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 $A = L U$ of the given matrix is:

$\begin{pmatrix} 1 & -1 & 4 \\ 2 & 9 & 8 \\ 6 & 5 & 2 \end{pmatrix} = \begin{pmatrix} 1 & 0 & 0\\ 2 & 1 & 0\\ 6 & 1 & 1 \end{pmatrix} \begin{pmatrix} 1 & -1 & 4\\ 0 & 11 & 0\\ 0 & 0 & -22 \end{pmatrix}$

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

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:

$\left(\begin{array}{ccc|ccc} 1 & 0 & 0 & 1 & 0 & 0\\ 2 & 1 & 0 & 0 & 1 & 0\\ 6 & 1 & 1 & 0 & 0 & 1 \end{array}\right) \sim (R_2 \rightarrow R_2- 2 R_1; R_3 \rightarrow R_3- 6 R_1) \sim$

$\left(\begin{array}{ccc|ccc} 1 & 0 & 0 & 1 & 0 & 0\\ 0 & 1 & 0 & -2 & 1 & 0\\ 0 & 1 & 1 & -6 & 0 & 1 \end{array}\right) \sim(R_3 \rightarrow R_3 - R_2) \sim \left(\begin{array}{ccc|ccc} 1 & 0 & 0 & 1 & 0 & 0\\ 0 & 1 & 0 & -2 & 1 & 0\\ 0 & 0 & 1 & -4 & -1 & 1 \end{array}\right)$

Inverse of U:

$\left(\begin{array}{ccc|ccc} 1 & -1 & 4 & 1 & 0 & 0\\ 0 & 11 & 0 & 0 & 1 & 0\\ 0 & 1 & -22 & 0 & 0 & 1 \end{array}\right) \sim (R_2 \rightarrow \frac{R_2}{11}; R_3 \rightarrow \frac{R_3}{22}) \sim \left(\begin{array}{ccc|ccc} 1 & -1 & 4 & 1 & 0 & 0\\ 0 & 1 & 0 & 0 & \frac{1}{11} & 0\\ 0 & 0 & 1 & 0 & 0 & -\frac{1}{22} \end{array}\right) \sim (R_1 \rightarrow R_1 + R_2 - 4R_3) \sim \left(\begin{array}{ccc|ccc} 1 & 0 & 0 & 1 & \frac{1}{11} & \frac{2}{11}\\ 0 & 1 & 0 & 0 & \frac{1}{11} & 0\\ 0 & 0 & 1 & 0 & 0 & -\frac{1}{22} \end{array}\right)$

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