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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 = In. We call B the inverse of A and denote it as A-1. Thus, AA-1 = A-1A = In. 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, ..., Ar are n x n nonsingular matrices, then A1A2...Ar is nonsingular an (A1A2...Ar)-1 = Ar-1Ar-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 AT is nonsingular and (A-1)T = (AT)-1.



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) $


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:
$ A = \left(\begin{array}{ccc}24&-12&-2\\5&3&-5\\-4&2&4\end{array}\right) $

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