(New page: =Inverse of a Matrix= Consider the <math>n</math>x<math>n</math> matrix <math>A</math>. The inverse of <math>A</math> is defined as <math>A^{-1}</math> such that <math>AA^{-1} = A^{-1}A=I_...) |
|||
| Line 11: | Line 11: | ||
<math>A^{-1} = \begin{bmatrix} -3 & 4 \\ 1 & -1 \end{bmatrix}</math> | <math>A^{-1} = \begin{bmatrix} -3 & 4 \\ 1 & -1 \end{bmatrix}</math> | ||
| + | |||
| + | This can be verified by matrix multiplication. | ||
| + | |||
| + | <math>AA^{-1}=\begin{bmatrix}1 & 4 \\1 & 3 \end{bmatrix}\begin{bmatrix} -3 & 4 \\ 1 & -1 \end{bmatrix}=\begin{bmatrix}1(-3)+4(1) & 1(4)+4(-1) \\1(-3)+3(1) & 1(4)+3(-1) \end{bmatrix}=\begin{bmatrix}1 & 0\\0 & 1 \end{bmatrix}</math> | ||
| + | |||
| + | Although not shown, the commutative case also works | ||
| + | |||
| + | ==Ways to Find the Inverse== | ||
| + | |||
| + | There is no trivial way of finding the inverse of an arbitrary matrix. In fact, if the matrix is not a square matrix it cannot have an inverse. Even if it is square, you need to make sure its reduced row echelon form is the identity matrix. Provided that the matrix is invertible, you can use one of the following methods for finding the inverse. | ||
| + | |||
| + | *'''Use the Determinant Trick''' | ||
| + | For simple 2x2 matrices, if the determinant is nonzero, then you can find the inverse of a matrix by flipping its upper left and lower right elements, multiplying the upper right and lower left by negative one, and dividing by the determinant. Observe: | ||
| + | |||
| + | Let | ||
| + | |||
| + | <math>A=\begin{bmatrix}a & b \\c & d \end{bmatrix}</math> | ||
| + | |||
| + | <math>A^{-1}=\frac{1}{ad-bc}\begin{bmatrix}d & -b \\-c & a \end{bmatrix} </math> | ||
| + | |||
| + | Then | ||
| + | |||
| + | <math>\begin{array}{lcl}AA^{-1}&=&\begin{bmatrix}a & b \\c & d \end{bmatrix}\frac{1}{ad-bc}\begin{bmatrix}d & -b \\-c & a \end{bmatrix}\\ | ||
| + | &=&\frac{1}{ad-bc}\begin{bmatrix}ad+b(-c) & a(-b)+b(a) \\cd+d(-c) & c(-b)+d(a) \end{bmatrix}\\ | ||
| + | &=&\begin{bmatrix}1 & 0\\0 & 1 \end{bmatrix}\end{array}</math> | ||
| + | |||
| + | Although not shown, the commutative case also works. | ||
| + | |||
| + | *'''Identity-Extended Matrix and Row Operations''' | ||
| + | |||
| + | You can attach an appropriately sized identity matrix to a matrix and perform row operations until the left half side is the identity matrix. The right half is now the inverse. | ||
| + | |||
| + | ==Purpose of Inverse== | ||
| + | Provided a matrix has an inverse, then any linear transformation characterized by the matrix has a unique solution for all elements in the codomain. | ||
| + | |||
| + | In other words, you can multiply both sides of the matrix equation by the inverse. On the side with the original matrix, they cancel out and leave the identity matrix (hurrah!) and on the other you get your solution. | ||
| + | |||
| + | Suppose we used the first example matrix above to define a linear transformation | ||
| + | |||
| + | <math>T(\vec v)=A\vec v =\vec b</math> | ||
| + | |||
| + | <math>\begin{bmatrix}1 & 4 \\1 & 3 \end{bmatrix} \vec v = \begin{bmatrix}1\\2\end{bmatrix}</math> | ||
| + | |||
| + | <math>\begin{bmatrix} -3 & 4 \\ 1 & -1 \end{bmatrix}\begin{bmatrix}1 & 4 \\1 & 3 \end{bmatrix} \vec v = \begin{bmatrix} -3 & 4 \\ 1 & -1 \end{bmatrix}\begin{bmatrix}1\\2\end{bmatrix}</math> | ||
| + | |||
| + | <math>\begin{bmatrix}1 & 0\\0 & 1 \end{bmatrix} \vec v = \begin{bmatrix}-3(1)+4(2)\\1(1)+-1(2)\end{bmatrix}</math> | ||
| + | |||
| + | <math>\vec v = \begin{bmatrix}5\\-1\end{bmatrix}</math> | ||
| + | |||
| + | So the inverses of matrices are really nice for solving linear equations. | ||
Revision as of 08:09, 24 February 2010
Inverse of a Matrix
Consider the $ n $x$ n $ matrix $ A $. The inverse of $ A $ is defined as $ A^{-1} $ such that $ AA^{-1} = A^{-1}A=I_n $ where $ I_n $ is the identity matrix.
For example let
$ A= \begin{bmatrix} 1 & 4 \\ 1 & 3 \end{bmatrix} $
Then the inverse of A is
$ A^{-1} = \begin{bmatrix} -3 & 4 \\ 1 & -1 \end{bmatrix} $
This can be verified by matrix multiplication.
$ AA^{-1}=\begin{bmatrix}1 & 4 \\1 & 3 \end{bmatrix}\begin{bmatrix} -3 & 4 \\ 1 & -1 \end{bmatrix}=\begin{bmatrix}1(-3)+4(1) & 1(4)+4(-1) \\1(-3)+3(1) & 1(4)+3(-1) \end{bmatrix}=\begin{bmatrix}1 & 0\\0 & 1 \end{bmatrix} $
Although not shown, the commutative case also works
Ways to Find the Inverse
There is no trivial way of finding the inverse of an arbitrary matrix. In fact, if the matrix is not a square matrix it cannot have an inverse. Even if it is square, you need to make sure its reduced row echelon form is the identity matrix. Provided that the matrix is invertible, you can use one of the following methods for finding the inverse.
- Use the Determinant Trick
For simple 2x2 matrices, if the determinant is nonzero, then you can find the inverse of a matrix by flipping its upper left and lower right elements, multiplying the upper right and lower left by negative one, and dividing by the determinant. Observe:
Let
$ A=\begin{bmatrix}a & b \\c & d \end{bmatrix} $
$ A^{-1}=\frac{1}{ad-bc}\begin{bmatrix}d & -b \\-c & a \end{bmatrix} $
Then
$ \begin{array}{lcl}AA^{-1}&=&\begin{bmatrix}a & b \\c & d \end{bmatrix}\frac{1}{ad-bc}\begin{bmatrix}d & -b \\-c & a \end{bmatrix}\\ &=&\frac{1}{ad-bc}\begin{bmatrix}ad+b(-c) & a(-b)+b(a) \\cd+d(-c) & c(-b)+d(a) \end{bmatrix}\\ &=&\begin{bmatrix}1 & 0\\0 & 1 \end{bmatrix}\end{array} $
Although not shown, the commutative case also works.
- Identity-Extended Matrix and Row Operations
You can attach an appropriately sized identity matrix to a matrix and perform row operations until the left half side is the identity matrix. The right half is now the inverse.
Purpose of Inverse
Provided a matrix has an inverse, then any linear transformation characterized by the matrix has a unique solution for all elements in the codomain.
In other words, you can multiply both sides of the matrix equation by the inverse. On the side with the original matrix, they cancel out and leave the identity matrix (hurrah!) and on the other you get your solution.
Suppose we used the first example matrix above to define a linear transformation
$ T(\vec v)=A\vec v =\vec b $
$ \begin{bmatrix}1 & 4 \\1 & 3 \end{bmatrix} \vec v = \begin{bmatrix}1\\2\end{bmatrix} $
$ \begin{bmatrix} -3 & 4 \\ 1 & -1 \end{bmatrix}\begin{bmatrix}1 & 4 \\1 & 3 \end{bmatrix} \vec v = \begin{bmatrix} -3 & 4 \\ 1 & -1 \end{bmatrix}\begin{bmatrix}1\\2\end{bmatrix} $
$ \begin{bmatrix}1 & 0\\0 & 1 \end{bmatrix} \vec v = \begin{bmatrix}-3(1)+4(2)\\1(1)+-1(2)\end{bmatrix} $
$ \vec v = \begin{bmatrix}5\\-1\end{bmatrix} $
So the inverses of matrices are really nice for solving linear equations.
