User:Rhuso

$$ \mathbf{A}^{-1} ={1 \over \begin{vmatrix}\mathbf{A}\end{vmatrix}}\left(\mathbf{C}_{ij}\right)^{\mathrm{T}} ={1 \over \begin{vmatrix}\mathbf{A}\end{vmatrix}}\left(\mathbf{C}_{ji}\right) ={1 \over \begin{vmatrix}\mathbf{A}\end{vmatrix}}

\begin{pmatrix} \mathbf{C}_{11} & \mathbf{C}_{21} & \cdots & \mathbf{C}_{j1} \\ \mathbf{C}_{12} & \mathbf{C}_{22} & \cdots & \mathbf{C}_{j2} \\ \vdots & \vdots & \ddots & \vdots \\ \mathbf{C}_{1i} & \mathbf{C}_{2i} & \cdots & \mathbf{C}_{ji} \\ \end{pmatrix} $$

$$ \sin x + \ln y a^{2 + 2} $$

$$ \bullet \frac{1}{2} $$ $$ \tfrac{2}{3} \to \| \mid $$ $$ \Rightarrow \ddagger \lVert \rVert \nexists $$

$$ \parallel $$

$$ a^2 a_2 $$

$$ \ \, $$

$$ \begin{matrix} x & y \\ z & v \end{matrix} $$

$$ \begin{vmatrix} x & y \\ z & v \end{vmatrix} $$

$$ \begin{Vmatrix} x & y \\ z & v \end{Vmatrix} $$

$$ \begin{bmatrix} x & y \\ z & v \end{bmatrix} $$

$$ \begin{array}{rcl} right & center & left \\ r & c & l \end{array} $$

$$ \begin{cases} 3x + 5y + z = a \\ 7x - 2y + 4z = b \\ -6x + 3y + 2z = c \end{cases} $$

$$ \left[ stuff \right] $$

General Matrix Inversion
A matrix that is square and not singular has a single inverse. The transpose of the inverse of the transpose of the matrix is also equal to the inverse. The formulae $$ A^{-1^{T}} = A^{-1^{T}} $$ and $$ A^{T^{-1^{T}}} = A^{-1} $$ say it all! This simple equation also justifies using a variant of the Gauss-Jordan method for inverting a matrix. Inverting the transpose is effectively the same as augmenting the matrix with an identity matrix below it and using column operations to reduce the original to identity. This form of inverting a matrix is immediately apparent in other discussions, so what follows is my description of a matrix inversion method that is simpler than those previously discussed, and much more capable.

Second title
$$ \left [ \cfrac{ \begin{matrix} 2 & 0 & 1 & | \\ 0 & -1 & 2 & | \\ 1 & 1 & 0 & | \end{matrix} \, \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix} }{ \begin{matrix} 1 & 0 & 0 & | \\ 0 & 1 & 0 & | \\ 0 & 0 & 1 & | \end{matrix} \begin{matrix} \quad\quad\, & \end{matrix} } \right] $$

Must modify the layout to make the type of display I need. Hmmmmm... I'll have to think about this a bit. rand.

Invertible Matrix

rand huso (talk) 21:38, 3 February 2008 (UTC)