Answered: Matrix System 3 -2 x' = -1 3 -2 X; X1 =…

College Algebra

1st Edition

ISBN:9781938168383

Author:Jay Abramson

Publisher:Jay Abramson

Chapter7: Systems Of Equations And Inequalities

Section7.6: Solving Systems With Gaussian Elimination

Problem 5SE: Can a matrix that has 0 entries for an entire row have one solution? Explain why or why not.

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Question

First verify that the given vectors are solutions of the given system. Then use the Wronskian to show that they are linearly independent. Finally, write the general solution of the system. find a particular solution of the indicatedlinear system that satisfies the below initial conditions

x₁(0) = 0 , x₂ (0) = 0, x₃(0)=4

Matrix System
3 -2
x' =
-1
3
-2
X; X1 = et
-1
3
X2 = e3t
X3 = e5t
-2

Expert Solution

Step 1

Given that, the system is-

$x' = [\begin{matrix} 3 & - 2 & 0 \\ - 1 & 3 & - 2 \\ 0 & - 1 & 3 \end{matrix}] x$ with

$x_{1} = e^{t} [\begin{matrix} 2 \\ 2 \\ 1 \end{matrix}]$ , $x_{2} = e^{3 t} [\begin{matrix} - 2 \\ 0 \\ 1 \end{matrix}]$ , $x_{3} = e^{5 t} [\begin{matrix} 2 \\ - 2 \\ 1 \end{matrix}]$ .

Step 2

Since,

First given vector is $x_{1} = e^{t} [\begin{matrix} 2 \\ 2 \\ 1 \end{matrix}]$ .

Now,

$x_{1}' = e^{t} [\begin{matrix} 2 \\ 2 \\ 1 \end{matrix}]$

and,

$\begin{array}{rcl} A x_{1} & = & e^{t} [\begin{matrix} 3 & - 2 & 0 \\ - 1 & 3 & - 2 \\ 0 & - 1 & 3 \end{matrix}] [\begin{matrix} 2 \\ 2 \\ 1 \end{matrix}] \\ = & e^{t} [\begin{matrix} 2 \\ 2 \\ 1 \end{matrix}] \\ = & x_{1}^{'} \end{array}$

So,

$x_{1}$ is a solution of the system.

Step 3

Since,

The second given vector is $x_{2} = e^{3 t} [\begin{matrix} - 2 \\ 0 \\ 1 \end{matrix}]$ .

Now,

$x_{2}' = 3 e^{3 t} [\begin{matrix} - 2 \\ 0 \\ 1 \end{matrix}]$

and,

$\begin{array}{rcl} A x_{2} & = & e^{3 t} [\begin{matrix} 3 & - 2 & 0 \\ - 1 & 3 & - 2 \\ 0 & - 1 & 3 \end{matrix}] [\begin{matrix} - 2 \\ 0 \\ 1 \end{matrix}] \\ = & e^{3 t} [\begin{matrix} - 6 \\ 0 \\ 3 \end{matrix}] \\ = & 3 e^{3 t} [\begin{matrix} - 2 \\ 0 \\ 1 \end{matrix}] \\ = & x_{2}^{'} \end{array}$

So,

$x_{2}$ is a solution of the system.

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Matrix System 3 -2 x' = -1 3 -2 X; X1 = et -1 3 X2 = e3t X3 = e5t -2