Your B inverse is incorrect. How did you get it?
the formula is well known
https://www.mathsisfun.com/algebra/matrix-inverse.html

A quick check is B*B^(-1) should be the identity matrix.

(edited 3 years ago)

Reply 5

Student 999

Original post by mqb2766

Your B inverse is incorrect. How did you get it?
the formula is well known
https://www.mathsisfun.com/algebra/matrix-inverse.html

A quick check is B*B^(-1) should be the identity matrix.

I’m sure my B^(-1) is correct anyhow if I were to use the identity matrix then it’ll leave me with A= the a 0 0 b which cant be possible since 1 doesn’t equal 0.

The best way for me to understand is if you posted your working out

Reply 6

mqb2766

Original post by Student 999

Your b and c terms are wrong in B^(-1). Check your notes / website and if you want some help, please upload your calculations. Forum rules are that we dont give answers.
You can verify by
B*B^(-1) = B^(-1)*B = I
the identity matrix. This has no relation to A. Again, it's covered in your notes / website

Reply 7

Student 999

Original post by mqb2766

I got a=1 b=3 but what should the wavelength symbol equal to as it does not matter what value I put it as

Reply 8

davros

Original post by Student 999

I got a=1 b=3 but what should the wavelength symbol equal to as it does not matter what value I put it as

I agree with a = 1 and b = 3. Unless I've made a mistake too, lambda (

\lambda

) can be any non-zero value since the result appears to be independent of lambda. If it were necessary to choose a value you could take lambda = 1 for simplicity - it does not affect the remaining parts of the question.

Reply 9

Student 999

On question 4 of 2 where I’m supposed to find A^100

So far I’ve got (1 0)
(? 3^100)

I understand that there is a sequence for ‘?’ however I don’t know how to express it as the sequence is 1,4,12,40 where it is adding 3^n to the previous value

Reply 10

ghostwalker

Original post by Student 999

You shouldn't be arriving at a sequence. Post your working.

You need to use part (ii) and a generalisation of part (iii) for this last part.

(Agree with others regarding a=1,b=3, lambda = anything non-zero)

Reply 11

Student 999

Original post by ghostwalker

you shouldn't be arriving at a sequence. Post your working.

You need to use part (ii) and a generalisation of part (iii) for this last part.

(agree with others regarding a=1,b=3, lambda = anything non-zero)

a= (1 0)
(1 3)

a^2=(1 0)
(4 9)

a^3=(1 0)
(12 27)

a^4=(1 0)
(40 81)

Reply 12

mqb2766

Original post by Student 999

a= (1 0)
(1 3)

a^2=(1 0)
(4 9)

a^3=(1 0)
(12 27)

a^4=(1 0)
(40 81)

You need part iii) to relate
[[a 0],[0,b]]^100
to A^100.

You've done it when its squared, what is the result when you raise it to the power of 100?
Then how can you reformulated the equation to get
A^100 = ....

Reply 13

Student 999

Original post by mqb2766

A^100

=(1 0)
( 3^100)

I’m stuck on finding c

Reply 14

mqb2766

Original post by Student 999

A^100

=(1 0)
( 3^100)

I’m stuck on finding c

The is not A^100. It's the diagonal matrix raised to 100.
Use iii) to express
A^2 =...
In terms of the diagonal matrix squared, then do the same when it's raised to the 100th power.

Reply 15

Student 999

Original post by mqb2766

The is not A^100. It's the diagonal matrix raised to 100.
Use iii) to express
A^2 =...
In terms of the diagonal matrix squared, then do the same when it's raised to the 100th power.

Using iii. A^2=B^(-1)A^2B

I don’t see how that will help

Reply 16

mqb2766

Original post by Student 999

Using iii. A^2=B^(-1)A^2B

I don’t see how that will help

Calling the diagonal matrix D (part ii, right hand side), you have (part iii):
D^2 = B^(-1)*A^2*B
Rearrange to get an expression for
A^2 = ...
You need to "invert" the Bs and get them over the other side and combine with the D^2.
Now do the same to get an expression for A^100 in terms of Bs and D^100.

It helps because you know B, B^(-1) and D^100 and you have to determine A^100.