## Multiplicative Inverse

$a*(a^{-1}) = 1$

There's no such thing as division

## Additive Inverse

$a+(-a) = 0$

## Inverse of a Matrix

Does Every Matrix have an inverse? NO

Does every nxn Matrix have an inverse? NO

If the matrix is LINEARLY DEPENDENT, it does not have an inverse partly because the determinant is 0. The primary reason is because if we attempt to augment a matrix with a determinant of 0 with the identity matrix and row reduce, we find we cannot

In any other situation, we find that we can row reduce this matrix until the left looks like the identity matrix and the right is now our $A^{-1}$.

This is because $A\vec{x} = \vec{b}$ can be changed by multiplying both sides by the inverse of a ON THE LEFT (matrix multiplication is "side sensitive") to make $A^{-1}A\vec{x} = A^{-1}\vec{b}$ and we know that a number (or matrix) multiplied by its inverse is the identity, in this case, $I_{nxn}$ for matrices so our equation becomes $\vec{x} = A^{-1}\vec{b}$ which we can easily multiply to find.

**Singular** - NOT Invertible. These matrices don't have inverse "partners," thus are singular

**Non-singular** - Invertible. These matrices have "Partners" and are thus non-singular.

## Elementary Matrices

Matrices that can replace our "row reductions."

For Example, $\begin{bmatrix} 1&0&0\\0&1&0\\2&0&1 \end{bmatrix} \begin{bmatrix} 1&0&1\\2&1&1\\0&1&2 \end{bmatrix} = \begin{bmatrix} 1&0&1\\2&1&1\\2&1&4 \end{bmatrix}$ is the same as the row reduction $r_3 = r_3+2r_1$ by our previous system. Therefore, it seems like an elementary matrix is a play on the identity matrix where the coefficient in the column corresponding to the row that we wish to add or subtract is also the multiplier of that row. In other words,

$\begin{bmatrix}1+f_1&f_2&f_3\\ f_4&1+f_5&f_6\\f_7&f_8&1+f_9 \end{bmatrix} \begin{bmatrix} a&b&c\\d&e&f\\g&h&i \end{bmatrix}$

where $f_1$ is the multiplier corresponding to $r_1 = r_1+f_1r_1$, $f_1$ corresponds to $r_1 = r_1+f_2r_2$ etc.