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 }

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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.

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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.

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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.

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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.

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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

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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 }

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Answer to Problem 1MCCP

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