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Chapter 6: Problem 4

Use the \(L D L^{t}\) Factorization Algorithm to find a factorizaton of the form \(A=L D L^{t}\) for the following matrices: a. \(A=\left[\begin{array}{rrr}2 & -1 & 0 \\ -1 & 2 & -1 \\ 0 & -1 & 2\end{array}\right]\) b. \(\quad A=\left[\begin{array}{rrrr}4 & 1 & 1 & 1 \\ 1 & 3 & -1 & 1 \\ 1 & -1 & 2 & 0 \\ 1 & 1 & 0 & 2\end{array}\right]\) c. \(\quad A=\left[\begin{array}{rrrr}4 & 1 & -1 & 0 \\ 1 & 3 & -1 & 0 \\ -1 & -1 & 5 & 2 \\ 0 & 0 & 2 & 4\end{array}\right]\) d. \(\quad A=\left[\begin{array}{rrrr}6 & 2 & 1 & -1 \\ 2 & 4 & 1 & 0 \\ 1 & 1 & 4 & -1 \\ -1 & 0 & -1 & 3\end{array}\right]\)

Short Answer

Expert verified
The \(A=L D L^{t}\) factorization for the given matrices are as stated in the content.

Step by step solution

01

Solving for A part (a)

By applying the Cholesky-Decomposition: The first step - Put \(a_{11}\) in \(d_{11}\).And set \(l_{11}\) = 1 by normalizing the factor. \(a_{21}\) and \(a_{31}\)stays in \(d_{21}\) and \(d_{31}\) The rest of the column goes in L. The mixed terms go in L. The result for part (a) is\[D=\left[\begin{array}{ccc}2 & 0 & 0 \0 & 4/3 & 0 \0 & 0 & 4/3\end{array}\right], L=\left[\begin{array}{ccc}1 & -1/2 & 0 \0 & 1 & -3/4 \0 & 0 & 1\end{array}\right]\]
02

Solving for A part (b)

Using same principle, the result for part (b) is\[D=\left[\begin{array}{cccc}4 & 0 & 0 & 0 \0 & 8/3 & 0 & 0 \0 & 0 & 4/3 & 0 \0 & 0 & 0 & 8/5\end{array}\right], L=\left[\begin{array}{cccc}1 & 1/4 & 1/4 & 1/4 \0 & 1 & -1/2 & 1/2 \0 & 0 & 1 & 1/2 \0 & 0 & 0 & 1\end{array}\right]\]
03

Solving for A part (c)

The result for part (c) is\[D=\left[\begin{array}{cccc}4 & 0 & 0 & 0 \0 & 44/15 & 0 & 0 \0 & 0 & 82/15 & 0 \0 & 0 & 0 & 4\end{array}\right], L=\left[\begin{array}{cccc}1 & 1/4 & -1/4 & 0 \0 & 1 & -1 & 1/3 \0 & 0 & 1 & 1/2 \0 & 0 & 0 & 1\end{array}\right]\]
04

Solving for A part (d)

The result for part (d) is\[D=\left[\begin{array}{cccc}6 & 0 & 0 & 0 \0 & 16/3 & 0 & 0 \0 & 0 & 18/5 & 0 \0 & 0 & 0 & 6\end{array}\right],L=\left[\begin{array}{cccc}1 & 2/3 & 1/3 & -1/3 \0 & 1 & 5/8 & 1/4 \0 & 0 & 1 & -1/2 \0 & 0 & 0 & 1\end{array}\right]\]

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Most popular questions from this chapter

Given the linear system $$ \begin{aligned} x_{1}-x_{2}+\alpha x_{3} &=-2 \\ -x_{1}+2 x_{2}-\alpha x_{3} &=3 \\ \alpha x_{1}+x_{2}+x_{3} &=2 \end{aligned} $$ a. Find value(s) of \(\alpha\) for which the system has no solutions. b. Find value(s) of \(\alpha\) for which the system has an infinite number of solutions. c. Assuming a unique solution exists for a given \(\alpha\), find the solution.

In Section \(3.6\) we found that the parametric form \((x(t), y(t))\) of the cubic Hermite polynomials through \((x(0), y(0))=\left(x_{0}, y_{0}\right)\) and \((x(1), y(1))=\left(x_{1}, y_{1}\right)\) with guide points \(\left(x_{0}+\alpha_{0}, y_{0}+\beta_{0}\right)\) and \(\left(x_{1}-\alpha_{1}, y_{1}-\right)\) \(\beta_{1}\) ), respectively, are given by $$ x(t)=\left(2\left(x_{0}-x_{1}\right)+\left(\alpha_{0}+\alpha_{1}\right)\right) t^{3}+\left(3\left(x_{1}-x_{0}\right)-\alpha_{1}-2 \alpha_{0}\right) t^{2}+\alpha_{0} t+x_{0} $$ and $$ y(t)=\left(2\left(y_{0}-y_{1}\right)+\left(\beta_{0}+\beta_{1}\right)\right) t^{3}+\left(3\left(y_{1}-y_{0}\right)-\beta_{1}-2 \beta_{0}\right) t^{2}+\beta_{0} t+y_{0} $$ The B茅zier cubic polynomials have the form $$ \hat{x}(t)=\left(2\left(x_{0}-x_{1}\right)+3\left(\alpha_{0}+\alpha_{1}\right)\right) t^{3}+\left(3\left(x_{1}-x_{0}\right)-3\left(\alpha_{1}+2 \alpha_{0}\right)\right) t^{2},+3 \alpha_{0} t+x_{0} $$ and $$ \hat{y}(t)=\left(2\left(y_{0}-y_{1}\right)+3\left(\beta_{0}+\beta_{1}\right)\right) t^{3}+\left(3\left(y_{1}-y_{0}\right)-3\left(\beta_{1}+2 \beta_{0}\right)\right) t^{2}+3 \beta_{0} t+y_{0} $$ a. Show that the matrix $$ A=\left[\begin{array}{rrrr} 7 & 4 & 4 & 0 \\ -6 & -3 & -6 & 0 \\ 0 & 0 & 3 & 0 \\ 0 & 0 & 0 & 1 \end{array}\right] $$ transforms the Hermite polynomial coefficients into the B茅zier polynomial coefficients. b. Determine a matrix \(B\) that transforms the B茅zier polynomial coefficients into the Hermite polynomial coefficients.

Find all values of \(\alpha\) so that the following linear system has no solutions. $$ \begin{aligned} 2 x_{1}-x_{2}+3 x_{3} &=5 \\ 4 x_{1}+2 x_{2}+2 x_{3} &=6 \\ -2 x_{1}+\alpha x_{2}+3 x_{3} &=4 \end{aligned} $$

Use the Gaussian Elimination Algorithm to solve the following linear systems, if possible, and determine whether row interchanges are necessary: a. \(\begin{aligned} x_{2}-2 x_{3} &=4 \\ x_{1}-x_{2}+x_{3} &=6 \\\ x_{1}-x_{3} &=2 . \end{aligned}\) b. \(\begin{aligned} \quad x_{1}-\frac{1}{2} x_{2}+x_{3} &=4, \\ 2 x_{1}-x_{2}-x_{3}+x_{4} &=5, \\ x_{1}+x_{2}+\frac{1}{2} x_{3} &=2, \\\ x_{1}-\frac{1}{2} x_{2}+x_{3}+x_{4} &=5 . \end{aligned}\) c. \(\begin{aligned} 2 x_{1}-x_{2}+x_{3}-x_{4} &=6, \\ x_{2}-x_{3}+x_{4} &=5, \\\ x_{4} &=5 \\ x_{3}-x_{4} &=3 . \end{aligned}\) d. \(\begin{aligned} x_{1}+x_{2}+x_{4} &=2, \\ 2 x_{1}+x_{2}-x_{3}+x_{4} &=1, \\\\-x_{1}+2 x_{2}+3 x_{3}-x_{4} &=4, \\ 3 x_{1}-x_{2}-x_{3}+2 x_{4} &=-3 . \end{aligned}\)

Find all \(\alpha>0\) and \(\beta>0\) so that the matrix $$ A=\left[\begin{array}{lll} 3 & 2 & \beta \\ \alpha & 5 & \beta \\ 2 & 1 & \alpha \end{array}\right] $$ is strictly diagonally dominant.
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