# Why can't Matrices be divided?

I understand that when it comes to a Matrix, Multiplication is not Commutative. But can someone tell me a Legit Reason why a Matrix cannot be divided by another Matrix? An Example would help too.

Thank You.

### 7 Answers

- spoon737Lv 61 decade agoFavorite Answer
The formal definition of division is actually given in terms of multiplication. a/b is defined to mean ab', where b' is the multiplicative inverse of b. The stuff the teach you early on in school about how division means "how many times one number goes into another number" is only the intuitive interpretation of division in the real numbers, and it makes no sense in other systems.

With this idea in mind, a logical matrix division A/B would be defined as AB^-1, where B^-1 is the inverse matrix of B. However, only square matrices with nonzero determinants have inverses, so this would be a very restricted operation. Fortunately, it doesn't matter. Division is only really a relevant operation in a field, and matrices pretty much never form a field, just vector spaces and modules.

- NovyLv 51 decade ago
This is false (kinda). It is true you cannot divide. However, it probably doesn't mean what you think it means.

Imagine I gave you this equation:

5x = 10

And I want you to solve for x. No problem, right? Well, what if I told you you cannot divide. What now?

Well, that is simple too. You can just multiply by the reciprocal. So instead of doing 10/5, you can do 10*(0.2). It is a trivial change. Your answer is still x = 2.

So what if I said:

Ax = B

Solve for x without dividing. Now you need to know what the reciprocal of A is. In general, the reciprocal of any number is A^(-1). So we get that:

x = B * A^(-1)

Great! Now we can say that:

[A]x = [B]

x = [A]^(-1) * [B] <-- must be in this order

But this isn't really the reciprocal. Because what I said earlier was kind of cheating. Because 1/B is still division, perse; I just wanted the notation of ^(-1). Sure, when applied to integers and such, it does give a reciprocal. However, when applied to matrices, it gives the inverse matrix.

But what is this inverse matrix nonsense? You know how if you multiply any number by its reciprocal you get 1? Well this inverse matrix is about the same; if you multiply a matrix by its inverse matrix, you get the identity matrix (where it's filled with 0's and 1's and where the 1's form a diagonal line). Similar, eh?

Inverse-matrix-finding is easy by hand if it is a 2x2. If it is 3x3 or larger (as it must be a square), it can be fairly difficult; your calculator can do it quickly though. But for a 2x2 matrix:

[a b]

[c d]

rewrite it as:

[d -b]

[-c a]

Then the determinant R = (d)(a) - (-c)(-b)

And the inverse matrix is:

[d/R -b/R]

[-c/R a/R]

Ya, I know, it's a lot of work. Because even after that, you still need to multiply the thing to another matrix.

- 1 decade ago
Three matrices A,B,C such that;

A*B=C

We can rearange this as so;

B = inv(A) C

remembering commutation. In this way we have 'divided' both sides by A. Its just that matrix 'division' is really taking about taking the inverse of a matrix and multiplying that. If a matrix Q takes a matrix X to X' under multiplication, then the inverse is defined as inv(Q)X' -> X. If you think of the definition of division then this is equivalent - its just that this isn't the same type of division we normally think of, because regular division is commutative. If we have X/Y then which comes first - how do you define commutation for this?

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- Anonymous1 decade ago
You're biased by the name multiplication, as you associate it with the "normal" multiplication. What would a "matrix division" do?

- 1 decade ago
we accept many things in maths without proving them they are called AXIOMS....similarly matrix is an array of real nos...on which only some operations are defined...and division of a matrix by other is not defined..........however you can divide it by a SCALAR(any real no.)......whosoever developed the concept of matrices ,he did not define is division -maybe because its of no use!