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Linear Algebra With Applications
Found in: Page 441
Linear Algebra With Applications

Linear Algebra With Applications

Book edition 5th
Author(s) Otto Bretscher
Pages 442 pages
ISBN 9780321796974

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Short Answer

Solve the system dxdt=[-1-22-1]xwithx(0)=[1-1] . Give the solution in real form. Sketch the solution.

The solution of the system isx(t)=et[cos(2t)+sin(2t)sin(2t)cos(2t)] and the graph is:

See the step by step solution

Step by Step Solution

Step 1: Find the Eigen values of the matrix

Consider the equation dxdt=[1221]xwith the initial valuex(0)=[11] .

Compare the equationsdxdt=[1221]x and dxdt=Axas follows.


Assume λis an Eigen value of the matrix [1221] implies |AλI|=0.

Substitute the values[1221] for Aand[1001] forI in the equation|AλI|=0 as follows.


Simplify the equation|[1221]λ[1001]|=0 as follows.


Further, simplify the equation as follows.


Further, simplify the equation as follows.


Therefore, the Eigen values of Aare λ=1±i2.

Step 2: Determine the Eigen vector corresponding to the Eigen value λ=-1+i2 

Substitute the values1+i2 forλ in the equation |[1λ221λ]|=0as follows.


AsE1+2i=ker[2i222i]=span[i1] , the values v+iwis defined as follows.


Therefore, the value ofS is .S=[1001]

Step 3: Determine the solution for  dx→dt=[-1-22-1]x→

The inverse of the matrixS=[1001] is defined as follows.


Asx(t)=eptS[cos(qt)sin(qt)sin(qt)cos(qt)]S1x0 , Substitute the value[1001] forS , [1001]forS1 , [11]for x0, 1for pand 2 for q in the equationx(t)=eptS[cos(qt)sin(qt)sin(qt)cos(qt)]S1x0 as follows.


Therefore, the solution of the system is x(t)=et[cos(2t)+sin(2t)sin(2t)cos(2t)].

Step 4: Sketch the solution

As λ1,2=1±2iand1<0 , draw the graph of the solutionx(t)=et[cos(2t)+sin(2t)sin(2t)cos(2t)] as follows.

Hence the solution of the system dxdt=[1221]xwith the initial valuex(0)=[11] is x(t)=et[cos(2t)+sin(2t)sin(2t)cos(2t)]and the graph of the solution is a spirals in counterclockwise direction toward the origin.


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