HI Im learning about matrices in school and am curious about how to divide them so i can find out different identity matrices for refecting shapes over linear functions
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$\begingroup$When you are dealing with regular numbers, you can usually just divide two numbers:
$ 1 / 9 \approx .111... $
But not if the denominator is 0:
$ 1 / 0 = \text{undef} $
In general, if we try to define division among matrices we'll have problems with many matrices being incapable of being used to divide other matrices.
The hint is this:
$ 1 / 9 = 1 * 9^{-1} $
In order to define something "like" division among the matrices, we'll just take the inverse of the divisor and multiply it on the right with the dividend.
$ A / B = A B^{-1} $.
But usually we just talk of multiplying on the (left/right) with the inverse of a matrix. The main reason for this is that most matrices are not invertible.
$\endgroup$ $\begingroup$One big difference with matrices and numbers is that matrices are not necessarily commutative (ie, it may be the case that$AB \neq BA$).
However, if a matrix has an inverse, you can use the inverse to 'divide' as in if $Ax = b$, then $A^{-1}A x = Ix = x = A^{-1} b$.
As with numbers, you cannot divide by zero. The corresponding value with a matrix is a number called its determinant (usually $\det$, sometimes $|\cdot|$). For the two by two $A$ below, $\det A = ac - bd$. If this number is non-zero, the matrix has an inverse.
The inverse of $A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$ is $A^{-1} = \frac{1}{a c -bd} \begin{bmatrix} d & -c \\ -b & d \end{bmatrix}$.
For a general square matrix, a similar rule (and generalization of $\det$) apply.
$\endgroup$ $\begingroup$Some $n\times n$ matrices $A$ are invertible. That means there exists a matrix $A^{-1}$ such that $A\cdot A^{-1}=A^{-1}\cdot A=I$, the identity matrix. Then you can think of $B\cdot A^{-1}$ as dividing $B$ by $A$. This agrees with your experience with real or complex numbers, where $\frac {x}{y}$, for $y\ne 0$, is equal to $x\cdot y^{-1}$.
However, it is actually not such a good idea to think of division of matrices at all. The reason is that matrices' sole reason for existence is to represent linear transformation and make computing things related to linear transformation (sometimes) easier.
A linear transformation is a function $T:V\to W$ between vector spaces that satisfy some properties. In particular, to linear transformations $T:V\to V$, where the dimension of $V$ is $n$, there correspond a matrix (actually many matrices) representing $T$. While a matrix is just a block of numbers, it might be tempting to consider division, but for a linear transformation $T:V\to V$ it does not. It makes sense to ask where two linear transformations $T,S:V\to V$ can be added (and that will correspond to addition of matrices), and it also makes sense to talk about the composition $T\circ S$ (which will correspond to matrix multiplication (and explains why matrix multiplication is defined the way it is)). It makes sense to ask whether $T:V\to V$ is invertible (and that will correspond to the inverse matrix I mentioned above).
I don't know if you had already seen linear transformations or not. It is very often preferred (wrongly) to introduce matrices before introducing linear transformations. The reason it is preferred is because a matrix is just a block of numbers, so it is easy. The reason it's wrong to do so is that there is absolutely no reason for any mathematician to wake up in the morning and just say "hey, it would be cool to have blocks of numbers, add that this way, multiply them that way, and play with these things". It portrays a wrong impression of what mathematics is. The reason to consider matrices is to study linear transformations. Questions you should ask about matrices should be related to linear transformations. That is why in mathematics we don't study, for instance, triangular blocks of numbers, or pentagonal blocks of numbers. They don't represent anything (we are aware of).
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