i just want to know if its possible to divide matrix?
ive run through many mathbooks and i found no matrix division just addition and multiplcation.. can someone enlightened my thoughts about this matrix division. thnks!
If $\displaystyle AB=C$ then $\displaystyle A=CB^{-1}$ and $\displaystyle B=A^{-1}C$ provided that these inverses exist.
Matrix Inverse -- from Wolfram MathWorld
no. that is not what HallsofIvy said.
if A is invertible, we can solve the matrix equation AX = B for X:
X = A^{-1}B.
if A is invertible, we can also solve the equation XA = B for X:
X = BA^{-1}.
these are two totally different situations, and should NOT be confused. A^{-1}B and BA^{-1} are MOST OF THE TIME different matrices.
matrix multiplication is NOT like "ordinary multiplication", we cannot say that AB and BA are the same (even if BOTH are "the same size", nxn).
contrast this with a field (like the reals, or the rationals), it does not matter if we write:
x/y = x(1/y) or x/y = (1/y)x, because these are the same.
it's even worse, because we can have an mxn matrix A, with an nxm matrix B with AB = I (where I is mxm), but NO nxm matrix C with CA = I (where I is nxn).
for example, if:
$\displaystyle A = \begin{bmatrix}2&3&0\\3&5&0 \end{bmatrix};\ B = \begin{bmatrix}5&-3\\-3&2\\0&0 \end{bmatrix}$
then:
$\displaystyle AB = \begin{bmatrix}1&0\\0&1 \end{bmatrix}$
but for ANY matrix:
$\displaystyle C = \begin{bmatrix}c_1&c_2\\c_3&c_4\\c_5&c_6 \end{bmatrix}$
$\displaystyle CA = \begin{bmatrix}2c_1+3c_2&3c_1+5c_2&0\\2c_3+3c_4&3c _3+5c_4&0\\2c_5+3c_6&3c_5+5c_6&0 \end{bmatrix}$
which is NOT the 3x3 identity matrix no matter HOW we choose the c's, since the last column is all zeros.
so a matrix can have a "right-inverse" without having a "left-inverse". in such an unhappy state of affairs, division doesn't even make sense as a CONCEPT.
matrices are NOT numbers. they are an entirely different BREED of animal.
Matrix division is the same as ordinary division if you observe order of multiplication (matrices do not commute, AB unequal BA).
a=b/c means ac=b has a solution for a if c unequal 0. a=bc^{-1}
A=B/C means AC=B has a solution for A if C unequal “0” where “0” = singular (|C| = 0). A=BC^{-1}
i'm sorry, could you explain to me just how you divide one matrix by another? and could you perhaps tell me which authors refer to BC^{-1} as B/C, and why this is not used for, say: C^{-1}B?
how is "matrix division" just like "ordinary division"? even in groups the process of multiplying by an inverse is not referred to as "division".
i cannot, in good conscience, mislead the original poster into thinking "there is matrix division". matrices are not field elements, nor even elements of a (non-commutative) division ring, except in very special cases. the notion of left-inverse and right-inverse are not even necessarily coincidental.
your glib reply gives the impression matrix arithmetic is somehow "just like ordinary arithmetic". nothing could be further from the truth.
Yes. Why?:
STEP 1
What does division by B mean? Multiply by 1/B
But what is 1/B? B^{-1}
But what is B^{-1}? Unique sol of BB^{-1}=I for B, which requires |B| not equal 0 ( B non-singular).
For numbers ab=ba. For matrices, AB generally unequal BA (don't commute).
STEP 2
Work out the details.
EXAMPLE 1
Divide ABC by B: (ABC)/B = ABCB^{-1} or B^{-1}ABC, B non-singular
But (ABC)(1/B) = ABCB^{-1}
EXAMPLE 2
Solve ABC = E for B: B=A^{-1}EC^{-1}, A&C non-singular
EXAMPLE 3
Solve ABC = E for AC: AC=EC^{-1}B^{-1}C, B&C non-singular. If matrices commuted, AC = EB^{-1 }, B non-singular.