mattrix equationes

Linear, quadratic, module, parametric equations

mattrix equationes

Postby Guest » Thu May 21, 2020 7:21 am

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
Guest
 

Re: mattrix equationes

Postby Guest » Tue May 26, 2020 4:41 pm

You say "these are the answers they got but i dont understand how they did it".

Since you are asking about matrices, I would think that you had already learned basic algebra. There are some ways in which "matrix algebras" differs from basic (numerical) algebra but they are minor.

1. X · B – I = X · A + A
 X · B - X · A = I + A

They subtracted X·I from both sides and added I to both sides, exactly what you would do with numbers instead of matrices.

 X (B – A) = I + A

The "distributive law" says that XB- XA= X(B- A). Again that would be true whether these were matrices or numbers.

X = (I + A) · (B – A)-1

Here IS one of the ways in which matrices differ from numbers. To get X by itself on the left side of the equation where you now have X(B- A) you would "divide both sides of the equation by B-A". With numbers you would have to be careful to say "if B- A is not 0" since every number except 0 has a "multiplicative inverse" (reciprocal) and "dividing" is the same as "multiplying by the multiplicative". With matrices we always say "multiply by the multiplicative inverse" and many matrices do not have "multiplicative inverses" Here you should say "IF B- A has a multiplicative inverse then multiply (on the right) by that multiplicative inverse, (B- A)^(-1). [
Two more points: Notice that "^". That indicates that the -1 is an exponent the -1 power or inverse, not "minus 1". Also notice that "on the right". Multiplication of matrices is NOT "commutative". X(B- A) is not the same as (B-A)X. In order to cancel the "B- A" in X(B- A) we must have X(B- A)(B- A)^(-1); (B- A)^(-1)X(B- A) wouldn't work.

2. X · A – X = B

 X · (A – I) = B[/quote]
Again, using the "distributive" law: X·A- X= X·A- X·I= X(A- I)
Then, IF A- I has an inverse,
 X = B · (A – I)-1

X= B· (A- I)^(-1) Again ^(-1) indicating the inverse, NOT "- 1" subtracting 1.

3. B · X – A = 4 · X

Subtract 4·X from both sides and add A to both sides.

 B · X – 4 · X = A

Since I is the identity matrix, 4·X is the same a 4·I·X= (4·I)·X
(Personally, I wouldn't use "·" when multiplying a number times a matrix, just 4X= (4I)·X.)

 (B – 4 · I) · X = A

And IF B- 4·I HAS an inverse, 

 (B- 4·I)-1(B- 4·I)· X = (B – 4 · I)-1 · A

It's interesting that they actually wrote the "(B- 4·I)-1(B- 4·I)". They didn't do that on the previous problems! First, that should be "^(-1)", NOT "- 1". That is the multiplicative inverse multiplied by the value itself so they cancel leaving X on the left. Also notice that here, in the original equation, X was on the right and (B- 4·I) on the left. In order to cancel that (B- 4·I)^(-1) must be multiplied on the left, not the right as before.

[quote]  X = (B – 4 · I)-1 · A [quote]
X= (B- 4·I)^(-1)·A
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