Astha Jain

Astha Jain

Answered question

2022-07-02

Answer & Explanation

user_27qwe

user_27qwe

Skilled2023-06-01Added 375 answers

To find a fundamental matrix of the given system and apply it to find a solution satisfying the initial conditions, we can follow the steps below:
Step 1: Writing the system in matrix form
The given system can be written in matrix form as:
x=[506111222]x
where x is the column vector [x1x2x3].
Step 2: Finding the eigenvalues and eigenvectors
To find the fundamental matrix, we need to find the eigenvalues and eigenvectors of the coefficient matrix.
The characteristic equation is given by:
det(AλI)=0
where A is the coefficient matrix, λ is the eigenvalue, and I is the identity matrix.
Substituting the given coefficient matrix A, we have:
det([506111222]λ[100010001])=0
Simplifying the determinant equation, we get:
det[5λ0611λ1222λ]=0
Expanding the determinant along the first row, we have:
(5λ)(|1λ122λ|)0+(6)(|11λ22|)=0
Simplifying further, we obtain:
(5λ)(λ2+3λ+2)6(2λ)=0
Expanding and rearranging the equation, we get:
λ3+6λ2+9λ+4=0
Factoring the equation, we find that λ=1 is a repeated root. So the characteristic equation becomes:
(λ+1)2(λ+4)=0
Hence, the eigenvalues are λ1=1 (with algebraic multiplicity 2) and λ2=4.
Next, we need to find the eigenvectors corresponding to these eigenvalues.
For λ1=1, we solve the equation (Aλ1I)v1=0:
[606101201][v11v21v31]=0
Row reducing the augmented matrix, we get:
[101000000][v11v21v31]=0
From the reduced row-echelon form, we can see that v21 and v31 are free variables. Let v21=t and v31=s (where t and s are arbitrary constants).
Therefore, the eigenvector corresponding to λ1=1 is:
v1=[v11v21v31]=[sts]
For λ2=4, we solve the equation (Aλ2I)v2=0:
[906131222][v12v22v32]=0
Row reducing the augmented matrix, we get:
[10230113000][v12v22v32]=0
From the reduced row-echelon form, we can see that v12, v22, and v32 are all free variables. Let v12=p, v22=q, and v32=r (where p, q, and r are arbitrary constants).
Therefore, the eigenvector corresponding to λ2=4 is:
v2=[v12v22v32]=[pqr]
Step 3: Constructing the fundamental matrix
The fundamental matrix Φ(t) is constructed using the eigenvectors v1 and v2 corresponding to the eigenvalues λ1 and λ2, respectively.
We have:
Φ(t)=[v1v2]=[sptqsr]
Step 4: Applying the initial condition
To find a solution satisfying the initial condition x(0)=[102], we use the formula:
x(t)=Φ(t)Φ(0)1x0
We are given x(0)=[102], so we substitute the values into the formula:
x(t)=[sptqsr][v1v2]1[102]
To find the inverse of the matrix [v1v2], we can use the formula for the inverse of a 2×2 matrix:
[abcd]1=1adbc[dbca]
Applying this formula, we find:
[v1v2]1=1sqpt[qprs]
Substituting this back into the formula for x(t), we get:
x(t)=[sptqsr]1sqpt[qprs][102]
Simplifying the expression, we have:
x(t)=1sqpt[sq+p(r)s(p)+p(s)tq+q(r)t(p)+q(s)sq+p(2r)s(p)+p(2s)][102]
Performing the matrix multiplication, we get:
x(t)=1sqpt[3p2rt(qp)2p+4r]
Therefore, the solution satisfying the initial condition x(0)=[102] is given by:
x(t)=1sqpt[3p2rt(qp)2p+4r]

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