The Inverse of a Matrix

Division by a matrix is not defined but the equivalent is obtained from the inverse of the matrix.

If the product of two square matrices, $A*B$

, equals to the identity matrix, $I$

is said to be the inverse of $A$

(and also

is the inverse of $B$

By multiplying each element with its cofactor to find the inverse of the matrix is not useful since $N^3$

multiplication and division are required for an N-dimensional matrix.

To find the inverse of matrix $A$

, use an elimination method.

We augment the $A$

matrix with the identity matrix of the same size and solve. The solution is $A^{-1}$ . Example;

$\begin{indisplay} A=\left[ \begin{array}{rrr} 1 & -1 & 2 \\ 3 & 0 & 1 \\ 1 & 0 & 2 \\ \end{array} \right] \rightarrow \end{indisplay}$

$\begin{indisplay} \left[ \begin{array}{rrrrrr} 1 & -1 & 2 & 1 &0 & 0\\ 3 & 0 & 1 & 0 &1 & 0\\ 1 & 0 & 2 & 0 &0 & 1\\ \end{array} \right] \end{indisplay}$

$\begin{indisplay} \left\Vert \begin{array}{r} \\ R_2-(3/1)R_1 \rightarrow \\ R_3-(1/1)R_1 \rightarrow \\ \end{array} \right\Vert \end{indisplay}$

$\begin{indisplay} \left[ \begin{array}{rrrrrr} 1 & -1 & 2 & 1 &0 & 0\\ 0 & 3 &-5 &-3 &1 & 0\\ 0 & 1 & 0 &-1&0 & 1\\ \end{array} \right] \end{indisplay}$

$\begin{indisplay} \underbrace{\left[ \begin{array}{rrrrrr} 1 & -1 & 2 & 1 &0 & ... ... 0 & 3 &-5 &-3 &1 & 0\\ \end{array}\right]}_{Row~Interchange} \end{indisplay}$

$\displaystyle \left\Vert \begin{array}{r} \\ \\ R_3-(3/1)R_2 \rightarrow \\ \end{array} \right\Vert$

Contd.

$\begin{indisplay} \left[ \begin{array}{rrrrrr} 1 & -1 & 2 & 1 &0 & 0\\ 0 & 1 & 0 &-1&0 & 1\\ 0 & 0 &-5 & 0 &1 &-3\\ \end{array}\right] \end{indisplay}$

$\displaystyle \left\Vert \begin{array}{r} \\ \\ R_3/(-5) \rightarrow \\ \end{array} \right\Vert$

$\displaystyle \left\Vert \begin{array}{r} R_1-(2/1)R_3 \rightarrow \\ \\ \\ \end{array} \right\Vert$

$\begin{indisplay} \left[ \begin{array}{rrrrrr} 1 & -1 & 0 & 1 &2/5&-6/5\\ 0 & 1 & 0 &-1&0 & 1\\ 0 & 0 &1 & 0 &-1/5&3/5\\ \end{array}\right] \end{indisplay}$

$\displaystyle \left\Vert \begin{array}{r} \\ R_2-(1/-1)R_1 \rightarrow \\ \\ \end{array} \right\Vert$

$\begin{indisplay} \left[ \begin{array}{rrrrrr} 1 & 0 & 0 & 0 &2/5&-1/5\\ 0 & 1 & 0 &-1&0 & 1\\ 0 & 0 &1 & 0 &-1/5&3/5\\ \end{array}\right] \end{indisplay}$

We confirm the fact that we have found the inverse by multiplication:

$\begin{indisplay} \underbrace{\left[ \begin{array}{rrr} 1 & -1 & 2 \\ 3 & 0 & 1 \\ 1 & 0 & 2 \\ \end{array} \right]}_A* \end{indisplay}$

$\begin{indisplay} \underbrace{\left[ \begin{array}{rrr} 0 &2/5&-1/5\\ -1&0 & 1\\ 0 &-1/5&3/5\\ \end{array} \right]}_{A^{-1}}= \end{indisplay}$

$\begin{indisplay} \underbrace{\left[ \begin{array}{rrr} 1 & 0 & 0\\ 0 & 1 & 0 \\ 0 & 0 &1 \\ \end{array} \right]}_I \end{indisplay}$

It is more efficient to use Gaussian elimination. We show only the final triangular matrix; we used pivoting:

$\begin{indisplay} \left[ \begin{array}{rrrrrr} 1 & -1 & 2 & 1 &0 & 0\\ 3 & 0 ... ...\ (0.333) & (0) & 1.667 & 0 &-0.333 & 1\\ \end{array} \right] \end{indisplay}$

After doing the back-substitutions, we get

$\begin{indisplay} \left[ \begin{array}{rrrrrr} 3 & 0 & 1 & 0 &0.4& -0.2\\ (0.... ...\ (0.333) & (0) & 1.667 & 0 &-0.2 & 0.6\\ \end{array} \right] \end{indisplay}$

If we have the inverse of a matrix, we can use it to solve a set of equations, $Ax=b$

because multiplying by $A^{-1}$ gives the answer ( $x$

$\begin{indisplay} \begin{array}{l} A^{-1}Ax=A^{-1}b \\ x=A^{-1}b\\ \end{array}\end{indisplay}$

Phyton Code:

    1 import numpy as np
    2 A = np.array([[1.0,-1.0,2.0],[3.0,0.0,1.0],[1.0,0.0,2.0]])
    3 b = np.array([[1.0,0.0,0.0],[0.0,1.0,0.0],[0.0,0.0,1.0]])
    4 x = np.linalg.solve(A, b)
    5 print("NumPy - Inverse Matrix: \n",x)
    6 from scipy import linalg
    7 x=linalg.solve(A,b)
    8 print("SciPy - Inverse Matrix: \n",x)