In Exercises 1–4 the given matrix represents an augmented matrix for a linear system. Write the corresponding set of linear equations for the system, and use Gaussian elimination to solve the linear system. Introduce free parameters as necessary. [ 3 − 1 0 4 1 2 0 3 3 − 1 ]

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Answer 1

Textbook Question

Chapter 1, Problem 1SE

In Exercises 1–4 the given matrix represents an augmented matrix for a linear system. Write the corresponding set of linear equations for the system, and use Gaussian elimination to solve the linear system. Introduce free parameters as necessary.

[ 3 − 1 0 4 1 2 0 3 3 − 1 ]

Expert Solution & Answer

To determine

The solution of the system corresponding to the augmented matrix [3−10412033−1] by Gaussian elimination method.

Answer to Problem 1SE

The solution of the system corresponding to the augmented matrix [3−10412033−1] is x1=−12−32s−32t, x2=−52−92s−12t, x3=s,x4=t_.

Explanation of Solution

The given augmented matrix is [3−10412033−1].

Reduce the matrix [3−10412033−1] to row echelon form by Gaussian elimination method as follows.

Multiply the first row by 13 as shown below.

[1−13043132033−1]

Add −2 times the first row to the second row and obtain the reduced matrix as follows.

[1−1304313023313−53]

Multiply the second row of the matrix by 32 as shown below.

[1−1304313019212−52]

The above matrix is in row echelon form.

The corresponding linear system of the row echelon matrix is,

x1−13x2+43x4=13x2+92x3+12x4=−52

Here x3 and x4 are free variables.

Now assign arbitrary values s, t (called parameter) to x3 and x4 respectively. Therefore,

x2+92x3+12x4=−52x2=−52−92s−12t

Obtain the value of x1 as follows.

x1−13x2+43x4=13x1−13(−52−92s−12t)+43t=13x1+56+96s+16t+43t=13x1=−12−32s−32t

Therefore, the solution of the corresponding to the augmented matrix [3−10412033−1] is x1=−12−32s−32t, x2=−52−92s−12t, x3=s,x4=t_.

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Answer 2

Textbook Question

Chapter 1, Problem 1SE

In Exercises 1–4 the given matrix represents an augmented matrix for a linear system. Write the corresponding set of linear equations for the system, and use Gaussian elimination to solve the linear system. Introduce free parameters as necessary.

[ 3 − 1 0 4 1 2 0 3 3 − 1 ]

Expert Solution & Answer

To determine

The solution of the system corresponding to the augmented matrix [3−10412033−1] by Gaussian elimination method.

Answer to Problem 1SE

The solution of the system corresponding to the augmented matrix [3−10412033−1] is x1=−12−32s−32t, x2=−52−92s−12t, x3=s,x4=t_.

Explanation of Solution

The given augmented matrix is [3−10412033−1].

Reduce the matrix [3−10412033−1] to row echelon form by Gaussian elimination method as follows.

Multiply the first row by 13 as shown below.

[1−13043132033−1]

Add −2 times the first row to the second row and obtain the reduced matrix as follows.

[1−1304313023313−53]

Multiply the second row of the matrix by 32 as shown below.

[1−1304313019212−52]

The above matrix is in row echelon form.

The corresponding linear system of the row echelon matrix is,

x1−13x2+43x4=13x2+92x3+12x4=−52

Here x3 and x4 are free variables.

Now assign arbitrary values s, t (called parameter) to x3 and x4 respectively. Therefore,

x2+92x3+12x4=−52x2=−52−92s−12t

Obtain the value of x1 as follows.

x1−13x2+43x4=13x1−13(−52−92s−12t)+43t=13x1+56+96s+16t+43t=13x1=−12−32s−32t

Therefore, the solution of the corresponding to the augmented matrix [3−10412033−1] is x1=−12−32s−32t, x2=−52−92s−12t, x3=s,x4=t_.

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Answer 3

Textbook Question

Chapter 1, Problem 1SE

In Exercises 1–4 the given matrix represents an augmented matrix for a linear system. Write the corresponding set of linear equations for the system, and use Gaussian elimination to solve the linear system. Introduce free parameters as necessary.

[ 3 − 1 0 4 1 2 0 3 3 − 1 ]

Expert Solution & Answer

To determine

The solution of the system corresponding to the augmented matrix [3−10412033−1] by Gaussian elimination method.

Answer to Problem 1SE

The solution of the system corresponding to the augmented matrix [3−10412033−1] is x1=−12−32s−32t, x2=−52−92s−12t, x3=s,x4=t_.

Explanation of Solution

The given augmented matrix is [3−10412033−1].

Reduce the matrix [3−10412033−1] to row echelon form by Gaussian elimination method as follows.

Multiply the first row by 13 as shown below.

[1−13043132033−1]

Add −2 times the first row to the second row and obtain the reduced matrix as follows.

[1−1304313023313−53]

Multiply the second row of the matrix by 32 as shown below.

[1−1304313019212−52]

The above matrix is in row echelon form.

The corresponding linear system of the row echelon matrix is,

x1−13x2+43x4=13x2+92x3+12x4=−52

Here x3 and x4 are free variables.

Now assign arbitrary values s, t (called parameter) to x3 and x4 respectively. Therefore,

x2+92x3+12x4=−52x2=−52−92s−12t

Obtain the value of x1 as follows.

x1−13x2+43x4=13x1−13(−52−92s−12t)+43t=13x1+56+96s+16t+43t=13x1=−12−32s−32t

Therefore, the solution of the corresponding to the augmented matrix [3−10412033−1] is x1=−12−32s−32t, x2=−52−92s−12t, x3=s,x4=t_.

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Answer 4

Textbook Question

Chapter 1, Problem 1SE

In Exercises 1–4 the given matrix represents an augmented matrix for a linear system. Write the corresponding set of linear equations for the system, and use Gaussian elimination to solve the linear system. Introduce free parameters as necessary.

[ 3 − 1 0 4 1 2 0 3 3 − 1 ]

Expert Solution & Answer

To determine

The solution of the system corresponding to the augmented matrix [3−10412033−1] by Gaussian elimination method.

Answer to Problem 1SE

The solution of the system corresponding to the augmented matrix [3−10412033−1] is x1=−12−32s−32t, x2=−52−92s−12t, x3=s,x4=t_.

Explanation of Solution

The given augmented matrix is [3−10412033−1].

Reduce the matrix [3−10412033−1] to row echelon form by Gaussian elimination method as follows.

Multiply the first row by 13 as shown below.

[1−13043132033−1]

Add −2 times the first row to the second row and obtain the reduced matrix as follows.

[1−1304313023313−53]

Multiply the second row of the matrix by 32 as shown below.

[1−1304313019212−52]

The above matrix is in row echelon form.

The corresponding linear system of the row echelon matrix is,

x1−13x2+43x4=13x2+92x3+12x4=−52

Here x3 and x4 are free variables.

Now assign arbitrary values s, t (called parameter) to x3 and x4 respectively. Therefore,

x2+92x3+12x4=−52x2=−52−92s−12t

Obtain the value of x1 as follows.

x1−13x2+43x4=13x1−13(−52−92s−12t)+43t=13x1+56+96s+16t+43t=13x1=−12−32s−32t

Therefore, the solution of the corresponding to the augmented matrix [3−10412033−1] is x1=−12−32s−32t, x2=−52−92s−12t, x3=s,x4=t_.

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Answer 5

Textbook Question

Answer 6

Textbook Question

Answer 7

Expert Solution & Answer

To determine

The solution of the system corresponding to the augmented matrix [3−10412033−1] by Gaussian elimination method.

Answer to Problem 1SE

The solution of the system corresponding to the augmented matrix [3−10412033−1] is x1=−12−32s−32t, x2=−52−92s−12t, x3=s,x4=t_.

Explanation of Solution

The given augmented matrix is [3−10412033−1].

Reduce the matrix [3−10412033−1] to row echelon form by Gaussian elimination method as follows.

Multiply the first row by 13 as shown below.

[1−13043132033−1]

Add −2 times the first row to the second row and obtain the reduced matrix as follows.

[1−1304313023313−53]

Multiply the second row of the matrix by 32 as shown below.

[1−1304313019212−52]

The above matrix is in row echelon form.

The corresponding linear system of the row echelon matrix is,

x1−13x2+43x4=13x2+92x3+12x4=−52

Here x3 and x4 are free variables.

Now assign arbitrary values s, t (called parameter) to x3 and x4 respectively. Therefore,

x2+92x3+12x4=−52x2=−52−92s−12t

Obtain the value of x1 as follows.

x1−13x2+43x4=13x1−13(−52−92s−12t)+43t=13x1+56+96s+16t+43t=13x1=−12−32s−32t

Therefore, the solution of the corresponding to the augmented matrix [3−10412033−1] is x1=−12−32s−32t, x2=−52−92s−12t, x3=s,x4=t_.

Answer 8

Expert Solution & Answer

To determine

The solution of the system corresponding to the augmented matrix [3−10412033−1] by Gaussian elimination method.

Answer to Problem 1SE

The solution of the system corresponding to the augmented matrix [3−10412033−1] is x1=−12−32s−32t, x2=−52−92s−12t, x3=s,x4=t_.

Explanation of Solution

The given augmented matrix is [3−10412033−1].

Reduce the matrix [3−10412033−1] to row echelon form by Gaussian elimination method as follows.

Multiply the first row by 13 as shown below.

[1−13043132033−1]

Add −2 times the first row to the second row and obtain the reduced matrix as follows.

[1−1304313023313−53]

Multiply the second row of the matrix by 32 as shown below.

[1−1304313019212−52]

The above matrix is in row echelon form.

The corresponding linear system of the row echelon matrix is,

x1−13x2+43x4=13x2+92x3+12x4=−52

Here x3 and x4 are free variables.

Now assign arbitrary values s, t (called parameter) to x3 and x4 respectively. Therefore,

x2+92x3+12x4=−52x2=−52−92s−12t

Obtain the value of x1 as follows.

x1−13x2+43x4=13x1−13(−52−92s−12t)+43t=13x1+56+96s+16t+43t=13x1=−12−32s−32t

Therefore, the solution of the corresponding to the augmented matrix [3−10412033−1] is x1=−12−32s−32t, x2=−52−92s−12t, x3=s,x4=t_.

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

To determine

The solution of the system corresponding to the augmented matrix [3−10412033−1] by Gaussian elimination method.

Answer 12

Answer to Problem 1SE

The solution of the system corresponding to the augmented matrix [3−10412033−1] is x1=−12−32s−32t, x2=−52−92s−12t, x3=s,x4=t_.

Answer 13

Explanation of Solution

The given augmented matrix is [3−10412033−1].

Reduce the matrix [3−10412033−1] to row echelon form by Gaussian elimination method as follows.

Multiply the first row by 13 as shown below.

[1−13043132033−1]

Add −2 times the first row to the second row and obtain the reduced matrix as follows.

[1−1304313023313−53]

Multiply the second row of the matrix by 32 as shown below.

[1−1304313019212−52]

The above matrix is in row echelon form.

The corresponding linear system of the row echelon matrix is,

x1−13x2+43x4=13x2+92x3+12x4=−52

Here x3 and x4 are free variables.

Now assign arbitrary values s, t (called parameter) to x3 and x4 respectively. Therefore,

x2+92x3+12x4=−52x2=−52−92s−12t

Obtain the value of x1 as follows.

x1−13x2+43x4=13x1−13(−52−92s−12t)+43t=13x1+56+96s+16t+43t=13x1=−12−32s−32t

Therefore, the solution of the corresponding to the augmented matrix [3−10412033−1] is x1=−12−32s−32t, x2=−52−92s−12t, x3=s,x4=t_.

Answer 14

In Exercises 1–4 the given matrix represents an augmented matrix for a linear system. Write the corresponding set of linear equations for the system, and use Gaussian elimination to solve the linear system. Introduce free parameters as necessary. [ 3 − 1 0 4 1 2 0 3 3 − 1 ]

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