Why can't you divide matrices?
I was just wondering that because one can multiply and add and subtract matrices, why can't one divide them?
Solution 1:
You CAN divide by matrices. To understand what division in the context of matrices mean, let us look at what division means in the context of real numbers.
$b/a$ in the context of real numbers denote the real number $x$ satisfying $x \cdot a=b$. Since multiplying two real numbers is commutative, the same real number $x$ also satisfies $a \cdot x = b$.
Similarly, in the context of matrices, $B/A$ means the matrix $X$ such that $X \cdot A = B$. However, in matrix algebra, multiplication is not commutative. Hence, in general, it is not true that $X \cdot A = A \cdot X$. Hence, you need to specify, whether you are dividing by a matrix on the right or on the left. Hence, if you are looking for a matrix $X$ such that $A \cdot X = B$, $X$ is denoted as $X = A\backslash B$.
$X = B/A$ means you are dividing by the matrix $A$ on the right and $X = A\backslash B$ means you are dividing by the matrix $A$ on the left.
Also, just like division by zero is not possible in the context of real numbers, you cannot divide by certain matrices, which are called singular matrices. Hence, the left division ($A\backslash B$) and right division ($B/A$) make sense only when $A$ is not singular.
Solution 2:
You sort of can. If a matrix $A$ is invertible, then there exists a matrix $A^{-1}$ such that $AA^{-1}=A^{-1}A=I$, where $I$ is the identity matrix. This is analogous to $a^{-1}=\frac1a$ for a nonzero real number, which satisfies $aa^{-1}=a^{-1}a=1$. Recall that dividing a real number $b$ by $a$ is the same as multiplying it by $a^{-1}$, i.e. $b/a=ba^{-1}$. Similarly, for matrices we can define "$B$ dividied by $A$" as $BA^{-1}$.
However, there are a few problems that arise:
Not all nonzero matrices are invertible. This is because the set of matrices, unlike real numbers, has zero divisors: there are nonzero matrices $A,B$ such that $AB=0$. If you could divide $B$ by $A$, you would get $B=0/A=0$, a contradiction.
Matrices do not commute, i.e. generally $AB\ne BA$. This means we have to make a distinction between dividing on the left ($A^{-1}B$) and dividing on the right ($BA^{-1}$), while for real numbers these are the same.
Solution 3:
Dividing by $x$ is undoing multiplication by $x$ via multiplication with $x$'s inverse. Further, with nonzero numbers, there is a unique way to do this.
With matrices, a lot of this breaks down. For some matrices $X$, there is no way to undo multiplication by $X$ via multiplication with a would-be $X^{-1}$ matrix, because multiplication by $X$ has destroyed too much of the original information.
Even when there is way to undo multiplication by $X$, there can be multiple would-be $X^{-1}$ matrices that do it.
For some matrices, an inverse matrix both exists and is unique, and only for those matrices is it OK to do division in the way you are trying to do it. In a linear algebra class you will learn that these matrices are the square matrices with determinant not equal to $0$.