Use the same notations as on Page 6 of the lecture notes: A is the coefficient matrix for each linearsystem, D is the diagonal matrix with diagonal value ai, and D-L is the lower triangular matrix ofA. For the following linear systems3x₁ + x₂ + x3 = 0x₁ + 3x₂x3 = 4(3x₁ + x₂ = 5x3 = -6(1)(x₁ + 3x₂ 3x3 = 2(3) x₁ + x₂ + x3 = 2(3x₁ + 3x₂ + x3 = 4(x₁ + 2x₂ - 2x3 = 1(2) x₁ + x₂ + x3 = 2(2x₁ + 2x₂ + x3 = 3(2x₁ - x₂ + x3 = -3(4) 2x₁ + 2x₂ + 2x3 = 4-x₁x₂ + 2x3 = 1a)For system (3), find T₁, TĠ, p(Tj), and p(TG). Does the Jacobi iterative method converge formethod converge for system (4)? Does Gauss-Seidel iterative method converge for system (4)?

Given: A is the coefficient matrix for each linear system, D is the diagonal matrix with diagonal value aii, and D-L is the lower triangular matrix of A. 3. x1+3x2-3x3=2x1+x2+x3=23x1+3x2+x3=4 4. 2x1-x2+x3=-32x1+2x2+2x3=4-x1-x2+2x3=1 To do: For system (3), find Tj, TG, ρ(Tj), ρ(TG). Does the Jacobi iterative method converge for system (4)? Does Gauss-Seidel iterative method converge for system (4)?We know that TJ=I-D-1Aρ(TJ)=max( eigenvalues of TJ)TG=I-(D-L)-1A ρ(TG)=max( eigenvalues of TG). Now, for system (3) A=13-3111331, D=100010001 So, Tj=I-D-1A=100010001-100010001-113-3111331=100010001-10001000113-3111331=100010001-13-3111331=0-33-10-1-3-30. Now, we will find the eigenvalues of above matrix TJ. A-λI=0-λ-33-1-λ-1-3-3-λ=0λ(λ2+3)=0λ=0,±3i Therefore, ρ(TJ)=maxλ=max(0,3i,-3i)=3. Let us now find TG. TG=I-(D-L)-1A. First, we will find (D-L), the lower triangular matrix of A. We will use Doolittle's method for LU decomposition. Let A=LU. 13-3111331=100l2110l31l321×u11u12u130u22u2300u3313-3111331=u11u12u13l21u11l21u12+u22l21u13+u23l31u11l31u12+l32u22l31u13+l32u23+u33 This implies u11=1u12=3u13=-3l21=1u22=-2u23=4l31=3l32=3u33=-2 Hence, lower triangular matrix D-L=100110331. Therefore, TG=I-(D-L)-1A.=100010001-100110331-113-3111331=100010001-100-1100-3113-3111331=100010001-13-30-2400-2=0-3303-4003 We know that the eigenvalue of upper triangular matrix are the diagonal entries of the matrix. Hence, eigenvalues are λ=0, 3, 3. Thus, ρ(TG)=max( eigenvalues of TG)ρ(TG)=max( 0, 3, 3)ρ(TG)=3.We need to figure out whether Jacobi iterative method converge for system (4). 4.2x1-x2+x3=-32x1+2x2+2x3=4-x1-x2+2x3=1 First, we rewrite the equations in the form: x1=12(-3+x2-x3)x2=12(4-2x1-2x3)x3=12(1+x1+x2) Initial guess be (0, 0 ,0). 1st Approximation: x1(1)=12(-3+0-9)=-1.5x2(1)=12(4)=2x3(1)=12(1)=0.5 2nd Approximation: x1(2)=12(-3+2-0.5)=12(-1.5)=-0.75x2(2)=12(4-2(-1.5)-2(0.5))=3x3(2)=12(1+(-1.5)+2)=0.75 3rd Approximation: x1(3)=12(-3+3-0.75)=12(-0.75)=-0.375x2(3)=12(4-2(-0.75)-2(0.75))=2x3(3)=12(1+(-0.75)+3)=1.625 On further iteration we get: Iteration x1 x2 x3 1 -1.5 2 0.5 2 -0.75 3 0.75 3 -0.375 2 1.625 4 -1.3125 0.75 1.3125 5 -1.7812 2 0.2188 6 -0.6094 3.5625 0.6094 7 -0.0234 2 1.9766 8 -1.4883 0.0469 1.4883 We can see in above table that the Jacobi method become progressively worse instead of better. Thus, we can conclude that the method diverges.We need to figure out whether Gauss-Seidel iterative method converge for system (4). 2x1-x2+x3=-32x1+2x2+2x3=4-x1-x2+2x3=1 First, we rewrite the equations in the form: x1=12(-3+x2-x3)x2=12(4-2x1-2x3)x3=12(1+x1+x2) Initial guess be (0, 0 ,0). 1st Approximation: x1(1)=12(-3+0-0)=-1.5x2(1)=12(4-2(-1.5)-2(0))=3.5x3(1)=12(1-1.5+3.5)=1.5 2nd Approximation: x1(2)=12(-3+3.5-1.5)=12(-1)=-0.5x2(2)=12(4-2(-0.5)-2(1.5))=1x3(2)=12(1+(-0.5)+1)=0.75 3rd Approximation: x1(3)=12(-3+1-0.75)=12(-2.75)=-1.375x2(3)=12(4-2(-1.375)-2(0.75))=2.625x3(3)=12(1+(-1.375)+2.625)=1.125 On further iteration we get: Iteration x1 x2 x3 1 -1.5 3.5 1.5 2 -0.5 1 0.75 3 -1.375 2.625 1.125 4 -0.75 1.625 0.9375 5 -1.1562 2.2188 1.0312 6 -0.9062 1.875 0.9844 7 -1.0547 2.0703 1.0078 8 -0.9688 1.9609 0.9961 9 -1.0176 2.0215 1.002 10 -0.9902 1.9883 0.999 11 -1.0054 2.0063 1.0005 12 -0.9971 1.9966 0.9998 13 -1.0016 2.0018 1.0001 14 -0.9991 1.999 0.9999 15 -1.0005 2.0005 1 16 -0.9998 1.9997 1 Hence, by Gauss Seidel method, the system (4) converges, and the solution is: x1≈-1x2≈2x3≈1.