Iterative Methods for Linear Systems Questions and Answers

Answer: C

After two Jacobi iterations the solution is \([2.64\ \ 1.70\ \ -1.12]^{T}\).

Solve each equation for its diagonal unknown and use only the previous iterate on the right:

\[ \begin{aligned} x &= (10 + y - z)/5 \\ y &= (12 - 2x)/4 \\ z &= (-1 - x - y)/5 \end{aligned} \]

From \((2, 3, 0)\): \(x^{(1)} = 13/5 = 2.6\), \(y^{(1)} = 8/4 = 2\), \(z^{(1)} = -6/5 = -1.2\). Then:

\[ \begin{aligned} x^{(2)} &= (10 + 2 + 1.2)/5 = 2.64 \\ y^{(2)} &= (12 - 5.2)/4 = 1.70 \\ z^{(2)} &= (-1 - 2.6 - 2)/5 = -1.12 \end{aligned} \]

All options agree on 2.64; the distractors flip the sign of \(y\) or of \(z\). \(y\) stays positive because \(12 - 2x\) is positive for \(x\) near 2.6, and \(z\) is negative because \(-1 - x - y\) is.

Answer: A

Both P and Q are TRUE.

P: Doolittle factorisation (unit diagonal in L):

\[ L = \begin{bmatrix} 1 & 0 \\ -2 & 1 \end{bmatrix},\ U = \begin{bmatrix} 2 & -1 \\ 0 & 1 \end{bmatrix} \]

since \(l_{21} = -4/2 = -2\) and \(u_{22} = 3 - (-2)(-1) = 1\). So trace(U) = 2 + 1 = 3.

Q: the Jacobi iteration matrix is \(-D^{-1}(L+U)\) built from the off-diagonal entries:

\[ J = \begin{bmatrix} 0 & 1/2 \\ 4/3 & 0 \end{bmatrix},\quad \lambda^{2} = \frac{2}{3} \]

Its spectral radius is \(\sqrt{2/3} \approx 0.82 \lt 1\), so the Jacobi iterates converge from every starting vector. (M is not diagonally dominant, so that sufficient condition would not settle it; the spectral radius does.)

Answer: A

Both I and II are TRUE.

In \(A + \varepsilon I_{3}\) each diagonal entry is \(2 + \varepsilon\), and the off-diagonal entries in each row have absolute values summing to \(1 + 1 = 2\). Since \(2 + \varepsilon \gt 2\) for \(\varepsilon \gt 0\), the matrix is strictly diagonally dominant by rows.

For a strictly diagonally dominant matrix both the Jacobi and the Gauss-Seidel iterations converge for every starting vector. So I and II both hold, and the options denying either are wrong.

Answer: C

Both S1 and S2 are TRUE.

The Jacobi scheme for the scaled system solves each equation for its diagonal unknown using the previous values:

\[ \begin{aligned} x_{1} &= 4 - x_{2} + 2x_{3} = 4 - 1 + 2 = 5 \\ x_{2} &= 4 - 3x_{1} - 2x_{3} = 4 - 3 - 2 = -1 \\ x_{3} &= 1 - x_{1} - x_{2} = 1 - 1 - 1 = -1 \end{aligned} \]

so S1 holds. The iteration matrix is \(B = I - D^{-1}A\), whose characteristic equation works out to \(\lambda^{3} - 3\lambda - 4 = 0\). It has a real root near 2.2, so the spectral radius exceeds 1 and the iteration diverges for a general starting vector (any whose error has a component along that eigenvector). So S2 holds too.