In Problems 1-4 , find a general solution for the system x ′ ( t ) = A x ( t ) , where A is given. A = [ 6 − 3 2 1 ]

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Answer 1

Textbook Question

Chapter 9.RP, Problem 1RP

In Problems 1-4, find a general solution for the system $x ′ ( t ) = A x ( t )$ , where $A$ is given.

$A = [ 6 − 3 2 1 ]$

Expert Solution & Answer

To determine

To solve: The system $x′(t)=Ax(t)$ , where $A=[6−321]$ .

Answer to Problem 1RP

Solution:

The solution of system $x′(t)=Ax(t)$ , where $A=[6−321]$ is $X(t)=c1e3t[11]+c2e4t[32]$ .

Explanation of Solution

Formula used:

The procedure to solve the $x′(t)=Ax(t)$ :

Step-1: Apply the condition $|A−λI|=0$ for matrix $A$ in order to calculate eigenvalues:

Step-2:Apply the $|A−λI|X=0$ in order to calculate eigenvectors $u1$ and $u2$ corresponding to eigenvalue $λ1$ and $λ2$ respectively:

Step-3:Write the general solution of system $x′(t)=Ax(t)$ as:

$X(t)=c1eλ1tu1+c2eλ2tu2$

Calculation:

Consider the system $x′(t)=Ax(t)$ ,

$A=[6−321]$

Apply the condition $|A−λI|=0$ for matrix $A$ in order to calculate eigenvalues:

$|[6−321]−λ[1001]|=0|6−λ−321−λ|=0$

Expand the above determinant:

$(6−λ)(1−λ)+6=06−λ−6λ+λ2+6=06−7λ+λ2+6=0λ2−7λ+12=0$

Factories the above quadratic equation:

$λ2−4λ−3λ+12=0λ(λ−4)−3(λ−3)=0(λ−4)(λ−3)=0$

Equate first factor to 0:

$(λ−4)=0λ=4$

Equate second factor to 0:

$(λ−3)=0 λ=3$

Apply the $|A−λI|X=0$ in order to calculate eigenvector corresponding to eigenvalue $λ=4$ :

$[6−4−321−4][x1x2]=[00][2−32−3][x1x2]=[00]$

Write above matrix notation in scalar form notation:

$2x1−3x2=02x1−3x2=0$

The eigenvector corresponding to eigenvalue $λ=4$ :

$u1=[32]$

Apply the $|A−λI|X=0$ in order to calculate eigenvector corresponding to eigenvalue $λ=3$ :

$[6−3−321−3][x1x2]=[00][3−32−2][x1x2]=[00]$

Write above matrix notation in scalar form notation:

$3x1−3x2=02x1−2x2=0$

The eigenvector corresponding to eigenvalue $λ=3$ :

$u2=[11]$

The general solution of system corresponding to eigenvector $u1$ and $u2$ :

$X(t)=c1e3t[11]+c2e4t[32]$

Conclusion: Therefore, the solution of system $x′(t)=Ax(t)$ , where $A=[6−321]$ is $X(t)=c1e3t[11]+c2e4t[32]$ .

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f E and F are disjoint events, P(E and F) =

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Answer 2

Textbook Question

Chapter 9.RP, Problem 1RP

In Problems 1-4, find a general solution for the system $x ′ ( t ) = A x ( t )$ , where $A$ is given.

$A = [ 6 − 3 2 1 ]$

Expert Solution & Answer

To determine

To solve: The system $x′(t)=Ax(t)$ , where $A=[6−321]$ .

Answer to Problem 1RP

Solution:

The solution of system $x′(t)=Ax(t)$ , where $A=[6−321]$ is $X(t)=c1e3t[11]+c2e4t[32]$ .

Explanation of Solution

Formula used:

The procedure to solve the $x′(t)=Ax(t)$ :

Step-1: Apply the condition $|A−λI|=0$ for matrix $A$ in order to calculate eigenvalues:

Step-2:Apply the $|A−λI|X=0$ in order to calculate eigenvectors $u1$ and $u2$ corresponding to eigenvalue $λ1$ and $λ2$ respectively:

Step-3:Write the general solution of system $x′(t)=Ax(t)$ as:

$X(t)=c1eλ1tu1+c2eλ2tu2$

Calculation:

Consider the system $x′(t)=Ax(t)$ ,

$A=[6−321]$

Apply the condition $|A−λI|=0$ for matrix $A$ in order to calculate eigenvalues:

$|[6−321]−λ[1001]|=0|6−λ−321−λ|=0$

Expand the above determinant:

$(6−λ)(1−λ)+6=06−λ−6λ+λ2+6=06−7λ+λ2+6=0λ2−7λ+12=0$

Factories the above quadratic equation:

$λ2−4λ−3λ+12=0λ(λ−4)−3(λ−3)=0(λ−4)(λ−3)=0$

Equate first factor to 0:

$(λ−4)=0λ=4$

Equate second factor to 0:

$(λ−3)=0 λ=3$

Apply the $|A−λI|X=0$ in order to calculate eigenvector corresponding to eigenvalue $λ=4$ :

$[6−4−321−4][x1x2]=[00][2−32−3][x1x2]=[00]$

Write above matrix notation in scalar form notation:

$2x1−3x2=02x1−3x2=0$

The eigenvector corresponding to eigenvalue $λ=4$ :

$u1=[32]$

Apply the $|A−λI|X=0$ in order to calculate eigenvector corresponding to eigenvalue $λ=3$ :

$[6−3−321−3][x1x2]=[00][3−32−2][x1x2]=[00]$

Write above matrix notation in scalar form notation:

$3x1−3x2=02x1−2x2=0$

The eigenvector corresponding to eigenvalue $λ=3$ :

$u2=[11]$

The general solution of system corresponding to eigenvector $u1$ and $u2$ :

$X(t)=c1e3t[11]+c2e4t[32]$

Conclusion: Therefore, the solution of system $x′(t)=Ax(t)$ , where $A=[6−321]$ is $X(t)=c1e3t[11]+c2e4t[32]$ .

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Answer 3

Textbook Question

Chapter 9.RP, Problem 1RP

In Problems 1-4, find a general solution for the system $x ′ ( t ) = A x ( t )$ , where $A$ is given.

$A = [ 6 − 3 2 1 ]$

Expert Solution & Answer

To determine

To solve: The system $x′(t)=Ax(t)$ , where $A=[6−321]$ .

Answer to Problem 1RP

Solution:

The solution of system $x′(t)=Ax(t)$ , where $A=[6−321]$ is $X(t)=c1e3t[11]+c2e4t[32]$ .

Explanation of Solution

Formula used:

The procedure to solve the $x′(t)=Ax(t)$ :

Step-1: Apply the condition $|A−λI|=0$ for matrix $A$ in order to calculate eigenvalues:

Step-2:Apply the $|A−λI|X=0$ in order to calculate eigenvectors $u1$ and $u2$ corresponding to eigenvalue $λ1$ and $λ2$ respectively:

Step-3:Write the general solution of system $x′(t)=Ax(t)$ as:

$X(t)=c1eλ1tu1+c2eλ2tu2$

Calculation:

Consider the system $x′(t)=Ax(t)$ ,

$A=[6−321]$

Apply the condition $|A−λI|=0$ for matrix $A$ in order to calculate eigenvalues:

$|[6−321]−λ[1001]|=0|6−λ−321−λ|=0$

Expand the above determinant:

$(6−λ)(1−λ)+6=06−λ−6λ+λ2+6=06−7λ+λ2+6=0λ2−7λ+12=0$

Factories the above quadratic equation:

$λ2−4λ−3λ+12=0λ(λ−4)−3(λ−3)=0(λ−4)(λ−3)=0$

Equate first factor to 0:

$(λ−4)=0λ=4$

Equate second factor to 0:

$(λ−3)=0 λ=3$

Apply the $|A−λI|X=0$ in order to calculate eigenvector corresponding to eigenvalue $λ=4$ :

$[6−4−321−4][x1x2]=[00][2−32−3][x1x2]=[00]$

Write above matrix notation in scalar form notation:

$2x1−3x2=02x1−3x2=0$

The eigenvector corresponding to eigenvalue $λ=4$ :

$u1=[32]$

Apply the $|A−λI|X=0$ in order to calculate eigenvector corresponding to eigenvalue $λ=3$ :

$[6−3−321−3][x1x2]=[00][3−32−2][x1x2]=[00]$

Write above matrix notation in scalar form notation:

$3x1−3x2=02x1−2x2=0$

The eigenvector corresponding to eigenvalue $λ=3$ :

$u2=[11]$

The general solution of system corresponding to eigenvector $u1$ and $u2$ :

$X(t)=c1e3t[11]+c2e4t[32]$

Conclusion: Therefore, the solution of system $x′(t)=Ax(t)$ , where $A=[6−321]$ is $X(t)=c1e3t[11]+c2e4t[32]$ .

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Answer 4

Textbook Question

Chapter 9.RP, Problem 1RP

In Problems 1-4, find a general solution for the system $x ′ ( t ) = A x ( t )$ , where $A$ is given.

$A = [ 6 − 3 2 1 ]$

Expert Solution & Answer

To determine

To solve: The system $x′(t)=Ax(t)$ , where $A=[6−321]$ .

Answer to Problem 1RP

Solution:

The solution of system $x′(t)=Ax(t)$ , where $A=[6−321]$ is $X(t)=c1e3t[11]+c2e4t[32]$ .

Explanation of Solution

Formula used:

The procedure to solve the $x′(t)=Ax(t)$ :

Step-1: Apply the condition $|A−λI|=0$ for matrix $A$ in order to calculate eigenvalues:

Step-2:Apply the $|A−λI|X=0$ in order to calculate eigenvectors $u1$ and $u2$ corresponding to eigenvalue $λ1$ and $λ2$ respectively:

Step-3:Write the general solution of system $x′(t)=Ax(t)$ as:

$X(t)=c1eλ1tu1+c2eλ2tu2$

Calculation:

Consider the system $x′(t)=Ax(t)$ ,

$A=[6−321]$

Apply the condition $|A−λI|=0$ for matrix $A$ in order to calculate eigenvalues:

$|[6−321]−λ[1001]|=0|6−λ−321−λ|=0$

Expand the above determinant:

$(6−λ)(1−λ)+6=06−λ−6λ+λ2+6=06−7λ+λ2+6=0λ2−7λ+12=0$

Factories the above quadratic equation:

$λ2−4λ−3λ+12=0λ(λ−4)−3(λ−3)=0(λ−4)(λ−3)=0$

Equate first factor to 0:

$(λ−4)=0λ=4$

Equate second factor to 0:

$(λ−3)=0 λ=3$

Apply the $|A−λI|X=0$ in order to calculate eigenvector corresponding to eigenvalue $λ=4$ :

$[6−4−321−4][x1x2]=[00][2−32−3][x1x2]=[00]$

Write above matrix notation in scalar form notation:

$2x1−3x2=02x1−3x2=0$

The eigenvector corresponding to eigenvalue $λ=4$ :

$u1=[32]$

Apply the $|A−λI|X=0$ in order to calculate eigenvector corresponding to eigenvalue $λ=3$ :

$[6−3−321−3][x1x2]=[00][3−32−2][x1x2]=[00]$

Write above matrix notation in scalar form notation:

$3x1−3x2=02x1−2x2=0$

The eigenvector corresponding to eigenvalue $λ=3$ :

$u2=[11]$

The general solution of system corresponding to eigenvector $u1$ and $u2$ :

$X(t)=c1e3t[11]+c2e4t[32]$

Conclusion: Therefore, the solution of system $x′(t)=Ax(t)$ , where $A=[6−321]$ is $X(t)=c1e3t[11]+c2e4t[32]$ .

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Answer 5

Textbook Question

Answer 6

Textbook Question

Answer 7

Expert Solution & Answer

To determine

To solve: The system $x′(t)=Ax(t)$ , where $A=[6−321]$ .

Answer to Problem 1RP

Solution:

The solution of system $x′(t)=Ax(t)$ , where $A=[6−321]$ is $X(t)=c1e3t[11]+c2e4t[32]$ .

Explanation of Solution

Formula used:

The procedure to solve the $x′(t)=Ax(t)$ :

Step-1: Apply the condition $|A−λI|=0$ for matrix $A$ in order to calculate eigenvalues:

Step-2:Apply the $|A−λI|X=0$ in order to calculate eigenvectors $u1$ and $u2$ corresponding to eigenvalue $λ1$ and $λ2$ respectively:

Step-3:Write the general solution of system $x′(t)=Ax(t)$ as:

$X(t)=c1eλ1tu1+c2eλ2tu2$

Calculation:

Consider the system $x′(t)=Ax(t)$ ,

$A=[6−321]$

Apply the condition $|A−λI|=0$ for matrix $A$ in order to calculate eigenvalues:

$|[6−321]−λ[1001]|=0|6−λ−321−λ|=0$

Expand the above determinant:

$(6−λ)(1−λ)+6=06−λ−6λ+λ2+6=06−7λ+λ2+6=0λ2−7λ+12=0$

Factories the above quadratic equation:

$λ2−4λ−3λ+12=0λ(λ−4)−3(λ−3)=0(λ−4)(λ−3)=0$

Equate first factor to 0:

$(λ−4)=0λ=4$

Equate second factor to 0:

$(λ−3)=0 λ=3$

Apply the $|A−λI|X=0$ in order to calculate eigenvector corresponding to eigenvalue $λ=4$ :

$[6−4−321−4][x1x2]=[00][2−32−3][x1x2]=[00]$

Write above matrix notation in scalar form notation:

$2x1−3x2=02x1−3x2=0$

The eigenvector corresponding to eigenvalue $λ=4$ :

$u1=[32]$

Apply the $|A−λI|X=0$ in order to calculate eigenvector corresponding to eigenvalue $λ=3$ :

$[6−3−321−3][x1x2]=[00][3−32−2][x1x2]=[00]$

Write above matrix notation in scalar form notation:

$3x1−3x2=02x1−2x2=0$

The eigenvector corresponding to eigenvalue $λ=3$ :

$u2=[11]$

The general solution of system corresponding to eigenvector $u1$ and $u2$ :

$X(t)=c1e3t[11]+c2e4t[32]$

Conclusion: Therefore, the solution of system $x′(t)=Ax(t)$ , where $A=[6−321]$ is $X(t)=c1e3t[11]+c2e4t[32]$ .

Answer 8

Expert Solution & Answer

To determine

To solve: The system $x′(t)=Ax(t)$ , where $A=[6−321]$ .

Answer to Problem 1RP

Solution:

The solution of system $x′(t)=Ax(t)$ , where $A=[6−321]$ is $X(t)=c1e3t[11]+c2e4t[32]$ .

Explanation of Solution

Formula used:

The procedure to solve the $x′(t)=Ax(t)$ :

Step-1: Apply the condition $|A−λI|=0$ for matrix $A$ in order to calculate eigenvalues:

Step-2:Apply the $|A−λI|X=0$ in order to calculate eigenvectors $u1$ and $u2$ corresponding to eigenvalue $λ1$ and $λ2$ respectively:

Step-3:Write the general solution of system $x′(t)=Ax(t)$ as:

$X(t)=c1eλ1tu1+c2eλ2tu2$

Calculation:

Consider the system $x′(t)=Ax(t)$ ,

$A=[6−321]$

Apply the condition $|A−λI|=0$ for matrix $A$ in order to calculate eigenvalues:

$|[6−321]−λ[1001]|=0|6−λ−321−λ|=0$

Expand the above determinant:

$(6−λ)(1−λ)+6=06−λ−6λ+λ2+6=06−7λ+λ2+6=0λ2−7λ+12=0$

Factories the above quadratic equation:

$λ2−4λ−3λ+12=0λ(λ−4)−3(λ−3)=0(λ−4)(λ−3)=0$

Equate first factor to 0:

$(λ−4)=0λ=4$

Equate second factor to 0:

$(λ−3)=0 λ=3$

Apply the $|A−λI|X=0$ in order to calculate eigenvector corresponding to eigenvalue $λ=4$ :

$[6−4−321−4][x1x2]=[00][2−32−3][x1x2]=[00]$

Write above matrix notation in scalar form notation:

$2x1−3x2=02x1−3x2=0$

The eigenvector corresponding to eigenvalue $λ=4$ :

$u1=[32]$

Apply the $|A−λI|X=0$ in order to calculate eigenvector corresponding to eigenvalue $λ=3$ :

$[6−3−321−3][x1x2]=[00][3−32−2][x1x2]=[00]$

Write above matrix notation in scalar form notation:

$3x1−3x2=02x1−2x2=0$

The eigenvector corresponding to eigenvalue $λ=3$ :

$u2=[11]$

The general solution of system corresponding to eigenvector $u1$ and $u2$ :

$X(t)=c1e3t[11]+c2e4t[32]$

Conclusion: Therefore, the solution of system $x′(t)=Ax(t)$ , where $A=[6−321]$ is $X(t)=c1e3t[11]+c2e4t[32]$ .

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

To determine

To solve: The system $x′(t)=Ax(t)$ , where $A=[6−321]$ .

Answer 12

Answer to Problem 1RP

Solution:

The solution of system $x′(t)=Ax(t)$ , where $A=[6−321]$ is $X(t)=c1e3t[11]+c2e4t[32]$ .

Answer 13

Explanation of Solution

Formula used:

The procedure to solve the $x′(t)=Ax(t)$ :

Step-1: Apply the condition $|A−λI|=0$ for matrix $A$ in order to calculate eigenvalues:

Step-2:Apply the $|A−λI|X=0$ in order to calculate eigenvectors $u1$ and $u2$ corresponding to eigenvalue $λ1$ and $λ2$ respectively:

Step-3:Write the general solution of system $x′(t)=Ax(t)$ as:

$X(t)=c1eλ1tu1+c2eλ2tu2$

Calculation:

Consider the system $x′(t)=Ax(t)$ ,

$A=[6−321]$

Apply the condition $|A−λI|=0$ for matrix $A$ in order to calculate eigenvalues:

$|[6−321]−λ[1001]|=0|6−λ−321−λ|=0$

Expand the above determinant:

$(6−λ)(1−λ)+6=06−λ−6λ+λ2+6=06−7λ+λ2+6=0λ2−7λ+12=0$

Factories the above quadratic equation:

$λ2−4λ−3λ+12=0λ(λ−4)−3(λ−3)=0(λ−4)(λ−3)=0$

Equate first factor to 0:

$(λ−4)=0λ=4$

Equate second factor to 0:

$(λ−3)=0 λ=3$

Apply the $|A−λI|X=0$ in order to calculate eigenvector corresponding to eigenvalue $λ=4$ :

$[6−4−321−4][x1x2]=[00][2−32−3][x1x2]=[00]$

Write above matrix notation in scalar form notation:

$2x1−3x2=02x1−3x2=0$

The eigenvector corresponding to eigenvalue $λ=4$ :

$u1=[32]$

Apply the $|A−λI|X=0$ in order to calculate eigenvector corresponding to eigenvalue $λ=3$ :

$[6−3−321−3][x1x2]=[00][3−32−2][x1x2]=[00]$

Write above matrix notation in scalar form notation:

$3x1−3x2=02x1−2x2=0$

The eigenvector corresponding to eigenvalue $λ=3$ :

$u2=[11]$

The general solution of system corresponding to eigenvector $u1$ and $u2$ :

$X(t)=c1e3t[11]+c2e4t[32]$

Conclusion: Therefore, the solution of system $x′(t)=Ax(t)$ , where $A=[6−321]$ is $X(t)=c1e3t[11]+c2e4t[32]$ .

Answer 14

In Problems 1-4 , find a general solution for the system x ′ ( t ) = A x ( t ) , where A is given. A = [ 6 − 3 2 1 ]

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Answer to Problem 1RP

Explanation of Solution

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