Latest revision as of 11:32, 11 December 2012

Inverse of a Matrix

An n x n matrix A is said to have an inverse provided there exists an n x n matrix B such that AB = BA = I_n. We call B the inverse of A and denote it as A^-1. Thus, AA^-1 = A^-1A = I_n. In this case, A is also called nonsingular.

Example.

$ A = \left(\begin{array}{cc}4&3\\3&2\end{array}\right) $
$ A^{-1} = \left(\begin{array}{cc}-2&3\\3&-4\end{array}\right) $

$ AA^{-1} = \left(\begin{array}{cc}4&3\\3&2\end{array}\right) $$ \left(\begin{array}{cc}-2&3\\3&-4\end{array}\right) = $ $ \left(\begin{array}{cc}1&0\\0&1\end{array}\right) $

and
$ A^{-1} = \left(\begin{array}{cc}-2&3\\3&-4\end{array}\right) $$ \left(\begin{array}{cc}4&3\\3&2\end{array}\right) = $$ \left(\begin{array}{cc}1&0\\0&1\end{array}\right) $

Theorem 1

The inverse of a matrix, if it exists, is unique

Theorem 2

If A and B are both nonsingular n x n matrices (i.e. invertible), then AB is nonsingular and (AB^)-1 = B^-1A^-1.

Corollary 1

If A₁, A₂, ..., A_r are n x n nonsingular matrices, then A₁A₂...A_r is nonsingular an (A₁A₂...A_r)^-1 = A_r^-1A_r-1^-1...A₁^-1.

Theorem 3

If A is a nonsingular matrix, then A^-1 is nonsingular and (A^-1)^-1 = A.

Theorem 4

If A is a nonsingular matrix, then A^T is nonsingular and (A^-1)^T = (A^T)^-1.

List of Equivalences

The following are equivalent for an n x n matrix A:
1. A is nonsingular.
2. Ax = 0 has only the trivial solution.
3. A is row(column equivalent to I. (The reduced row echelon form of A is I).
4. The linear system Ax = b has a unique solution for every n x 1 matrix b.
5. A is a product of elementary matrices.

Methods for determining the inverse of a matrix

1. Shortcut for determining the inverse of a 2 x 2 matrix

If $ A = \left(\begin{array}{cc}a&b\\c&d\end{array}\right) $ then the inverse of matrix A can be found using:
$ A^{-1} = \frac{1}{detA}\left(\begin{array}{cc}d&-b\\-c&a\end{array}\right) = \frac{1}{ad - bc}\left(\begin{array}{cc}d&-b\\-c&a\end{array}\right) $

Example

$ A = \left(\begin{array}{cc}1&2\\3&4\end{array}\right) $
$ detA = ad - bc = 1 \times 4 - 2 \times 3 = -2 $
$ A^{-1} = \frac{1}{-2}\left(\begin{array}{cc}4&-2\\-3&1\end{array}\right) $
$ A^{-1} = \left(\begin{array}{cc}-2&1\\\frac{3}{2}&\frac{-1}{2}\end{array}\right) $

2. Augmented Matrix Method (Can be used for any n x n matrix)

Use Gauss-Jordan Elimination to transform [ A | I ] into [ I | A^-1 ].

$ \left(\begin{array}{ccc|ccc}a&b&c&1&0&0\\d&e&f&0&1&0\\g&h&i&0&0&1\end{array}\right) $

Perform elementary row operations to transform the A matrix into the identity matrix. Whatever row operations are carried out on the A matrix, should also be carried out on the augmented Identity matrix. The final state of the matrix on the left is the inverse of the A matrix.

Theorem 1

An n x n matrix A is singular if and only if A is row equivalent to a matrix B that has a row of zeros. (That is, the reduced row echelon form of A has a row of zeros.)

Theorem 2

If A and B are n x n matrices such that AB = I, then BA = I. Thus B = A^-1.

Example. Find A^-1 of
$ A = \left(\begin{array}{cc}1&2\\3&4\end{array}\right) $

$ \left(\begin{array}{cc|cc}1&2&1&0\\3&4&0&1\end{array}\right)\longrightarrow\left(\begin{array}{cc|cc}1&2&1&0\\0&-2&-3&1\end{array}\right)\longrightarrow\left(\begin{array}{cc|cc}1&2&1&0\\0&1&\frac{3}{2}&\frac{-1}{2}\end{array}\right)\longrightarrow \left(\begin{array}{cc|cc}1&0&-2&1\\0&1&\frac{3}{2}&\frac{-1}{2}\end{array}\right) $

so
$ A^{-1} = \left(\begin{array}{cc}-2&1\\\frac{3}{2}&\frac{-1}{2}\end{array}\right) $

3. Adjoint Matrix Method (Can be used for any n x n matrix)

$ A^{-1} = \frac{1}{detA}(adjoint(A)) = \frac{1}{detA}(cofactor(A^{T})) $

Example. Find A^-1 of
$ A = \left(\begin{array}{ccc}1&2&3\\0&4&5\\1&0&6\end{array}\right) $
$ detA = 4 \times (6 - 3) - 5 \times (0 - 2) = 12 + 10 = 22 $
Adjoint Matrix:
$ adjoint(A) = \left(\begin{array}{ccc}24&-12&-2\\5&3&-5\\-4&2&4\end{array}\right) $
$ A^{-1} = \frac{1}{22}\left(\begin{array}{ccc}24&-12&-2\\5&3&-5\\-4&2&4\end{array}\right) = \left(\begin{array}{ccc}\frac{12}{11}&\frac{-6}{11}&\frac{-1}{11}\\\frac{5}{22}&\frac{3}{22}&\frac{-5}{11}\\\frac{-2}{11}&\frac{1}{11}&\frac{2}{11}\end{array}\right) $