# The Matrix Division

*Independent thought by Kapil Krishnan*

**If A and B are diagonal matrices of the same order, then there will be always a unique soltuion exists when A/B or B/A.**

We know that a matrix is an arrangement of elements in rows and columns. We can add, multiply and subtract matrices. But why we can't divide matrices? This is the first question I asked myself when I started to study about matrices.

Suppose A and B are two matrices of the same order, then AB not equals to BA always, so we will get multiple results if we apply the reverse operation of multiplication on matrices. But in some cases we will get a unique solution.

Let us the consider the following example.

In the above example, you can see that AxB=BxA. Also the division can be easily performed by directly dividing the elements in the same row and column.

Conclusion: Division between diagonal matrices of the same order is always possible.

**Comments**

**
By Schedler Travis, Postdoctoral Fellow, Algebra, Noncommutative Geometry, MIT. USA
Email: trasched@math.mit.edu.**

*" Hi Kapil, yes, that's correct: provided that A and B are diagonal matrices with all nonzero entries, then A/B makes sense (i.e., it equals both A B^(-1) and B^(-1) A), and has the property that B(A/B) = (A/B)B = A.*

This is because, for any two diagonal matrices A and B (even with nonzero entries), AB=BA, always. If you restrict to any subset of matrices with the property that, for any two matrices A and B in that subset, AB=BA, then it will also be true that if A and B are invertible, then AB^(-1) = B^(-1)A. This is because we can multiply both sides of AB=BA by B^(-1).

For example, other than the subset of diagonal matrices, we could consider the subset of 2x2 rotation matrices in all 2x2 matrices, i.e., matrices of the form

( cos x -sin x )

( sin x cos x )

where x is an angle (or a real number). Any two such, A and B, commute: AB=BA. We could more generally consider matrices of the form

( r cos x -r sin x )

( r sin x r cos x )

where r is any number. These also all commute (and they are invertible if r is nonzero). "

This is because, for any two diagonal matrices A and B (even with nonzero entries), AB=BA, always. If you restrict to any subset of matrices with the property that, for any two matrices A and B in that subset, AB=BA, then it will also be true that if A and B are invertible, then AB^(-1) = B^(-1)A. This is because we can multiply both sides of AB=BA by B^(-1).

For example, other than the subset of diagonal matrices, we could consider the subset of 2x2 rotation matrices in all 2x2 matrices, i.e., matrices of the form

( cos x -sin x )

( sin x cos x )

where x is an angle (or a real number). Any two such, A and B, commute: AB=BA. We could more generally consider matrices of the form

( r cos x -r sin x )

( r sin x r cos x )

where r is any number. These also all commute (and they are invertible if r is nonzero). "

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