Help please with 33 **Problem 33: Characteristic Equation of a Matrix**Consider the coefficient matrix \( \mathbf{A} \) of the system given by:\[\mathbf{x}' = \begin{bmatrix} 3 & -4 & 1 & 0 \\ 4 & 3 & 0 & 1 \\ 0 & 0 & 3 & -4 \\ 0 & 0 & 4 & 3 \end{bmatrix} \mathbf{x}\]The characteristic equation of this matrix \( \mathbf{A} \) is:\[\phi(\lambda) = (\lambda^2 - 6\lambda + 25)^2 = 0.\]This equation is derived from finding the eigenvalues of matrix \( \mathbf{A} \), which are the solutions to \( \det(\mathbf{A} - \lambda \mathbf{I}) = 0 \). The expression represents a polynomial equation whose roots correspond to these eigenvalues. Therefore, A has the repeated complex conjugate pair \(3 \pm 4i\) of eigenvalues. First, show that the complex vectors \[\mathbf{v_1} = \begin{bmatrix} 1 \\ i \\ 0 \\ 0 \end{bmatrix}^T \quad \text{and} \quad \mathbf{v_2} = \begin{bmatrix} 0 \\ 0 \\ 1 \\ i \end{bmatrix}^T\]form a length 2 chain \(\{\mathbf{v_1}, \mathbf{v_2}\}\) associated with the eigenvalue \(\lambda = 3 - 4i\). Then calculate the real and imaginary parts of the complex-valued solutions \[\mathbf{v_1} e^{\lambda t} \quad \text{and} \quad (\mathbf{v_1}t + \mathbf{v_2}) e^{\lambda t}\]to find four independent real-valued solutions of \(x' = Ax\).

Name: Help please with 33 **Problem 33: Characteristic Equation of a Matrix*
Uploaded: 2024-03-20T07:07:12.000Z
Channel: Bartleby
Description: Help please with 33 **Problem 33: Characteristic Equation of a Matrix**Consider the coefficient matrix \( \mathbf{A} \) of the system given by:\[\mathbf{x}' = \begin{bmatrix} 3 & -4 & 1 & 0 \\ 4 & 3 & 0 & 1 \\ 0 & 0 & 3 & -4 \\ 0 & 0 & 4 & 3 \end{bma

The linear combination of the vectors v1 and v2 is c1v1+c2v2, where c1, c2 are arbitrary constants. Let u=c1v1+c2v2=c1c1ic2c2iT. A-(3-4i)I=3-4104301003-40043-(3-4i)0000(3-4i)0000(3-4i)0000(3-4i)=4i-41044i01004i-40044i Now, (A-(3-4i)I)u=4i-41044i01004i-40044ic1c1ic2c2i=c2c2i00 Here, (A-(3-4i)I)2=4i-41044i01004i-40044i4i-41044i01004i-40044i=-32-32i8i-832i-3288i00-32-32i0032i-32 Also, (A-(3-4i)I)2u=-32-32i8i-832i-3288i00-32-32i0032i-32c1c1ic2c2i=0000 Thus, u serves as a generalized eigenvector for eigenvalue, λ=3-4i which forms a chain of length 2, since, two multiplications by (A-(3-4i)I) to get to zero. Case I: When λ=3-4i, Now, eλt=e3t-4ti=e3t·e-4ti=e3t[cos(-4t)+i sin(-4t)]=e3t[cos(4t)-i sin(4t)]. Thus, v1eλt=1i00e3t·(cos(4t)-i sin(4t))=e3t·(cos(4t)-i sin(4t))e3t·(sin(4t)+i cos(4t))00=e3t·cos(4t)e3t·sin(4t)00+i·-e3t·sin(4t)e3t·cos(4t)00 So, Real(v1eλt)=e3t·cos(4t)e3t·sin(4t)00 and Imaginary(v1eλt)=-e3t·sin(4t)e3t·cos(4t)00. Also, (v1t+v2)eλt=tt·i1ie3t(cos(4t)-i sin(4t))=t·e3t·(cos(4t)-i sin(4t))t·e3t(sin(4t)+i cos(4t))e3t(cos(4t)-i sin(4t))e3t(sin(4t)+i cos(4t))=t·e3t·cos(4t)t·e3t·sin(4t)e3t·cos(4t)e3t·sin(4t)+i·-t·e3t· sin(4t)t·e3tcos(4t)-e3tsin(4t)e3t cos(4t) So, Real((v1t+v2)eλt)=t·e3t·cos(4t)t·e3t·sin(4t)e3t·cos(4t)e3t·sin(4t) and Imaginary((v1t+v2)eλt)=-t·e3t·sin(4t)t·e3tcos(4t)-e3tsin(4t)e3tcos(4t). Case II: When λ=3+4i, Now,eλt=e3t+4ti=e3t·e4ti=e3t[cos(4t)+i sin(4t)]=e3t[cos(4t)+i sin(4t)]. Thus, v1eλt=1i00e3t·(cos(4t)+i sin(4t))=e3t·(cos(4t)+i sin(4t))e3t·(-sin(4t)+i cos(4t))00=e3t·cos(4t)-e3t·sin(4t)00+i·e3t·sin(4t)e3t·cos(4t)00 So, Real(v1eλt)=e3t·cos(4t)-e3t·sin(4t)00 and Imaginary(v1eλt)=e3t·sin(4t)e3t·cos(4t)00. Also, (v1t+v2)eλt=tt·i1ie3t(cos(4t)+i sin(4t))=t·e3t·(cos(4t)+i sin(4t))t·e3t(-sin(4t)+i cos(4t))e3t(cos(4t)+i sin(4t))e3t(-sin(4t)+i cos(4t))=t·e3t·cos(4t)-t·e3t·sin(4t)e3t·cos(4t)-e3t·sin(4t)+i·t·e3t· sin(4t)t·e3tcos(4t)e3tsin(4t)e3t cos(4t) So, Real((v1t+v2)eλt)=t·e3t·cos(4t)-t·e3t·sin(4t)e3t·cos(4t)-e3t·sin(4t) and Imaginary((v1t+v2)eλt)=t·e3t·sin(4t)t·e3tcos(4t)e3tsin(4t)e3tcos(4t). Thus, the four possible real-valued solutions of x'=Ax are Real(v1eλt)=e3t·cos(4t)e3t·sin(4t)00; Real((v1t+v2)eλt)=t·e3t·cos(4t)t·e3t·sin(4t)e3t·cos(4t)e3t·sin(4t) ; Real(v1eλt)=e3t·cos(4t)-e3t·sin(4t)00 and Real((v1t+v2)eλt)=t·e3t·cos(4t)-t·e3t·sin(4t)e3t·cos(4t)-e3t·sin(4t).