In Exercises 1-5, use matrices to find the complete solution to each system of equations, or show that none exists. { x + 2 y − 3 z = − 7 3 x − y + 2 z = 8 2 x − y + z = 5 }

Question

Want to see more full solutions like this?

Answer 1

Textbook Question

Chapter 6, Problem 1MCCP

In Exercises 1-5, use matrices to find the complete solution to each system of equations, or show that none exists.

{ x + 2 y − 3 z = − 7 3 x − y + 2 z = 8 2 x − y + z = 5 }

Expert Solution & Answer

To determine

The solution to the system of equations.

Answer to Problem 1MCCP

Solution: x=1,y=−1,z=2.

Explanation of Solution

Given: The system of equations:

x+2y−3z=−73x−y+2z=82x−y+z=5

Consider the system of equations:

x+2y−3z=−73x−y+2z=82x−y+z=5

The augmented matrix is:

[12−3−73−1282−115]

Now, let us perform the elementary row operations to find the echelon form of the matrix.

R2→R2+(−3)R1, R3→R3+(−2)R1 gives,

[12−3−70−711290−5719]

R2→1−7R2 gives,

[12−3−701−117−2970−5719]

R3→R3+5R2 gives,

[12−3−701−117−29700−67−127]

R3→−76R3 gives,

[12−3−701−117−2970012]

Now, the desired matrix is in row echelon form. Write the system of linear equations corresponding to the echelon form of matrix, that is:

x+2y−3z=−7(I)

y−117z=−297 (II)

z=2 (III)

Apply the back-substitution method:

Equation (III) gives,

z=2.

Substitute the value of z in equation (II)

y−117(2)=−297

Simplify,

y=−1.

Substitute the values of y and z in equation (I)

x+2(−1)−3(2)=−7

Simplify

x=2

Hence, x=1,y=−1,z=2

Conclusion: Therefore, x=1,y=−1,z=2.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Students have asked these similar questions

Solutions of inequalitie Google Classroom Mic Is (-3, 2) a solution of 7x+9y > -3? Choose 1 answer: A Yes B No Related content ▶6:06 Testing solutions to inequalities 2 of 4

Are natural logarithms used in real life ? How ? Can u give me two or three ways we can use them. Thanks

?

Answer 2

Textbook Question

Chapter 6, Problem 1MCCP

In Exercises 1-5, use matrices to find the complete solution to each system of equations, or show that none exists.

{ x + 2 y − 3 z = − 7 3 x − y + 2 z = 8 2 x − y + z = 5 }

Expert Solution & Answer

To determine

The solution to the system of equations.

Answer to Problem 1MCCP

Solution: x=1,y=−1,z=2.

Explanation of Solution

Given: The system of equations:

x+2y−3z=−73x−y+2z=82x−y+z=5

Consider the system of equations:

x+2y−3z=−73x−y+2z=82x−y+z=5

The augmented matrix is:

[12−3−73−1282−115]

Now, let us perform the elementary row operations to find the echelon form of the matrix.

R2→R2+(−3)R1, R3→R3+(−2)R1 gives,

[12−3−70−711290−5719]

R2→1−7R2 gives,

[12−3−701−117−2970−5719]

R3→R3+5R2 gives,

[12−3−701−117−29700−67−127]

R3→−76R3 gives,

[12−3−701−117−2970012]

Now, the desired matrix is in row echelon form. Write the system of linear equations corresponding to the echelon form of matrix, that is:

x+2y−3z=−7(I)

y−117z=−297 (II)

z=2 (III)

Apply the back-substitution method:

Equation (III) gives,

z=2.

Substitute the value of z in equation (II)

y−117(2)=−297

Simplify,

y=−1.

Substitute the values of y and z in equation (I)

x+2(−1)−3(2)=−7

Simplify

x=2

Hence, x=1,y=−1,z=2

Conclusion: Therefore, x=1,y=−1,z=2.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 3

Textbook Question

Chapter 6, Problem 1MCCP

In Exercises 1-5, use matrices to find the complete solution to each system of equations, or show that none exists.

{ x + 2 y − 3 z = − 7 3 x − y + 2 z = 8 2 x − y + z = 5 }

Expert Solution & Answer

To determine

The solution to the system of equations.

Answer to Problem 1MCCP

Solution: x=1,y=−1,z=2.

Explanation of Solution

Given: The system of equations:

x+2y−3z=−73x−y+2z=82x−y+z=5

Consider the system of equations:

x+2y−3z=−73x−y+2z=82x−y+z=5

The augmented matrix is:

[12−3−73−1282−115]

Now, let us perform the elementary row operations to find the echelon form of the matrix.

R2→R2+(−3)R1, R3→R3+(−2)R1 gives,

[12−3−70−711290−5719]

R2→1−7R2 gives,

[12−3−701−117−2970−5719]

R3→R3+5R2 gives,

[12−3−701−117−29700−67−127]

R3→−76R3 gives,

[12−3−701−117−2970012]

Now, the desired matrix is in row echelon form. Write the system of linear equations corresponding to the echelon form of matrix, that is:

x+2y−3z=−7(I)

y−117z=−297 (II)

z=2 (III)

Apply the back-substitution method:

Equation (III) gives,

z=2.

Substitute the value of z in equation (II)

y−117(2)=−297

Simplify,

y=−1.

Substitute the values of y and z in equation (I)

x+2(−1)−3(2)=−7

Simplify

x=2

Hence, x=1,y=−1,z=2

Conclusion: Therefore, x=1,y=−1,z=2.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 4

Textbook Question

Chapter 6, Problem 1MCCP

In Exercises 1-5, use matrices to find the complete solution to each system of equations, or show that none exists.

{ x + 2 y − 3 z = − 7 3 x − y + 2 z = 8 2 x − y + z = 5 }

Expert Solution & Answer

To determine

The solution to the system of equations.

Answer to Problem 1MCCP

Solution: x=1,y=−1,z=2.

Explanation of Solution

Given: The system of equations:

x+2y−3z=−73x−y+2z=82x−y+z=5

Consider the system of equations:

x+2y−3z=−73x−y+2z=82x−y+z=5

The augmented matrix is:

[12−3−73−1282−115]

Now, let us perform the elementary row operations to find the echelon form of the matrix.

R2→R2+(−3)R1, R3→R3+(−2)R1 gives,

[12−3−70−711290−5719]

R2→1−7R2 gives,

[12−3−701−117−2970−5719]

R3→R3+5R2 gives,

[12−3−701−117−29700−67−127]

R3→−76R3 gives,

[12−3−701−117−2970012]

Now, the desired matrix is in row echelon form. Write the system of linear equations corresponding to the echelon form of matrix, that is:

x+2y−3z=−7(I)

y−117z=−297 (II)

z=2 (III)

Apply the back-substitution method:

Equation (III) gives,

z=2.

Substitute the value of z in equation (II)

y−117(2)=−297

Simplify,

y=−1.

Substitute the values of y and z in equation (I)

x+2(−1)−3(2)=−7

Simplify

x=2

Hence, x=1,y=−1,z=2

Conclusion: Therefore, x=1,y=−1,z=2.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 5

Textbook Question

Answer 6

Textbook Question

Answer 7

Expert Solution & Answer

To determine

The solution to the system of equations.

Answer to Problem 1MCCP

Solution: x=1,y=−1,z=2.

Explanation of Solution

Given: The system of equations:

x+2y−3z=−73x−y+2z=82x−y+z=5

Consider the system of equations:

x+2y−3z=−73x−y+2z=82x−y+z=5

The augmented matrix is:

[12−3−73−1282−115]

Now, let us perform the elementary row operations to find the echelon form of the matrix.

R2→R2+(−3)R1, R3→R3+(−2)R1 gives,

[12−3−70−711290−5719]

R2→1−7R2 gives,

[12−3−701−117−2970−5719]

R3→R3+5R2 gives,

[12−3−701−117−29700−67−127]

R3→−76R3 gives,

[12−3−701−117−2970012]

Now, the desired matrix is in row echelon form. Write the system of linear equations corresponding to the echelon form of matrix, that is:

x+2y−3z=−7(I)

y−117z=−297 (II)

z=2 (III)

Apply the back-substitution method:

Equation (III) gives,

z=2.

Substitute the value of z in equation (II)

y−117(2)=−297

Simplify,

y=−1.

Substitute the values of y and z in equation (I)

x+2(−1)−3(2)=−7

Simplify

x=2

Hence, x=1,y=−1,z=2

Conclusion: Therefore, x=1,y=−1,z=2.

Answer 8

Expert Solution & Answer

To determine

The solution to the system of equations.

Answer to Problem 1MCCP

Solution: x=1,y=−1,z=2.

Explanation of Solution

Given: The system of equations:

x+2y−3z=−73x−y+2z=82x−y+z=5

Consider the system of equations:

x+2y−3z=−73x−y+2z=82x−y+z=5

The augmented matrix is:

[12−3−73−1282−115]

Now, let us perform the elementary row operations to find the echelon form of the matrix.

R2→R2+(−3)R1, R3→R3+(−2)R1 gives,

[12−3−70−711290−5719]

R2→1−7R2 gives,

[12−3−701−117−2970−5719]

R3→R3+5R2 gives,

[12−3−701−117−29700−67−127]

R3→−76R3 gives,

[12−3−701−117−2970012]

Now, the desired matrix is in row echelon form. Write the system of linear equations corresponding to the echelon form of matrix, that is:

x+2y−3z=−7(I)

y−117z=−297 (II)

z=2 (III)

Apply the back-substitution method:

Equation (III) gives,

z=2.

Substitute the value of z in equation (II)

y−117(2)=−297

Simplify,

y=−1.

Substitute the values of y and z in equation (I)

x+2(−1)−3(2)=−7

Simplify

x=2

Hence, x=1,y=−1,z=2

Conclusion: Therefore, x=1,y=−1,z=2.

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

To determine

The solution to the system of equations.

Answer 12

Answer to Problem 1MCCP

Solution: x=1,y=−1,z=2.

Answer 13

Explanation of Solution

Given: The system of equations:

x+2y−3z=−73x−y+2z=82x−y+z=5

Consider the system of equations:

x+2y−3z=−73x−y+2z=82x−y+z=5

The augmented matrix is:

[12−3−73−1282−115]

Now, let us perform the elementary row operations to find the echelon form of the matrix.

R2→R2+(−3)R1, R3→R3+(−2)R1 gives,

[12−3−70−711290−5719]

R2→1−7R2 gives,

[12−3−701−117−2970−5719]

R3→R3+5R2 gives,

[12−3−701−117−29700−67−127]

R3→−76R3 gives,

[12−3−701−117−2970012]

Now, the desired matrix is in row echelon form. Write the system of linear equations corresponding to the echelon form of matrix, that is:

x+2y−3z=−7(I)

y−117z=−297 (II)

z=2 (III)

Apply the back-substitution method:

Equation (III) gives,

z=2.

Substitute the value of z in equation (II)

y−117(2)=−297

Simplify,

y=−1.

Substitute the values of y and z in equation (I)

x+2(−1)−3(2)=−7

Simplify

x=2

Hence, x=1,y=−1,z=2

Conclusion: Therefore, x=1,y=−1,z=2.

Answer 14

Ch. 6.1 - A rectangular array of numbers, arranged in rows...Ch. 6.1 - Consider the matrix [ 111342211|246 ]. We can...Ch. 6.1 - Prob. 3CVC Ch. 6.1 - Using Gauss-Jordan diminution to solve the system...Ch. 6.1 - True or false: Back-substitution is required to...Ch. 6.1 - True or false: Back-substitution is required to...Ch. 6.1 - Prob. 1E Ch. 6.1 - In Exercises 18, write the augmented matrix for...Ch. 6.1 - In Exercises 1-8, write the augmented matrix for...Ch. 6.1 - In Exercises 1-8, write the augmented matrix for...

In Exercises 1-5, use matrices to find the complete solution to each system of equations, or show that none exists. { x + 2 y − 3 z = − 7 3 x − y + 2 z = 8 2 x − y + z = 5 }

Concept explainers

Videos

Answer to Problem 1MCCP

Explanation of Solution

Want to see more full solutions like this?

Chapter 6 Solutions