# Is division of matrices possible?

Is it possible to divide a matrix by another? If yes, What will be the result of $\dfrac AB$ if $$A = \begin{pmatrix} a & b \\ c & d \\ \end{pmatrix}, B = \begin{pmatrix} w & x \\ y & z \\ \end{pmatrix}?$$

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see for example here. You'll have to distinguish between $AB^{-1}$ and $B^{-1}A$. – Raymond Manzoni Nov 3 '12 at 16:43
To add various types of brackets around entries of a matrix, you can use pmatrix or bmatrix or other variants. See tutorial at meta. – Martin Sleziak Nov 3 '12 at 16:49

For ordinary numbers $\frac{a}{b}$ means the solution to the equation $xb=a$. This is the same as $bx=a$, but since matrix multiplication is not commutative, there are two different possible generalizations of "division" to matrices.

If $B$ is invertible, then you can form $AB^{-1}$ or $B^{-1}A$, but these are not in general the same matrix. They are the solutions to $XB=A$ and $BX=A$ respectively.

If $B$ is not invertible, then $XB=A$ and $BX=A$ may have solutions, but the solutions will not be unique. So in that situation speaking of "matrix division" is even less warranted.

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Normaly matrix division is defined as $\frac{A}{B}=AB^{-1}$ where $B^{-1}$ stands for inverse matrix of $B$. In case when inverse doesn't exist so called pseudoinverse may be used.

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We can say $$\frac{A}{B}=A\times B^{-1}$$ where $B^{-1}$ is inverse matrice of $B$

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if $\ \cfrac AB = A \times B^{-1}$ then what is $B^{-1} \times A$? – user31280 Nov 3 '12 at 18:29
That is an other possibility since product of matrices is not commutative.So division of matrices can not be defined in an unique way – Adi Dani Nov 3 '12 at 19:02

There is a way to performa sort of division , but I am not sure if it is the way you are looking for. For motivation ,consider the ordinary real numbers $\mathbb{R}$ . We have that for two real numbers, $x/y$ is really the same as multiplying x and $y^{-1}=1/y$. We call $y^{-1}$ the inverse of y, and note that it has the property that $yy^{-1}=1.$

The same goes for different algebraic structures. That is, for two elements x,y in this algebraic structure we define $x/y$ as $xy^{-1}$ (under some operation). Most notably, we have a notion of division in any division ring (hence the name!) . It turns out that if you consider invertible $n \times n$ matrices with addition and ordinary matrix multiplication, there is a sensible way to define division since every invertible matrix has well, an inverse. So just to help you grip what an inverse is, say that you have a 2x2 matrix $$A= \begin{bmatrix} a & b \\ c & d \end{bmatrix}.$$ The inverse of A is then given by $$A^{-1} = \dfrac{1}{(ad-bc)} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$$ and you should check that $AA^{-1}=E$, the identity matrix. Now, for two matrices $B$ and $A$, $B/A = BA^{-1}$.

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I know this is not the most eloquent answer, but why the downvote? – Dedalus Apr 2 '14 at 13:02

There are two issues: first, that matrices have divisors of zero; second, that matrix multiplication is in general not commutative.

To give meaning to $A/B$, you need to give meaning to $I/B$ (because then $A/B=A(I/B)$. Now, no one ever writes $I/B$, people actually write $B^{-1}$. Anyway, what is $B^{-1}$? It should be a matrix such that multiplied by $B$ gives you the identity. Now, there exist nonzero matrices $C$, $B$ with $BC=0$. If $B$ had an inverse $B^{-1}$, we would have $$0=B^{-1}0=B^{-1}BC=C,$$ a contradiction. So such a matrix $B$ cannot have an inverse, i.e. "$I/B$" does not make sense.

The invertible matrices are exactly those with nonzero determinant. So, if $\det B\ne0$, then $AB^{-1}$ does make sense.

In you case, that would be the condition $wz-yx\ne0$. In that case, $$\begin{bmatrix}w&x\\ y&z\end{bmatrix}^{-1}=\frac1{wz-yx}\begin{bmatrix}z&-x\\ -y&w\end{bmatrix}$$

The second issue is a non-issue, because it can be proven that, for matrices, if $B^{-1}A=I$, then $AB^{-1}=I$.

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## protected by Community♦Apr 1 '14 at 11:51

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