GATE 2023 | Maths | Problems for DA Paper

Question

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 \(\text{trace}(U) = 3\).

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

Then
 

(A) both 𝑃 and 𝑄 are TRUE

(B) 𝑃 is FALSE and 𝑄 is TRUE

(C) 𝑃 is TRUE and 𝑄 is FALSE

(D) both 𝑃 and 𝑄 are FALSE

Answer 1

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

When a matrix is decomposed into LU form, the diagonal elements of the lower triangular matrix \(L\) are always set to 1 to facilitate the forward elimination process. In this case, if all diagonal elements of \(L\) are 1, then the diagonal elements of \(M\) will be identical to the diagonal elements of the upper triangular matrix \(U\). Therefore, the trace of \(U\) (sum of diagonal elements) will be equal to the trace of \(M\). The trace of \(M\) is \(2 + 3 + 1 = 6\). So, statement \(P\) is FALSE.

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

The Jacobi iteration method only guarantees convergence for diagonally dominant matrices, meaning the absolute value of any diagonal element must be greater than the sum of the absolute values of the other elements in its row. Let's check if \(M\) is diagonally dominant:
\[ |2| > |-1| + |-4| \]
\[ |3| > |-1| \]
Both conditions are met. Therefore, statement \(Q\) is TRUE.

Therefore, the correct answer is (C) \(P\) is FALSE and \(Q\) is TRUE.