91Ó°ÊÓ

First, we need to remind ourselves of the Doolittle's algorithm. The algorithm can be described as follows: 1. Find L and U such that A = L * U, where L is a lower triangular matrix with ones on its diagonal, and U is an upper triangular matrix. 2. For each element a_ij in A, find the corresponding l_ij and u_ij according to the following formulas: - \(u_{ij} = a_{ij} - \sum_{k=1}^{i-1} l_{ik}u_{kj}\) for all j ≥ i - \(l_{ij} = \frac{a_{ij} - \sum_{k=1}^{j-1} l_{ik}u_{kj}}{u_{jj}}\) for all j < i Now, let us follow these steps.

Following the Doolittle's algorithm, we will calculate the elements of L and U matrices one by one. We have 8 unknowns to calculate (4 elements in L and 4 in U). The calculations for these unknowns are as follows: For i = 1: \(u_{11} = a_{11} = 2\) For i = 2: \(l_{21} = \frac{a_{21}}{u_{11}} = -\frac{1}{2}; u_{22} = a_{22} - l_{21}u_{12} = 2 - (-\frac{1}{2})(-1) = \frac{3}{2}\) For i = 3: \(l_{31} = 0, l_{32} = \frac{a_{32}}{u_{22}} = -\frac{2}{3}; u_{33} = a_{33} - l_{32}u_{23} = 2 - (-\frac{2}{3})(-1) = \frac{4}{3}\) For i = 4: \(l_{41} = 0, l_{42} = 0, l_{43} = \frac{a_{43}}{u_{33}} = -\frac{3}{4}; u_{44} = a_{44} - l_{43}u_{34} = 2 - (-\frac{3}{4})(-1) = \frac{5}{4}\) Now, we can construct the L and U matrices: \(L = \left(\begin{array}{cccc} 1 & 0 & 0 & 0 \\ -\frac{1}{2} & 1 & 0 & 0 \\ 0 & -\frac{2}{3} & 1 & 0 \\ 0 & 0 & -\frac{3}{4} & 1 \end{array}\right)\) \(U = \left(\begin{array}{cccc} 2 & -1 & 0 & 0 \\ 0 & \frac{3}{2} & -1 & 0 \\ 0 & 0 & \frac{4}{3} & -1 \\ 0 & 0 & 0 & \frac{5}{4} \end{array}\right)\)

Since we have calculated the LU factorization, we can check the pivots of the U matrix, which are its diagonal elements, to see if A is positive definite. All the pivots \(u_{11}, u_{22}, u_{33}\), and \(u_{44}\) are positive, which implies that A is positive definite.

Using Doolittle's algorithm, we have calculated the LU factorization of matrix A: \(L = \left(\begin{array}{cccc} 1 & 0 & 0 & 0 \\ -\frac{1}{2} & 1 & 0 & 0 \\ 0 & -\frac{2}{3} & 1 & 0 \\ 0 & 0 & -\frac{3}{4} & 1 \end{array}\right)\) \(U = \left(\begin{array}{cccc} 2 & -1 & 0 & 0 \\ 0 & \frac{3}{2} & -1 & 0 \\ 0 & 0 & \frac{4}{3} & -1 \\ 0 & 0 & 0 & \frac{5}{4} \end{array}\right)\) Since all the pivots of the U matrix are positive, we can conclude that A is a positive definite.