FP1 - matrices exam two days help please.

Rules of matrix.

anyone explain why

(AB)^-1 = B^-1*A^-1

wondering why its not A^-1*B^-1

and also what are the other rules for multiplying and adding matrices?

Reply 1

ian.slater

12

Original post by Freakonomics123

Rules of matrix.

anyone explain why

(AB)^-1 = B^-1*A^-1

wondering why its not A^-1*B^-1

and also what are the other rules for multiplying and adding matrices?

B^-1*A^-1*A*B = B^-1*I*B = B^-1*B = I

so your inverse must be OK - you could also check left inverse = right inverse

Reply 2

ian.slater

12

Original post by Freakonomics123

Rules of matrix.

anyone explain why

(AB)^-1 = B^-1*A^-1

wondering why its not A^-1*B^-1

and also what are the other rules for multiplying and adding matrices?

You also need to know multiplication is associative so (AB)C = A(BC) for all A,B,C and so it's fine to write just ABC.

Multiplication is not commutative so AB ¬= BA for all A,B just as rotations and reflections do not commute.

You can only multiply matrices if the number of columns in the first equals the number of rows in the second.

(edited 13 years ago)

Reply 3

H.C. Chinaski

12

Original post by Freakonomics123

Rules of matrix.

anyone explain why

(AB)^-1 = B^-1*A^-1

wondering why its not A^-1*B^-1

and also what are the other rules for multiplying and adding matrices?

starting with (AB)^-1

hence (AB)(AB)^-1 = I

so A^-1 (AB)(AB)^-1 = (A^-1) I

so (B)(AB)^-1 = (A^-1) I

so (B^-1)(B)(AB)^-1 = (B^-1)(A^-1) I

so (AB)^-1 = (B^-1)(A^-1) I

Hence (AB)^-1 = (B^-1)(A^-1)

Reply 4

Freakonomics123

OP

10

Original post by H.C. Chinaski

starting with (AB)^-1

hence (AB)(AB)^-1 = I

so A^-1 (AB)(AB)^-1 = (A^-1) I

so (B)(AB)^-1 = (A^-1) I

so (B^-1)(B)(AB)^-1 = (B^-1)(A^-1) I

so (AB)^-1 = (B^-1)(A^-1) I

Hence (AB)^-1 = (B^-1)(A^-1)

oh ok thank you very much that helped alot =)

one more question though, the matrix (.5(root)2; .5(root)2; -.5(root)2, .5(root)2)

to find angle of rotation i use sin^-1 and cos-1, however when i use sin i get -1/4 pi, and when i use cos i get 1/4 pi, when i use sin is correct but not sure why?

Reply 5

ilyking

Original post by Freakonomics123

oh ok thank you very much that helped alot =)

one more question though, the matrix (.5(root)2; .5(root)2; -.5(root)2, .5(root)2)

to find angle of rotation i use sin^-1 and cos-1, however when i use sin i get -1/4 pi, and when i use cos i get 1/4 pi, when i use sin is correct but not sure why?

are you sure that's FP1? Goes beyond the scope of my understanding :s-smilie:

Reply 6

Freakonomics123

OP

10

Original post by ilyking

are you sure that's FP1? Goes beyond the scope of my understanding :s-smilie:

ha nah i found out why anyway, if your interested read on :tongue:

think of matrix as a b
c d

if c is negative its clockwise rotation if b is negative anti clockwise ;p

Reply 7

ilyking

Original post by Freakonomics123

ha nah i found out why anyway, if your interested read on :tongue:

think of matrix as a b
c d

if c is negative its clockwise rotation if b is negative anti clockwise ;p

no im talking about the sins/ cos in matrix algebra. are you sure that's FP1

Reply 8

Farhan.Hanif93

18

Original post by ilyking

no im talking about the sins/ cos in matrix algebra. are you sure that's FP1

For OCR FP1, we need to know that the matrix which represents a rotation of theta degrees anticlockwise about the origin is given by:

R_{\theta} = \begin{pmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{pmatrix}

We also need to know that the matrix which represents a reflection in the line

y=x\tan \theta

is given by:

M_\theta = \begin{pmatrix} \cos 2\theta & \sin 2\theta \\ \sin 2\theta & -\cos 2\theta \end{pmatrix}

Reply 9

ian.slater

12

Original post by Freakonomics123

oh ok thank you very much that helped alot =)

one more question though, the matrix (.5(root)2; .5(root)2; -.5(root)2, .5(root)2)

to find angle of rotation i use sin^-1 and cos-1, however when i use sin i get -1/4 pi, and when i use cos i get 1/4 pi, when i use sin is correct but not sure why?

You need to find an angle where cos(theta) = sqrt(1/2) and sin(theta) = -sqrt(1/2)

It's true that cos(pi/4) = sqrt(1/2), but sin(pi/4) isn't -sqrt(1/2)

so you need to find another angle.

cos(-pi/4) = sqrt(1/2) as well, and now sin(-pi/4) = -sqrt(1/2)

That gives you -pi/4 anticlock, which is pi/4 clockwise.

FP1 - matrices exam two days help please.

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