Shirley Thompson

Answered

2021-12-16

Is division of matrices possible?
Is it possible to divide a matrix by another? If yes, What will be the result of $\frac{A}{B}$ if
$A=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$

$B=\left(\begin{array}{cc}w& x\\ y& z\end{array}\right)?$

Answer & Explanation

Marcus Herman

Expert

2021-12-17Added 41 answers

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 R . We have that for two real numbers, $\frac{x}{y}$ is really the same as multiplying x and ${y}^{-1}=\frac{1}{y}.$ We call ${y}^{-1}$ the inverse of y, and note that it has the property that $y-1=1.$
The same goes for different algebraic structures. That is, for two elements x, y in this algebraic structure we define $\frac{x}{y}$ as $x{y}^{-1}$ (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 $n×n$ 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 $2×2$ matrix
The inverse of A is then given by
${A}^{-1}=\frac{1}{\left(ad-bc\right)}\left[\begin{array}{cc}d& -b\\ -c& a\end{array}\right]$
and you should check that $A{A}^{-1}=E$, the identity matrix. Now, for two matrices B and A, $\frac{B}{A}=B{A}^{-1}$

Chanell Sanborn

Expert

2021-12-18Added 41 answers

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 $A{B}^{-1}\phantom{\rule{1em}{0ex}}\text{or}\phantom{\rule{1em}{0ex}}{B}^{-1}A$, 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.

Jeffrey Jordon

Expert

2021-12-26Added 2575 answers

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

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