Question

Find the inverse of the matrix


1 −1 4
2 9 8
6 5 2

 using LU decomposition method.

Answer 1

The LU decomposition $A=LU$  of the given matrix is:

$A=\begin{bmatrix} 1 & -1&4 \\ 2 &9&8\\ 6&5&2 \end{bmatrix}= \begin{bmatrix} 1 & 0&0 \\ 2 & 1&0\\ 6&a&1 \end{bmatrix}\begin{bmatrix} 1 & -1&4 \\ 0 & b&0\\ 0&0&c \end{bmatrix}$

-2+b=9 -6+11a=5 24+c=2

b=11 a=1 c=-22

$A= \begin{bmatrix} 1 & 0&0 \\ 2 & 1&0\\ 6&1&1 \end{bmatrix}\begin{bmatrix} 1 & -1&4 \\ 0 & 11&0\\ 0&0&-22 \end{bmatrix}=LU$

Hence, the inverse of the matrix is $A^{-1}=U^{-1}L^{-1}$

Since U and L are already in upper and lower triangular form, it is easy to find their inverses using Gauss-Jordan elimination.

Inverse of L

$\begin{bmatrix} 1 & 0&0 |1&0&0\\ 2 &1&0|0&1&0\\ 6&1&1|0&0&1 \end{bmatrix}$

IIr+Ir(-2)

IIIr+Ir(-6)

$\begin{bmatrix} 1 &0&0|\quad1&0&0\\ 0 &1&0|-2&1&0\\ 0&1&1|-6&0&1 \end{bmatrix}$

IIIr+IIr(-1)

$\begin{bmatrix} 1 &0&0 |1&0&0\\ 0 &1&\quad0 |-2&1&0\\ 0&0&\quad1|-4&-1&1 \end{bmatrix}$

Inverse of U:

$\begin{bmatrix} 1 & -1&\quad4 |1&0&0\\ 0 &11&\quad0 |0&1&0\\ 0&1&-22|0&0&1 \end{bmatrix}$

IIr$\cdot(\frac{1}{11})$

$\begin{bmatrix} 1 & -1&\quad4 |1&0&0\\ 0 &1&\quad0 |0&\frac{1}{11}&0\\ 0&1&-22|0&0&1 \end{bmatrix}$

IIIr+IIr(-1)

$\begin{bmatrix} 1 & -1&\quad4 |\frac{1}{11}&-\frac{1}{11}&\frac{2}{11}\\ 0 &1&\quad0 |0&\frac{1}{11}&0\\ 0&0&-22|0&-\frac{1}{11}&1 \end{bmatrix}$

IIIr$\cdot(-\frac{1}{22})$

Ir+IIr+IIIr(-4)

$\begin{bmatrix} 1 & 0&\quad0 |1&\frac{1}{11}&\frac{2}{11}\\ 0 &1&\quad0 |0&\frac{1}{11}&0\\ 0&0&1|0&0&-\frac{1}{22} \end{bmatrix}$

Hence,

$A^{-1}=U^{-1}L^{-1}=\\=\begin{bmatrix} 1&\frac{1}{11}&\frac{2}{11}\\ 0&\frac{1}{11}&0\\ 0&0&-\frac{1}{22} \end{bmatrix} \begin{bmatrix} 1&0&0\\ -2&1&0\\ -4&-1&1 \end{bmatrix}=\\ = \begin{bmatrix} \frac{1}{11}&-\frac{1}{11}&\frac{2}{11}\\ -\frac{2}{11}&\frac{1}{11}&0\\ \frac{2}{11}&\frac{1}{22}&-\frac{1}{22} \end{bmatrix}$