Question:medium

Which of the following statements is/are true?
(I) Gauss-Seidel method for solving a linear system converges faster than the Gauss-Jacobi method.
(II) In the Gauss-Jordan method for solving a linear system, the coefficient matrix is transformed into an upper triangular matrix.

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Recall that Gauss-Jordan eliminates entries above the pivot too, going all the way to a diagonal matrix.
Updated On: Jul 3, 2026
  • Only (I)
  • Only (II)
  • Both (I) and (II)
  • Neither (I) nor (II)
Show Solution

The Correct Option is A

Solution and Explanation

This can be confirmed using the matrix splitting used to define each method. Write $A = D - L - U$, where $D$ is diagonal and $L, U$ contain the strictly lower and upper parts of $A$ with sign reversed.
The Jacobi iteration matrix is $T_J = D^{-1}(L+U)$, while the Gauss-Seidel iteration matrix is $T_{GS} = (D-L)^{-1}U$. For a wide class of matrices met in practice, such as tridiagonal or consistently ordered matrices, it can be shown that $\rho(T_{GS}) = \rho(T_J)^2 < \rho(T_J)$ whenever $\rho(T_J) < 1$, where $\rho(\cdot)$ denotes the spectral radius. A smaller spectral radius means the error shrinks faster per iteration, so Gauss-Seidel converges faster, confirming statement (I).
For statement (II), track what Gauss-Jordan elimination does to a coefficient matrix $A$. At each pivot step $k$, it does not just zero out entries below the pivot as Gaussian elimination does, it also zeroes out the entries above the pivot in that column. After all $n$ pivot steps on an $n\times n$ matrix, every off-diagonal entry has been eliminated, leaving a diagonal matrix, not an upper triangular one. So statement (II) is false, giving the same conclusion by an independent route.
\[\boxed{\text{Only (I)}}\]
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