matrices equation

raven2k7

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Aug 25, 2019
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hi to all , i really need some help with these matrix equationes. i dont understand how to resolve it when there is a unknown variable on both sides. the equations are below

1. X · B – I = X · A + A
2. X · A – X = B
3. B · X – A = 4 · X

( these are the answers they got but i dont understand how they did it . any help will be apreciated thanks. )
1. X · B – I = X · A + A  X · B - X · A = I + A  X (B – A) = I + A 
X = (I + A) · (B – A)-1
2. X · A – X = B  X · (A – I) = B  X = B · (A – I)-1
3. B · X – A = 4 · X  B · X – 4 · X = A  (B – 4 · I) · X ·= A 
 (B – 4 · I)-1 · (B – 4 · I) · X = (B – 4 · I)-1 · A  X = (B – 4 · I)-1 · A
 
These are pretty much basic rules of algebra. The fact that these are matrices is largely irrelevant!

1. X · B – I = X · A + A
Get X on one side of the equation by subtracting XA from both sides:
XB- XA- I= A
Add I to both sides of the equation:
XB- XA= A+ I
Use the "distributive law" a(x+ y)= ax+ ay. Here, my "a" is your X, my "x" is your B, and my "y" is your A.
XB- XA= X(B- A)= A+ I.

Now, the fact that these are matrices does become relevant. If these were numbers I would say "as long as B- A is not 0, we can divide both sides by B-A". But "division" is not defined for matrices. Instead I say "If B- A is "invertible" multiply both sides by its inverse, (B- A)^-1":
X= (A+ I)(B- A)^-1.
(Notice the "^". That indicates a "power" or "exponent". Just writing (B- A)-1 as you have it looks like "subtract 1" while we mean "to the -1 power" or "inverse".

2 X · A – X = B
Again, "distributive law"- XA- X= X(A- I)= B. ("I", not "1". This is the "identity matrix", not the number "1".)
Now multiply both sides by the inverse of A- I (if it has an inverse! Many matrices do not have an inverse): X= B(A- I)^-1.
(If A- I does not have an inverse [is not "invertible"] then these equation would not have a solution- NO matrix X would satisfy it).

3. B · X – A = 4 · X
Add A to both sides:
BX= 4X+ A.
Subtract 4X from both sides:
BX- 4X= A
Distributive law: (B- 4I)X= A
(Notice the "4I". 4A, 4 times a matrix, is defined but B- 4, a matrix minus the number 4 is not defined. We have to write 4A as 4IA (which is, of course, true since IA= A) and then can "factor" it as (4I)A.)
Finally, IF B- 4I is invertible, multiply by its inverse:
X= (B- 4I)^-1A

Another way in which matrices are different from numbers- multiplication of matrices is NOT "commutative", AB is not, in general, the same as BA. Since (B- 4I) is on the left of A we have to multiply by its inverse on the left so we have to multiply A by (B- 4I)^-1 on the left. In problems 1 and 2, we multiplied on the right.

I said earlier that "division is not defined for matrices". Of course, for numbers, "dividing by x" is the same as multiplying by x inverse, \(\displaystyle x^{-1}\). But for numbers multiplying "on the left" or "on the right" gives the same thing. For matrices we have to specify left multiplication or right multiplication so we don't say just "divide".
 
wow , that was great, thanks alot, is there any short way to arrive at the same answer? or other method?
 
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