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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 systemdxdt=[01-40]x withx(0)=[10] . Give the solution in real form. Sketch the solution.

The solution of the system is 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=[0140]xwith the initial valuex(0)=[10] .

Compare the equations dxdt=[0140]xanddxdt=Ax as follows.


Assumeλ is an Eigen value of the matrix [0140]implies|AλI|=0 .

Substitute the values [0140]forA and [1001]forI in the equation|AλI|=0 as follows.


Simplify the equation |[0140]λ[1001]|=0as follows.


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

Step 2: Determine the Eigen vector corresponding to the Eigen valueλ=i2 .

Substitute the valuesi2 forλ in the equation|[λ14λ]|=0 as follows.


As E1+2i=ker[2i142i]=span[2i1], the values v+iwis defined as follows.


Therefore, the value of SisS=[2001] .

Step 3: Determine the solution for dx→dt=[01-40]x→

The inverse of the matrixS=[2001] is defined as follows.


As x(t)=eptS[cos(qt)sin(qt)sin(qt)cos(qt)]S1x0, Substitute the value[2001] forS , [12001]for S1, [10]for x0,0 forp and2 forq in the equation

x(t)=eptS[cos(qt)sin(qt)sin(qt)cos(qt)]S1x0as follows.


Further, simplify the equation as follows.


Therefore, the solution of the system isx(t)=[cos(2t)sin(2t)2] .

 Step 4: Sketch the solution.

Asλ1,2=±2i , draw the graph of the solutionx(t)=[cos(2t)sin(2t)2] as follows.

Hence, the solution of the systemdxdt=[0140]x with the initial valuex(0)=[10] isx(t)=[cos(2t)sin(2t)2] and the graph of the solution is an ellipse in counterclockwise direction.

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