Question

For the iterative solution to linear algebraic equations$\mathbf{Ax}\mathbf{-}\mathbf{y}$, the $\mathbf{D}$matrix in the Jacobi method is the ________ portion of A, whereas D for Gauss-Siedel is the ________ portion of A.

Short answer

Therefore, for the iterative solution to linear algebraic equations $\mathrm{Ax}-\mathrm{y}$, the $D$matrix in the Jacobi method is the diagonal portion of $A$, whereas $D$for Gauss-Siedel is the lower triangular portion of A.

Step-by-step solution

Determine the formula of linear equation.  

Write the formula of linear algebraic equation.

$\mathbf{Ax}\mathbf{-}\mathbf{y}\mathbf{=}\mathbf{0}$ ……. (1)

Here, $X$and $y$are $N$vectors and $A$is $N\times N$square matrix.

Determine the answer.

Write equation (1) in matrix form.

$$\begin{gathered} \left[\begin{array}{ccc}{A}_{11}& \dots & A_{1N}\\ ‐& ‐& ‐\\ ‐& ‐& ‐\\ A_{N1}& \dots & A_{NN}\end{array}\right]\left[\begin{array}{c}{x}_1\\ ‐\\ ‐\\ x_N\end{array}\right]-\left[\begin{array}{c}{y}_1\\ ‐\\ ‐\\ y_N\end{array}\right]=0 \\ Here,A=\left[\begin{array}{ccc}{A}_{11}& \dots & A_{1N}\\ ‐& ‐& ‐\\ ‐& ‐& ‐\\ A_{N1}& \dots & A_{NN}\end{array}\right] \end{gathered}$$

In the equations (1), A , X and y may be real or complex.

Write the D matrix for Jacobi method.

$D=\left[\begin{array}{cccc}{A}_{11}& 0& 0& 0\\ 0& A_{22}& 0& 0\\ 0& 0& \dots & 0\\ 0& 0& 0& A_{NN}\end{array}\right]$

That is, for Jacobi, matrix consists of a diagonal elements of A matrix.

Write the D matrix for Gauss Siedel method.

$D=\left[\begin{array}{cccc}{A}_{11}& 0& 0& 0\\ A_{21}& A_{22}& 0& 0\\ ‐& ‐& \dots & 0\\ A_{N1}& A_{N2}& ‐& A_{NN}\end{array}\right]......\left(4\right)$

That is, for Gauss Siedel, D matrix consists of the lower triangular portion of matrix.

Therefore, for the iterative solution to linear algebraic equations $\mathrm{Ax}-\mathrm{y}$, the D matrix in the Jacobi method is the diagonal portion of A , whereas D for Gauss-Siedel is the lower triangular portion of A.