141Notes Pivoting of a matrix Yox.vn Homework 05 Math 141 e chegg Study Guided S

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141Notes Pivoting of a matrix Yox.vn Homework 05 Math 141 e chegg Study Guided Sc e141Notes OPivotingOf x × . Secure https://www.webassign.net/web/Student/Assignment-Responses/submit?dep-17964107 : Apps @ Howdy My eCampus DTV D Nursing Twerk eaming Catalytics Math 141 OChen AA STAT HW Other bookmarks 6 -1.11 polnts oi6 Submisslons Used My Notes Solve the following system of equations. (Enter your answers as a comma-separated list. If there are infinitely many solutions, enter a parametric solution using t and/or s If there is no solution, enter NONE.) X1+2X2 + 6x3 = 8 X1 + X2 + 3x3 = 4 x1 , x2,X3) = 7. 0/1.11 points| Previous Answers 2/6 Submissions Used My Notes Solve the following system of equations. (Enter your answers as a comma-separated list. If there are infinitely many solutions, enter a parametric solution using t an If there is no solution, enter NONE.) calcPad - Operations Functions X1-2x2 + x3 =-1 2x1 + x2-2x3 = x1 + 3x2-3x3 = VI!Vectors TI Greek hele Subrrit Answer 8. 1.11 points 0i6 Submissions Used 11308 PM Type here to search 2/1/2018

Explanation / Answer

6. The augmented matrix for the given linear system is A =

1

2

6

8

1

1

3

4

To solve the given linear system of equations, we will reduce A to its RREF as under:

Add -1 times the 1st row to the 2nd row

Multiply the 2nd row by -1

Add -2 times the 2nd row to the 1st row

Then the RREF of A is

1

0

0

0

0

1

3

4

Thus, the given linear system of equations is equivalent to x1 = 0 and x2+3x3= 4 or, x2 = 4-3x3. Hence (x1,x2,x3) = (0,4-3x3,x3) = (0,4,0)+t(0,-3,1), where x3 = t is an arbitrary real number.

7. The augmented matrix for the given linear system is A =

1

-2

1

-1

2

1

-2

2

1

3

-3

3

To solve the given linear system of equations, we will reduce A to its RREF as under:

Add -2 times the 1st row to the 2nd row

Add -1 times the 1st row to the 3rd row

Multiply the 2nd row by 1/5

Add -5 times the 2nd row to the 3rd row

Add 2 times the 2nd row to the 1st row

Then the RREF of A is

1

0

-3/5

3/5

0

1

-4/5

4/5

0

0

0

0

Thus, the given linear system of equations is equivalent to x1-(3/5)x3 = 3/5 or, x1 =3/5 +3x3/5 and x2 –(4/5)x3 = 4/5 or, x2 =4/5 +4x3/5. Then (x1,x2,x3) = (3/5 +3x3/5, 4/5 +4x3/5,x3)= 1/5(3,4,0)+t(3,4,5) where x3 = 5t is an arbitrary real number.

Note:

A reduced row echelon calculator is needed here. If is available on line. If TI-84 has this feature, you can use it to reduce the augmented matrix to its RREF.

1

2

6

8

1

1

3

4

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