1. Consider the following matrix A = \begin{bmatrix} 1 & -1 \\ -1 & 5 \end{bmatrix} (a) (5 points) Demonstrate that A admits a Cholesky decomposition. (b) (5 points) Derive Cholesky decomposition of A. (Note: imaginary numbers are possible.) 2. (10 points) Consider the following matrix A = \begin{bmatrix} 1 & -1 & 3 \\ -1 & 2 & 5 \\ 3 & 5 & 1 \end{bmatrix} Given the elementary matrices G_1 = \begin{bmatrix} 1 & 0 & 0 \\ 1 & 1 & 0 \\ -3 & 0 & 1 \end{bmatrix}, G_2 = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & -8 & 1 \end{bmatrix}, and G = G_2G_1 = \begin{bmatrix} 1 & 0 & 0 \\ 1 & 1 & 0 \\ -11 & -8 & 1 \end{bmatrix}, check that the matrix GA is upper triangular and use this to find the LU decomposition of A.
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To determine if A admits Cholesky decomposition, we need to check if A is symmetric and positive definite. Symmetry: A matrix is symmetric if it is equal to its transpose. In this case, A = A^T, so A is symmetric. Positive Definite: A matrix is positive definite Show more…
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