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Could someone pls explain question 9 (e)?

9. Consider the set of matrices F = a) Show that AB BA for all A, B E F b) Show that every A E F\ {0} is invertible and compu

That is, p respects both the addition and multiplication in C and F. It is straightforward to see that p is a bijection. Henc

9. Consider the set of matrices F = a) Show that AB BA for all A, B E F b) Show that every A E F\ {0} is invertible and compute A-. c) Show that F is a field d) Show that F can be identified with C e) What form of matrix in F corresponds to the modđulus-argument form of a complex number Comment on the geometric significance. Solution a) Let A = and Then ai-up -(BA+ au ai-uB Bi+au - = BA AB = - - b) The inverse is -1 1 A-= c) We need to check the three axioms for a field (they appear in the solutions to Question 5). Since Mat(R) is a vector space, and F a subspace, it is clear that (F, + is an additive abelian group Moreover, the (non-zero) identity matrix in F is a multiplicative identity and matrix multiplication satisfies the associative law and distributive laws, and it is commutative. We have already checked in part (b) that every non-zero matrix A in F has an inverse. Hence, all of the field axioms hold in F, so F is a field d) Define a map p : C F by for a, b E R pla+ib)= [;. V-E C. Using part (a) it is easy to check that if a + ib,c + id E C then Here i p((aib)(c id)) = p(a +ib) + pc + id) and p((aib(cid)) = p(a + ib)p(c + id) MATH2922: Tutorial 3 (Week 3)-Solutions Page 6 of 8 MATH2922 That is, p respects both the addition and multiplication in C and F. It is straightforward to see that p is a bijection. Hence, p defines an isomorphism of fields - we have not defined homomorphisms, or isomorphisms, for fields but the obvious definition is the correct one! e) The modulus argument form corresponds to cos A r -sin sin 1 cos where r corresponds to the modulus and the matrix is the argument part. given a matrix
That is, p respects both the addition and multiplication in C and F. It is straightforward to see that p is a bijection. Hence, p defines an isomorphism of fields - we have not defined homomorphisms, isomorphisms, for fields but the obvious definition is the correct one! e) The modulus argument form corresponds to or -sin cost A r sin cos where r corresponds to the modulus and the matrix is the argument part. given a matrix We define rVa? + p?. Then it is possible to find a unique E (-r, x] Ssuch that and =sint r -=COS = 1 The reason for that is that Matrices of the above form describe a rotation about the origin of angle t in the anti-clockwise direction, and multiplication by r is a dilation
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Answer #1

7(e) part is just parametric representation of the matrices in F.

2 \e\ A siot Ct xsist zcet sin cestsint 1A Since daterminants madulsarguemmant ferm also the dame cattasponds to COt Scanned

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