Consider the linear system 𝑀𝑥 = 𝑏, where 𝑀 = \(\begin{bmatrix}2&-1\\\ -4&3\end{bmatrix}\) and b = \(\begin{bmatrix}-2\\\ 5\end{bmatrix}\).

Suppose 𝑀 = 𝐿𝑈, where 𝐿 and U are lower triangular and upper triangular square matrices, respectively. Consider the following statements:

𝑃: If each element of the main diagonal of 𝐿 is 1, then 𝑡𝑟𝑎𝑐𝑒(𝑈) = 3.

𝑄: For any choice of the initial vector 𝑥⁽⁰⁾ , the Jacobi iterates 𝑥^(𝑘) , 𝑘 = 1,2,3 … converge to the unique solution of the linear system 𝑀𝑥 = 𝑏.

Then 

Concept:

A square matrix (a_ij) is called a diagonally dominant matrix if |a_ii| ≥ \(\sum_{j\neq i}|a_{ij}|\) for all i

Explanation:

For P,

M = \(\begin{bmatrix}2&-1\\\ -4&3\end{bmatrix}\)

Let M = LU, where 𝐿 and U are lower triangular and upper triangular square matrices

Each element of the main diagonal of 𝐿 is 1

Let L = \(\begin{bmatrix}1&0\\a&1\end{bmatrix}\) and U = \(\begin{bmatrix}b&c\\0&d\end{bmatrix}\) 

Then

\(\begin{bmatrix}2&-1\\\ -4&3\end{bmatrix}\) = \(\begin{bmatrix}1&0\\a&1\end{bmatrix}\)\(\begin{bmatrix}b&c\\0&d\end{bmatrix}\)

⇒ \(\begin{bmatrix}2&-1\\\ -4&3\end{bmatrix}\) = \(\begin{bmatrix}b&c\\ab&ac+d\end{bmatrix}\)

Comparing both sides

b = 2, c = -1, ab = -4 and ac + d = 3

Substituting b = 2 in ab = -4 we get a = -2

Again substituting a = -2 and c = -1 in ac + d = 3 we get

(-2)(-1) + d = 3 ⇒  2 + d = 3 ⇒ d = 1

So U = \(\begin{bmatrix}2&-1\\0&1\end{bmatrix}\)

Hence trace(U) = 1 + 2 = 3

P is TRUE

For Q,

M = \(\begin{bmatrix}2&-1\\\ -4&3\end{bmatrix}\)

M is not a diagonally dominant matrix as 3 \(\ngeq\) |-4|

Then H_Jacobi = D^-1(L + U)  where

D is diagonal matric ie..e, \(\begin{bmatrix}2&0\\\ 0&3\end{bmatrix}\) and L + U = \(\begin{bmatrix}0&-1\\\ -4&0\end{bmatrix}\)

So, D^-1 = \(\begin{bmatrix}\frac12&0\\\ 0&\frac13\end{bmatrix}\)

So  HJacobi = \(\begin{bmatrix}\frac12&0\\\ 0&\frac13\end{bmatrix}\)\(\begin{bmatrix}0&-1\\\ -4&0\end{bmatrix}\) = \(\begin{bmatrix}0&-\frac12\\\ -\frac43&0\end{bmatrix}\)

Hence eigenvalues re given by

λ² - 0λ - 2/3 = 0 

⇒ λ = \(\pm\sqrt{\frac23}\)

Since |λ| < 1

Hence for any choice of the initial vector 𝑥(0) , the Jacobi iterates 𝑥(𝑘) , 𝑘 = 1,2,3 … converge to the unique solution of the linear system 𝑀𝑥 = 𝑏.

Q is TRUE

both 𝑃 and 𝑄 are TRUE

(1) is correct