Please just answer e,f Determine whether lim A exists for each of the following matrices818A, and compute the limit if it exists.(a) (0.7 0.1)(a) (-1-1.8 4.8-0.8 2.2(b)(e)-1.4 0.8-2.4 1.840.89)3(c) (0.6(f)0.4 0.70.6 0.32.0 -0.53.0 -0.5

Given Matric are as follows : e) A=-2-143 f) A=2.0-0.53.0-0.5 To Find : Whther thelimm→∞Am exist for the given matrix and also compute the limit if exist .e) Matrix A=-2-143 limm→∞Am exist for the given matrix, if and only if all the eigenvalues are in S=λ∈C:λ<1 or λ=1 . If 1 is the eigenvalue for A, then the dimension of eigenspace corresponding to eigenvalue 1 is equal to that of the multiplicity of eigenvalue. Now, we have to find the eigenvalue of A as follows : A-λI=-2-λ-143-λA-λI=det -2-λ-143-λλ-2λ+1=0λ=2 ,-1 Thus the eigenvalues are λ=2 ,-1 As λ=2>1 , Thus thelimm→∞Am does not exist for the given matrix.f) Matrix A=2.0-0.53.0-0.5 limm→∞Am exist for the given matrix, if and only if all the eigenvalues are in S=λ∈C:λ<1 or λ=1 If 1 is the eigenvalue for A, then the dimension of eigenspace corresponding to eigenvalue 1 is equal to that of the multiplicity of eigenvalue. Now, we have to find the eigenvalue of A as follows : A-λI=2.0-λ-0.53.0-0.5-λA-λI=det 2.0-λ-0.53.0-0.5-λλ-1λ-12=0λ=1/2 ,1 Thus the eigenvalues are λ=1/2 ,1 As λ=1/2<1 or λ=1 , Thus thelimm→∞Am exist for the given matrix. Now we have to find the matrix P as follows : For λ1=1/2 the corresponding eigenvector can be calculated as follows : A-λ1Iv=A-0.5= 2.0-0.5-0.53.0-0.5-0.5ab= 1.5-0.53.0-1ab1.5a -0.5b=03a-b=0⇒a=13 ,b=1 For λ1=1/2 , the corresponding eigenvector v1=131 For λ2=1 the corresponding eigenvector can be calculated as follows : A-λ1v=A-1I= 2.0-1-0.53.0-0.5-1ab= 1-0.53.0-1.5aba -0.5b=03a-1.5b=0⇒a=12 ,b=1 For λ2=1 the corresponding eigenvector v2=121. Therefore, matrix P is as follows : P=v1v2P=1/31/211P-1=-636-2 And the diagonal matrix D which has eigenvalues as the diagonal elements can be calculated as follows : D=λ100λ2D=1/2001 Now we have to find the limit by diagonalizing the matrix which is as follows : A=PDP-1Am=PDmP-1limn→∞Am=limn→∞PDmP-1limn→∞Am=limn→∞1/31/2111/2001m-636-2limn→∞Am=1/31/2110001-636-2limn→∞Am=3-162 Therefore, limn→∞Am=3-162