Answered: Matrix System -4 -2 1 X; Xі — е t -12…

Algebra and Trigonometry (MindTap Course List)

4th Edition

ISBN:9781305071742

Author:James Stewart, Lothar Redlin, Saleem Watson

Publisher:James Stewart, Lothar Redlin, Saleem Watson

Chapter11: Matrices And Determinants

Section11.CR: Chapter Review

Problem 2CC: What is the row-echelon form of a matrix? What is a leading entry?

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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

Matrix System
-4
-2
1
X; Xі — е t
-12
-1
-6
-4
0 -1
X2 = e=t
et
X4 = e
-2

Expert Solution

Step 1

Given : $x' = [\begin{matrix} 1 & - 4 & 0 & - 2 \\ 0 & 1 & 0 & 0 \\ 6 & - 12 & - 1 & - 6 \\ 0 & - 4 & 0 & - 1 \end{matrix}] x; x_{1} = e^{- t} [\begin{matrix} 1 \\ 0 \\ 0 \\ 1 \end{matrix}]; x_{2} = e^{- t} [\begin{matrix} 0 \\ 0 \\ 1 \\ 0 \end{matrix}]; x_{3} = e^{t} [\begin{matrix} 0 \\ 1 \\ 0 \\ - 2 \end{matrix}]; x_{4} = e^{t} [\begin{matrix} 1 \\ 0 \\ 3 \\ 0 \end{matrix}]$

(a) To verify: these vectors are solutions of given system

(b) To Show: $W r o n s k i a n$ is nonzero and Solution of the system

Step 2

$x' = [\begin{matrix} 1 & - 4 & 0 & - 2 \\ 0 & 1 & 0 & 0 \\ 6 & - 12 & - 1 & - 6 \\ 0 & - 4 & 0 & - 1 \end{matrix}] x = P x x_{1} = e^{- t} [\begin{matrix} 1 \\ 0 \\ 0 \\ 1 \end{matrix}]; x_{2} = e^{- t} [\begin{matrix} 0 \\ 0 \\ 1 \\ 0 \end{matrix}]; x_{3} = e^{t} [\begin{matrix} 0 \\ 1 \\ 0 \\ - 2 \end{matrix}]; x_{4} = e^{t} [\begin{matrix} 1 \\ 0 \\ 3 \\ 0 \end{matrix}]$

To verify that the solution the vector function $x_{1}, x_{2}, x_{3} & x_{4}$

$x_{1} = e^{- t} [\begin{matrix} 1 \\ 0 \\ 0 \\ 1 \end{matrix}]; x_{2} = [\begin{matrix} 0 \\ 0 \\ e^{- t} \\ 0 \end{matrix}]; x_{3} = [\begin{matrix} 0 \\ e^{t} \\ 0 \\ - 2 e^{t} \end{matrix}]; x_{4} = e^{t} [\begin{matrix} 1 \\ 0 \\ 3 \\ 0 \end{matrix}] x_{1}^{'} = e^{- t} [\begin{matrix} - e^{- t} \\ 0 \\ 0 \\ - e^{- t} \end{matrix}]; x_{2}^{'} = [\begin{matrix} 0 \\ 0 \\ - e^{- t} \\ 0 \end{matrix}]; x_{3}^{'} = [\begin{matrix} 0 \\ e^{t} \\ 0 \\ - 2 e^{t} \end{matrix}]; x_{4}^{'} = [\begin{matrix} e^{t} \\ 0 \\ 3 e^{t} \\ 0 \end{matrix}]$

are both solution of matrix differential equation with coefficient matrix P . we need only Calculate .

$P x_{1} = [\begin{matrix} 1 & - 4 & 0 & - 2 \\ 0 & 1 & 0 & 0 \\ 6 & - 12 & - 1 & - 6 \\ 0 & - 4 & 0 & - 1 \end{matrix}] [\begin{matrix} e^{- t} \\ 0 \\ 0 \\ e^{- t} \end{matrix}] = [\begin{matrix} - e^{- t} \\ 0 \\ 0 \\ - e^{- t} \end{matrix}] = x_{1}^{'} P x_{2} = [\begin{matrix} 1 & - 4 & 0 & - 2 \\ 0 & 1 & 0 & 0 \\ 6 & - 12 & - 1 & - 6 \\ 0 & - 4 & 0 & - 1 \end{matrix}] [\begin{matrix} 0 \\ 0 \\ e^{- t} \\ 0 \end{matrix}] = [\begin{matrix} 0 \\ 0 \\ - e^{- t} \\ 0 \end{matrix}] = x_{2}^{'} P x_{3} = [\begin{matrix} 1 & - 4 & 0 & - 2 \\ 0 & 1 & 0 & 0 \\ 6 & - 12 & - 1 & - 6 \\ 0 & - 4 & 0 & - 1 \end{matrix}] [\begin{matrix} 0 \\ e^{t} \\ 0 \\ - 2 e^{t} \end{matrix}] = [\begin{matrix} 0 \\ e^{t} \\ 0 \\ - 2 e^{t} \end{matrix}] = x_{3}^{'} P x_{4} = [\begin{matrix} 1 & - 4 & 0 & - 2 \\ 0 & 1 & 0 & 0 \\ 6 & - 12 & - 1 & - 6 \\ 0 & - 4 & 0 & - 1 \end{matrix}] [\begin{matrix} e^{t} \\ 0 \\ 3 e^{t} \\ 0 \end{matrix}] = [\begin{matrix} e^{t} \\ 0 \\ 3 e^{t} \\ 0 \end{matrix}] = x_{4}^{'}$

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