Y8= Y8=28-1=
Y16= Y16=216-1
Yn=2n-1Y formula valid for all natural numbers N; = {1,2,3,...} as it is shown in examples.
-
In the case of (X+Y)= 2I
(X+Y)= = =2I
(X+Y)n= n = n = = 2n
In this case we have two slightly different formulas, the first one includes the sum of (X+Y) and the second one includes the identify matrix, i.e. I just factorized 2n.
= 2n
Testing with a random number and checking it with GDC.
(X+Y)10= =
(X+Y)10= 210=
(X+Y)15= 215=
(X+Y)15= =
2nformula valid for all natural numbers N; = {1,2,3,...} as it is shown in examples.
A=aX and B=bY
X=, Y=
Find A2, A3, A4; B2, B3, B4 using different values of a and b.
a= 2 a=5
A=2XA= A=5XA=
A2=2= A2= 2=
A3=3= A3= 3=
A4= 4= A4= 4=
b=3 b=4
B=3YB= B=4YB=
B2=2= B2= 2=
B3=3= B3= 3=
B4=4= B4= 4=
Find expressions for An , Bn, (A+B)n
- As I’m raising powers by one, the entries increase double the constant is.
For example:
A= 6X A=
A2= A3= A4=
As it is visible, every entry increases by 12, i.e. double the constant a is.
From this pattern I derived formula that:
An= anXn
Testing with random numbers:
A2=62X2
A2=36=
A=10X A= A3=103X3
A3= 3= A3=
The same pattern goes for B=bY.
B=-3Y
B2=2=
B3=3=
B4=4= The entries increase by 6, i.e. the double b is.
B=Y
B≈
B2≈ 2≈
The entries increase by app. 6.26, i.e. the double
Therefore the derived formula is
Bn= bnYn
B5= 355
B5=
B=10Y B= B3=103Y3
B3= 3= A3=
(A+B)= 6X + 4Y
(A+B)= + =
(A+B)= -6X + 4Y
(A+B)= + =
From the examples solved, I derived formulas
(A+B)n=n
(A+B)n= (aX + bY)n= anXn + bnYn
Testing with different constant values:
(A+B)5= 5=5=
(A+B)5= 255 + (-4)55=
(A+B)5= 5= 5=
(A+B)4≈ 1.544 + 3.644≈
(A+B)4≈ 4≈ 4≈
M=
M=A+B, Mn=An+Bn
M=A+B and M2=A2+B2 I showed in the previous exercise and I will prove Mn=An+Bn by induction that I’ve learnt at HL Math classes.
XY== =0
Therefore we know AB=BA=0
Mn=An+Bn
If conjecture is true it must work for x=1
(A+B)1=A+B
If it is true it needs to work for any real number k, x=k
*(A+B)k=Ak+Bk
Again it should also work for any real number k + 1
*(A+B)k(A+B)= (Ak+Bk)(A+B)= AkA + BkB + AkB + BkA
We showed that AB=BA=0, therefore AkB + BkA= 0
It follows that
AkA + BkB + AkB + BkA= AkA + BkB
AkA + BkB= Ak+1 + Bk+1
Mk+1= Ak+1 + Bk+1
By doing and deriving formulas (especially in the second exercise) I found that the general formula is:
Mn=anXn + bnYn
Mn=(aX+ bY)n
M= M=
M2=
2= 2
2= n
2= 2
Using GDC:
M2=
M= M=
5= 5
5= 5
M5=
In the conclusion I can say that I found the scope of the statement.
|A|= ab-cd
|X|= |Y|=
|X|= 1-1= 0 |Y|= 1-1= 0
Since the starting matrices have determinant 0 and thus they don’t have inverse, in my conclusion I can say that n€. Also I showed in examples above that constants a and b can be rational and irrational numbers (, therefore a, b€R. I would limit my general statement on the set of real numbers this is the only set of number that we learnt in matrix unit.
