# Maths matrices

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#3

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I’ve tried to multiply out and equate but it doesn’t work

**Student 999**)I’ve tried to multiply out and equate but it doesn’t work

For c) just expand and use the B - B inverse .

Last edited by mqb2766; 1 day ago

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**mqb2766**)

Can you upload what you've tried for ii)?

For c) just expand and use the B - B inverse .

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#5

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.

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.

Last edited by mqb2766; 1 day ago

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(Original post by

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.

**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.

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

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#7

(Original post by

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

**Student 999**)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

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

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(Original post by

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 /

**mqb2766**)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 /

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#9

(Original post by

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

**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

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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

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

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#11

(Original post by

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

**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

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)

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(Original post by

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)

**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)

(1 3)

a^2=(1 0)

(4 9)

a^3=(1 0)

(12 27)

a^4=(1 0)

(40 81)

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#13

[[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 = ....

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(Original post by

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 = ....

**mqb2766**)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 = ....

=(1 0)

( 3^100)

I’m stuck on finding c

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#15

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.

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(Original post by

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.

**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.

I don’t see how that will help

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#17

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.

Last edited by mqb2766; 1 day ago

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