Question

LR-Zerlegung dieser 4x4-Matrix:

\( \begin{pmatrix} 1 & 3 & 2 & -1 \\ -1 & 2 & -2 & -1 \\ 1 & 1 & 0 & -1 \\ 3 & 0 & -1 & -1,5 \end{pmatrix} \)

Answer 1

T(a,b) : Tauschmatrix Zeile/Spalte a,b 

An = Pivotelement a_nn betragsgrößtes Spalten-Element

p1:=SPvt(A,1)=4
A1:= T(1,p1) A
A2:=L1 T(1,p1) A =
\(\small \, \left(\begin{array}{rrrr}1&0&0&0\\\frac{1}{3}&1&0&0\\-\frac{1}{3}&0&1&0\\\frac{1}{3}&0&0&1\\\end{array}\right) T(1,p1)\cdot A= \, \left(\begin{array}{rrrr}3&0&-2&\frac{3}{2}\\0&2&-\frac{8}{3}&-\frac{1}{2}\\0&1&\frac{2}{3}&-\frac{3}{2}\\0&3&\frac{4}{3}&-\frac{1}{2}\\\end{array}\right)\)

p2:=SPvt(A2,2)=4
A3:=L2 T(2, p2) (L1 T(1,p1) A) =

\(\small  \, \left(\begin{array}{rrrr}1&0&0&0\\0&1&0&0\\0&-\frac{1}{3}&1&0\\0&-\frac{2}{3}&0&1\\\end{array}\right)\; T(2, p2) \; (L1\;T(1,p1)\;A)= \, \left(\begin{array}{rrrr}3&0&-2&\frac{3}{2}\\0&3&\frac{4}{3}&-\frac{1}{2}\\0&0&\frac{2}{9}&-\frac{4}{3}\\0&0&-\frac{32}{9}&-\frac{1}{6}\\\end{array}\right)\)

p3:=SPvt(A3,3)=4   
R:=L3 T(3, p3) L2 T(2, p2) L1 T(1, p1) A=

\(\small \, \left(\begin{array}{rrrr}1&0&0&0\\0&1&0&0\\0&0&1&0\\0&0&\frac{1}{16}&1\\\end{array}\right) T(3, p3)\; L2 \;T(2, p2)\; L1\; T(1, p1)\; A=   \, \left(\begin{array}{rrrr}3&0&-2&\frac{3}{2}\\0&3&\frac{4}{3}&-\frac{1}{2}\\0&0&-\frac{32}{9}&-\frac{1}{6}\\0&0&0&-\frac{43}{32}\\\end{array}\right)\)

Abgleich

R:={L3}  {T(3, p3) L2} {T(2, p2) L1} {T(1, p1)} A
R={L_3}        {L_2}             {L_1}           {P} A =
{L3} {T(3, p3) L2 T(3, p3)} {T(3, p3) T(2, p2) L1 T(2, p2) T(3, p3)} {T(3, p3) T(2, p2) T(1, p1)} A

L:=( L3 T(3, p3) L2 T(3, p3) T(3, p3) T(2, p2) L1 T(2, p2) T(3, p3) )^(-1)=
\(\small L \, :=  \, \left(\begin{array}{rrrr}1&0&0&0\\-\frac{1}{3}&1&0&0\\-\frac{1}{3}&\frac{2}{3}&1&0\\\frac{1}{3}&\frac{1}{3}&-\frac{1}{16}&1\\\end{array}\right)\)

L R = P A = {T(3, p3) T(2, p2) T(1, p1)} A