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1996 Paper 3 Q6
D: 1674.0 B: 1529.9
groups matrix group axioms closure singular matrix determinant proof matrix multiplication

  1. Let \(S\) be the set of matrices of the form \[ \begin{pmatrix}a & a\\ a & a \end{pmatrix}, \] where \(a\) is any real non-zero number. Show that \(S\) is closed under matrix multiplication and, further, that \(S\) is a group under matrix multiplication.
  2. Let \(G\) be a set of \(n\times n\) matrices which is a group under matrix multiplication, with identity element \(\mathbf{E}.\) By considering equations of the form \(\mathbf{BC=D}\) for suitable elements \(\mathbf{B},\) \(\mathbf{C}\) and \(\mathbf{D}\) of \(G\), show that if a given element \(\mathbf{A}\) of \(G\) is a singular matrix (i.e. \(\det\mathbf{A}=0\)), then all elements of \(G\) are singular. Give, with justification, an example of such a group of singular matrices in the case \(n=3.\)

Solution:

  1. Let $\mathbf{A} = \begin{pmatrix}1 & 1\\ 1 & 1 \end{pmatrix}\(, then we need to show that \)(a\mathbf{A})(b\mathbf{A})\( is of the form \)cA\( where \)a, b, c \neq 0$. Since $\mathbf{A}^2 = \begin{pmatrix}2 & 2\\ 2 & 2 \end{pmatrix} = 2\mathbf{A}\( this is certainly the case, since \)(a\mathbf{A})(b\mathbf{A}) = 2ab\mathbf{A}$. To check that we have a group be need to check:
    • Closure (done)
    • Associativity (inherited from matrix multiplication)
    • Identity (\(\frac12 \mathbf{A}\))
    • Inverses the inverse of \(a\mathbf{A}\) is \(\frac{1}{4a}\mathbf{A}\)
  2. Suppose \(\mathbf{A}\) is singular (ie \(\det\mathbf{A}=0\)), then \(\mathbf{AA^{-1}B=B}\) (where inverse is the group inverse rather than the matrix inverse) for any matrix \(\mathbf{B}\). Taking determinants we have: \(\det(\mathbf{AA^{-1}B}) = \det(B) \Rightarrow \det(A) \det(A^{-1}B) = \det(B) \Rightarrow 0 = \det(B)\), ie all matrices are singular. Consider the set of non-zero multiples of \(\begin{pmatrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{pmatrix}\), then the same logic as part (i) will suffice

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1995 Paper 3 Q7
D: 1654.7 B: 1516.0
groups group axioms closure identity element inverse isomorphism complex numbers modular arithmetic matrix rotation matrix unit modulus

Consider the following sets with the usual definition of multiplication appropriate to each. In each case you may assume that the multiplication is associative. In each case state, giving adequate reasons, whether or not the set is a group.

  1. the complex numbers of unit modulus;
  2. the integers modulo 4;
  3. the matrices \[ \mathrm{M}(\theta)=\begin{pmatrix}\cos\theta & -\sin\theta\\ \sin\theta & \cos\theta \end{pmatrix}, \] where \(0\leqslant\theta<2\pi\);
  4. the integers \(1,3,5,7\) modulo 8;
  5. the \(2\times2\) matrices all of whose entries are integers;
  6. the integers \(1,2,3,4\) modulo 5.
In the case of each pair of groups above state, with reasons, whether or not they are isomorphic.

Solution:

  1. \(\{ z \in \mathbb{C} : |z| = 1\}\) is a group.
    1. (Closure) \(|z_1z_2| = |z_1||z_2| = 1\). Set is closed under multiplication
    2. (Associativity) Multiplication of complex numbers is associative
    3. (Identity) \(|1| = 1\)
    4. (Inverses) \(| \frac{1}{z} | = \frac{1}{|z|} = \frac{1}{1} = 1\), the set contains inverses
  2. the integers \(\pmod{4}\) are not a group under multiplication, \(2\) has no inverse, since \(0 \times k \equiv 0 \pmod{4}\)
  3. The set of rotation matrices is a group:
    1. (Closure) \begin{align*} \begin{pmatrix}\cos\theta_1 & -\sin\theta_1\\ \sin\theta_1 & \cos\theta_1 \end{pmatrix} \begin{pmatrix}\cos\theta_2 & -\sin\theta_2\\ \sin\theta_2 & \cos\theta_2 \end{pmatrix} &= {\scriptscriptstyle\begin{pmatrix}\cos\theta_1 \cos \theta_2 - \sin \theta_1\sin \theta_2 & -\sin\theta_1\ \cos \theta_1 - \sin \theta_2\cos\theta_1\\ \sin\theta_1\ \cos \theta_1 + \sin \theta_2\cos\theta_1 & \cos\theta_1 \cos \theta_2 - \sin \theta_1\sin \theta_2 \end{pmatrix}} \\ &= \begin{pmatrix}\cos(\theta_1+\theta_2) & -\sin(\theta_1+\theta_2)\\ \sin(\theta_1+\theta_2) & \cos(\theta_1+\theta_2) \end{pmatrix} \end{align*} Since \(\cos, \sin\) are periodic with period \(2\pi\), we can find \(\theta_3 = \theta_1 + \theta_2 + 2k\pi\) such that \(0 \leq \theta_3 < 2 \pi\), so our set is closed
    2. (Associativity) Matrix multiplication is associative
    3. (Identity) Consider \(\theta = 0\)
    4. (Inverses) Consider \(2\pi - \theta\)
  4. \(\{1, 3, 5, 7\} \pmod{8}\) is a group:
    1357
    11357
    33175
    55713
    77531
    1. (Closure) See Cayley table
    2. (Associativity) Integer multiplication is associative
    3. (Identity) \(1\)
    4. (Inverses) \(x \mapsto x\) (See Cayley table)
  5. \(2\times2\) matrices are not a group, consider $0 = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix}\(, then \)\mathbf{0}\mathbf{M} = \mathbf{0}$ for all other matrices.
  6. 1234
    11234
    22413
    33142
    44321
    1. (Closure) See Cayley table
    2. (Associativity) Integer multiplication is associative
    3. (Identity) \(1\)
    4. (Inverses) \(1 \mapsto 1, 2 \mapsto 3, 3 \mapsto 2, 4 \mapsto 4\) (See Cayley table)
(i)(iii)(iv)(vi)
(i)\(\checkmark\)\(\checkmark\) consider \(z \mapsto \begin{pmatrix} \cos \arg (z)- \sin \arg(z)
\sin \arg(z)\cos \arg(z) \end{pmatrix}\)not finitenot finite
(iii)\(\checkmark\)not finitenot finite
(iv)\(\checkmark\)no element order \(4\)
(vi)\(\checkmark\)

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1987 Paper 2 Q10
D: 1500.0 B: 1500.0
groups group axioms binary operation closure associativity identity element inverse proof abstract algebra

The set \(S\) consists of \(N(>2)\) elements \(a_{1},a_{2},\ldots,a_{N}.\) \(S\) is acted upon by a binary operation \(\circ,\) defined by \[ a_{j}\circ a_{k}=a_{m}, \] where \(m\) is equal to the greater of \(j\) and \(k\). Determine, giving reasons, which of the four group axioms hold for \(S\) under \(\circ,\) and which do not. Determine also, giving reasons, which of the group axioms hold for \(S\) under \(*\), where \(*\) is defined by \[ a_{j}*a_{k}=a_{n}, \] where \(n=\left|j-k\right|+1\).

Solution:

  1. (Closure) This operation is clearly closed by construction
  2. (Associative) \(a_j \circ (a_k \circ a_l) = a_j \circ a_{\max(k,l)} = a_{\max(j,k,l)} = a_{\max(j,k)} \circ a_l = (a_j \circ a_k) \circ a_l\), so it is associative
  3. (Identity) \(a_1 \circ a_k = a_{\max(1,k)} = a_k = a_{\max(k,1)} = a_k \circ a_1\) so \(a_1\) is an identity.
  4. (Inverses) There is no inverse, since \(a_N \circ a_k = a_N\) for all \(k\), and hence \(a_N\) can have no inverse.
  1. (Closure) \(n = |j-k|+1 \geq 1\) so we need to show that \(n \leq N\) to ensure closure. This is true since the largest \(j-k\) can be is if \(j = N\) and \(k = 1\), and this also satisfies \(|j-k| + 1 \leq N\). Hence the operation is closed.
  2. (Associative) \(a_j * (a_k * a_l) = a_j * (a_{|k-l|+1}) = a_{|j-|k-l|-1|+1}\). \((a_j * a_k) * a_l = a_{|j-k|+1}*a_l = a_{|l-|j-k|-1|+1}\). \(a_2 * (a_2 * a_3) = a_2 * a_2 = a_1\). \((a_2 *a_2)*a_3 = a_1 * a_3 = a_3 \neq a_2\) therefore this isn't associative for any \(N > 2\)
  3. (Identity) \(a_1\) is an identity, since \(a_1 * a_k = a_{|k-1|+1} = a_{k-1+1} = a_k\).
  4. (Inverse) Every element is self-inverse since \(a_k * a_k = a_{|k-k|+1} = a_1\)

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