Question

3.10. This problem considers the question, why not use UL instead of LU? You can assume that A is a 2 ×2 matrix. (a) Find a Doolittle version of the factorization A = UL, where L is a unit lower triangular matrix and U is upper triangular. You can assume pivoting is not necessary. (b) Describe the resulting algorithm for solving Ax = b. (c) Is there any connection between the Doolittle factorization of AT and the one you found in part (a)?

Answer 1

Answer:

a

The Doolittle version of factorization A =UL is

                     $\left[\begin{array}{ccc}a_{11}&a_{12\\a_{21}&a_{22}\end{array}\right] =\left[\begin{array}{ccc}a_{11}-a_{12}*\frac{a_{21}}{a_{22}} &a_{12}\\0&a_{22}\end{array}\right]\left[\begin{array}{ccc}1&0\\\frac{a_{21}}{a_{22}} &1\end{array}\right]$

b

           Algorithm

1)  Obtain the values of the individual elements of U and L

            $U_{22} = a_{22}$

            $l_{21} = \frac{a_[21} {U_{22}}}$

            $U_{12} = a_{12}$

            $U_{11} = a_{11} - U_{12}*l_{21}$

2) Obtain the value of z for the expression $Uz = b$ 1.e

                       $\left[\begin{array}{ccc}U_{11}&U_{12\\0&U_{22}\end{array}\right] \left[\begin{array}{ccc}z_1\\z_2\end{array}\right] =\left[\begin{array}{ccc}b_1\\b_2\end{array}\right]$

 $z_2 =\frac{b_2}{U_22}$

  $z_1 = \frac{1}{U_[11}} [b_1 - U_{12}\ z_2]$

 3   Obtain the value of x for the expression $Lx = z$ 1.e

              $\left[\begin{array}{ccc}1&0\\l_{21}&1\end{array}\right] \left[\begin{array}{ccc}x_1\\x_2\end{array}\right] = \left[\begin{array}{ccc}z_1\\z_2\end{array}\right]$

       $x_1 = z_1$

        $x_2 = z_2 -l_{21}x_1$

c

  Let say that  $A^T =LU$ is the Doolittle factorization of $A^T$

where L is a unit lower triangular matrix and U is upper triangular matrix

     Hence $A = U^T L^T$

Here  $L^T$ is a unit upper  triangular matrix and $U^T$ lower is triangular  

In the question that we are given  A = UL meaning that  $A^T = L^TU^T$ looking at this  U is upper triangular matrix and L is unit lower triangular matrix. So we can say that the Doolittle version given behave like Doolittle factorization only for A and there is no connection with $A^T$

Step-by-step explanation:

From the question let assume that          $A =\left[\begin{array}{ccc}a_{11}&a_{12\\a_{21}&a_{22}\end{array}\right]$

The objective is to find the Doolittle version of the factorized  A = UL and this can be expressed in matrix as

                          $\left[\begin{array}{ccc}a_{11}&a_{12\\a_{21}&a_{22}\end{array}\right] =\left[\begin{array}{ccc}U_{11}&U_{12\\0&U_{22}\end{array}\right]\left[\begin{array}{ccc}1&0\\l_{21}&1\end{array}\right]$

Looking at this in form of matrix multiplication

                             $a_{11} =U_{11} +U_{12} l_{21}$

                             $a_{12} = U_{12}$

                            $a_{21} = U_{22} l_{21}$

                            $a_{22} = U_{22}$

           =>      $l_{21} = \frac{a_{21}}{U_{22}} = \frac{a_{21}}{a_{22}}$

    and           $u_{11} = a_{11} - U_{12}l_{21} = a_{11} -a_{12} *\frac{a_{21}}{a_{22}}$

   Hence            $U = \left[\begin{array}{ccc}a_{11}-a_{12}*\frac{a_{21}}{a_{22}} &a_{12}\\0&a_{22}\end{array}\right]$

                       

                          $L = \left[\begin{array}{ccc}1&0\\\frac{a_{21}}{a_{22}} &1\end{array}\right]$

 b)   From the question we are given

                 $Ax =b$

And objective is to describe the algorithm to solve it

                $Ax =b$

                $UL x = b$

Let assume that $z = Lx$

               So

                           $U z = b$

   Given that  U is upper triangular  matrix then y can be obtained by back substitution

       Now that we are able to obtain y, given that  L is a unit lower triangular matrix we can obtain x by forward substitution

The algorithm is given in the answer section