Question : Apply LU decomposition method to solve 2x+4y+2z=1,x+5y+2z=1,4x-y+9z=2
Answer :
Writing system in the Matrix AX= b
\[\left[\begin{matrix}2&4&2\\1&5&2\\4&-1&9\\\end{matrix}\right]\left(\begin{matrix}x\\y\\z\\\end{matrix}\right)=\left(\begin{matrix}1\\1\\2\\\end{matrix}\right)\]
To find LU decomposition of A
\[A=\left[\begin{matrix}2&4&2\\1&5&2\\4&-1&9\\\end{matrix}\right]\]Step 1
A=LU
Where L is lower triangular matrix and U is upper triangular matrix respectively
\[L=\left[\begin{matrix}l_{11}&0&0\\l_{21}&l_{22}&0\\l_{31}&l_{32}&l_{33}\\\end{matrix}\right],
U=\left[\begin{matrix}u_{11}&u_{12}&u_{13}\\0&u_{22}&u_{23}\\0&0&u_{33}\\\end{matrix}\right]\]Therefore
\[A=LU=\left[\begin{matrix}l_{11}&0&0\\l_{21}&l_{22}&0\\l_{31}&l_{32}&l_{33}\\\end{matrix}\right],\left[\begin{matrix}u_{11}&u_{12}&u_{13}\\0&u_{22}&u_{23}\\0&0&u_{33}\\\end{matrix}\right]\]
\[\left[\begin{matrix}2&4&2\\1&5&2\\4&-1&9\\\end{matrix}\right]=\left[\begin{matrix}u_{11}&u_{12}&u_{13}\\l_{21}u_{11}&{l_{21}u_{12}+u}_{22}&l_{21}u_{13}+u_{23}\\l_{31}u_{11}&l_{31}u_{12}+l_{32}u_{22}&{l_{31}u_{13}+l_{32}u_{23}+u}_{33}\\\end{matrix}\right]\]
\[u_{11}=2,\ \ u_{12}=4,\ \ \ u_{13}=2\]
\[l_{21}u_{11}=1\ ,\ {l_{21}u_{12}+u}_{22}=5\ ,\ l_{21}u_{13}+u_{23}=2\ ,\ l_{31}u_{11}=4\ ,\ l_{31}u_{12}+l_{32}u_{22}=-1,\ {l_{31}u_{13}+l_{32}u_{23}+u}_{33}=9\]
Solving these equations we get
\[l_{11}=1,l_{21}=\frac{1}{2}\ ,l_{22}=10,l_{31}=2,l_{32}=-3,l_{33}=1\]Hence
\[L=\left[\begin{matrix}1&0&0\\\frac{1}{2}\ &10&0\\2&-3&1\\\end{matrix}\right],\ U=\left[\begin{matrix}2&4&2\\0&3&1\\0&0&8\\\end{matrix}\right]\]Hence \[\left[\begin{matrix}2&4&2\\1&5&2\\4&-1&9\\\end{matrix}\right]=\left[\begin{matrix}1&0&0\\\frac{1}{2}\ &10&0\\2&-3&1\\\end{matrix}\right]*\left[\begin{matrix}2&4&2\\0&3&1\\0&0&8\\\end{matrix}\right] \]is the required LU decomposition of A.
Now to solve the equation Ax = b
\[AX=b\]\[\left[\begin{matrix}1&0&0\\\frac{1}{2}\ &10&0\\2&-3&1\\\end{matrix}\right]*\left[\begin{matrix}2&4&2\\0&3&1\\0&0&8\\\end{matrix}\right]\left(\begin{matrix}x\\y\\z\\\end{matrix}\right)=\left(\begin{matrix}1\\1\\2\\\end{matrix}\right)\]
Let \[\left[\begin{matrix}2&4&2\\0&3&1\\0&0&8\\\end{matrix}\right]\left(\begin{matrix}x\\y\\z\\\end{matrix}\right)=\left(\begin{matrix}y_1\\y_2\\y_3\\\end{matrix}\right)=Y\]Now we shall first solve
\[\left[\begin{matrix}1&0&0\\\frac{1}{2}\ &10&0\\2&-3&1\\\end{matrix}\right]\left(\begin{matrix}y_1\\y_2\\y_3\\\end{matrix}\right)=\left(\begin{matrix}1\\1\\2\\\end{matrix}\right)\]Solving this matrix problem we obtained
\[y_1=1\ ,\ y_2=\frac{1}{2}\ ,y_3=\frac{3}{2}\]Now solving UX=Y
\[\left[\begin{matrix}2&4&2\\0&3&1\\0&0&8\\\end{matrix}\right] \left(\begin{matrix}x\\y\\z\\\end{matrix}\right)=\left(\begin{matrix}1\\\frac{1}{2}\\\frac{3}{2}\\\end{matrix}\right)\]By solving this linear system we obtained
\[x=\frac{5}{48}\ ,y=\frac{5}{48}\ ,\ z=\frac{3}{16}\]Hence required solution of the given system of equations is
\[\left(\begin{matrix}x\\y\\z\\\end{matrix}\right)=\left(\begin{matrix}\frac{5}{48}\\\frac{5}{48}\\\frac{3}{16}\\\end{matrix}\right)\]