4. Suppose \( x'(t) = Ax(t) \) where \( A = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \), \( a, b, c, d \in \mathbb{R} \) and \( x(t) = \begin{pmatrix} x_1(t) \\ x_2(t) \end{pmatrix} \). Use the definition of linear independence to justify each of the following. Recall that if \( v \in \mathbb{R}^n = \vec{0} \), then each component of \( v \) must equal 0.(a) Suppose the eigenvalues of \( A \) are distinct and real-valued. Show that the terms in the solution \( x(t) \) are linearly independent.(b) Suppose there is only one eigenvalue of \( A \). Show that the terms in the solution \( x(t) \) are linearly independent. (Hint: There are two cases here - one case when there are two linearly independent eigenvectors associated with that eigenvalue and the second case is when we cannot find two linearly independent eigenvalues.)(c) Suppose the eigenvalues of \( A \) are complex conjugates. Show that the terms in the solution \( x(t) \) are linearly independent.

Suppose x'(t)=Ax(t) Where A= abcd, a, b, c, d∈ℝ and x(t)=x1(t)x2(t). Use the definition of linear independence Justify to following. Recall that if v∈ℝn=0→,. then each component of v must equal 0. (a) Given, the eigenvalues of A are distinct and real valued. To prove that the terms x(t) are linearly independent. Let λ1 and λ2 be distinct eigenvalue A. To find α1, α2 be eigenvector corresponding eigenvalues λ1 and λ2. x1(t)=α1eλ1t x2(t)=α2eλ2t are the Soultion of differential equation. Then, we have, λ1α1 = Aα1 Using this and defination of x'(t), we obtain dx1(t)dt=λ1α1eλ1t=A(α1eλ1t)=Ax1(t) ⇒ dx1(t)dt=Ax1(t) Which states that x1(t) satisfies the vector differential equation. Similarly x2(t) satisfies vector differential equation. W (x1(t), x2(t)) = α1eλ1tα2eλ2tλ1α1eλ1tλ2α2eλ2t =eλ1t.eλ2tα1α2λ1α1λ2α2 =e(λ1+λ2)tα1α2λ1α1λ2α2≠0 ⇒W (x1(t), x2(t))≠0 Hence, the soultions x1(t) and x2(t) are linearly independent. (b) Given, there is only one eigenvalue of A. To prove that the terms in the solution x(t) are linearly independent. Let λ be eigenvalue of A, and x(t)=(αt+β)eλt . ..(1) is solution of the differential equation. Differentiate (1) w.r.t. 't'and substitute into x'(t)=Ax(t), we obtain dx(t)dt=(αt+β)λeλt+αeλt and (αt+β)λeλt+αeλt=A(αt+β)eλt⇒ (αt+β)λ+αeλt=A(αt+β)eλt⇒ (αt+β)λ+α=A(αt+β)⇒ αλt+βλ+α-Aαt+Aβ=0⇒ (λα-Aα)t+(βλ+α+Aβ)=0Which is hold for for all t∈[a, b] iff λα-Aα=0 and βλ+α+Aβ=0⇒ λα=Aα⇒ (A-λI)α =0Let γ be the eigenvector corresponding the eigenvalue λ. (A-λI)β =γ ...(2)is the equation for the determination of β in (1). Thus in the assumed solution the vector α is , in fact a eigenvector of A corresponding to neigen value λ, and the vector β is determined from (2). Moreover it can be shown that these two solution are obtained αeλt and (αt+β)eλt are linearly independent. (c) Given, the eigenvalues are4 complex conjugate. To prove that term in the solution x(t) are linearly independent. Let λ1=a+ib and λ2=a-ib are eigenvalues of A. Let α1 and α2 be eigenvectors corresponding the eigenvalues of λ1 and λ2. So, the corresponding linearly independent solution are α1e(a+ib)t and α2e(a-ib)t are complex solutions. Hence proved.