⋮−4−4−3−5−1⋮−4361⋮−2]⇒[480⋮−8−3−5−1⋮−4361⋮−2]

Hence the resultant matrix is [480⋮−8−3−5−1⋮−4361⋮−2]

Multiply R1 by 1/4 .

[4/48/40/4⋮−8/4−3−5−1⋮−4361⋮−2]⇒[120⋮−2−3−5−1⋮−4361⋮−2]

Hence, the resultant matrix is [120⋮−2−3−5−1⋮−4361⋮−2]

Add R3 to R2 .

[120⋮−2−3+3−5+6−1+1⋮−4−2361⋮−2]⇒[120⋮−2010⋮−6361⋮−2]

Hence, the resultant matrix is [120⋮−2010⋮−6361⋮−2]

Add −3 times R1 to R3 .

[120⋮−2010⋮−63−36−61−0⋮−2+6] ⇒[120⋮−2010⋮−6001⋮4]

Hence, the resultant matrix is [120⋮−2010⋮−6001⋮4]

Add −2 times R2 to R1 .

[1−02−20−0⋮−2+12010⋮−6001⋮4] ⇒[100⋮10010⋮−6001⋮4]

Hence, the resultant matrix is [100⋮10010⋮−6001⋮4]

To determine

Write and solve the system of linear equations.

Expert Solution

Answer to Problem 40E

x=10,y=−6,z=4

Explanation of Solution

Given information:

Write and solve the system of linear equations (in variables x,y and z , if applicable) represented by the augmented matrix.

[7131⋮−4−3−5−1⋮−4361⋮−2]

Calculation:

Solve the equations,

7x+13y+z=−4.....(1)−3x−5y−z=−4......(2)3x+6y+z=−2.....(3)

Eliminate x by adding equation (2) and (3) ,

−3x−5y−z=−43x+6y+z=−2y=−6

Now eliminate z .

7x+13yz=−4−3x−5y−z=−44x+8y=−8

Substitute y=−6 in the equation, 4x+8y=−8

4x+8(−6)=−84x−48=−84x=40x=10

Now put y=−6 and x=10 in equation (3) we get,

3x+6y+z=−23(10)+6(−6)+z=−230−36+z=−2z=−2+6z=4

Hence, the solution is x=10,y=−6,z=4 .

To determine

Identify the best method.

Expert Solution

Answer to Problem 40E

Row opration method

Explanation of Solution

Given information:

Compare the two solution methods. Which do you prefer?

[7131⋮−4−3−5−1⋮−4361⋮−2]

Calculation:

Both the methods are best applicable, but row opration is the easy metod,

Hence we prefer Row opration method .

Chapter 8.1, Problem 39E

Chapter 8.1, Problem 41E

Chapter 8 Solutions

EBK PRECALCULUS W/LIMITS

Ch. 8.1 - Prob. 1E Ch. 8.1 - Prob. 2E Ch. 8.1 - Prob. 3E Ch. 8.1 - Prob. 4E Ch. 8.1 - Prob. 5E Ch. 8.1 - Prob. 6E Ch. 8.1 - Prob. 7E Ch. 8.1 - Prob. 8E Ch. 8.1 - Prob. 9E Ch. 8.1 - Prob. 10E

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ISBN:9781285741550
Author:James Stewart
Publisher:Cengage Learning
Thomas' Calculus (14th Edition)
Calculus
ISBN:9780134438986
Author:Joel R. Hass, Christopher E. Heil, Maurice D. Weir
Publisher:PEARSON
Calculus: Early Transcendentals (3rd Edition)
Calculus
ISBN:9780134763644
Author:William L. Briggs, Lyle Cochran, Bernard Gillett, Eric Schulz
Publisher:PEARSON
Calculus: Early Transcendentals
Calculus
ISBN:9781319050740
Author:Jon Rogawski, Colin Adams, Robert Franzosa
Publisher:W. H. Freeman
Precalculus
Calculus
ISBN:9780135189405
Author:Michael Sullivan
Publisher:PEARSON
Calculus: Early Transcendental Functions
Calculus
ISBN:9781337552516
Author:Ron Larson, Bruce H. Edwards
Publisher:Cengage Learning