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Question
Let S be the set of matrices given by
\(\left[ \begin{array}{l}
a\\
c
\end{array} \right.\left. \begin{array}{l}
b\\
d
\end{array} \right]\) ; \(a,b,c,d \in \mathbb{R}\), \(ad – bc = 1\)
The relation \(R\) is defined on \(S\) as follows. Given \(\boldsymbol{A}\) , \(\boldsymbol{B} \in S\) , \(\boldsymbol{ARB}\) if and only if there exists \(\boldsymbol{X} \in S\) such that \(\boldsymbol{A} = \boldsymbol{BX}\) .
Show that \(R\) is an equivalence relation.
The relationship between \(a\) , \(b\) , \(c\) and \(d\) is changed to \(ad – bc = n\) . State, with a reason, whether or not there are any non-zero values of \(n\) , other than \(1\), for which \(R\) is an equivalence relation.
Answer/Explanation
Markscheme
since \(\boldsymbol{A} = \boldsymbol{AI}\) where \(\boldsymbol{I}\) is the identity A1
and \(\det (\boldsymbol{I}) = 1\) , A1
\(R\) is reflexive
\(\boldsymbol{ARB} \Rightarrow \boldsymbol{A} = \boldsymbol{BX}\) where \(\det (\boldsymbol{X}) = 1\) M1
it follows that \(\boldsymbol{B} = \boldsymbol{A}{\boldsymbol{X}^{ – 1}}\) A1
and \(\det ({\boldsymbol{X}^{ – 1}}) = \det{(\boldsymbol{X})^{ – 1}} = 1\) A1
\(R\) is symmetric
\(\boldsymbol{ARB}\) and \(\boldsymbol{BRC} \Rightarrow \boldsymbol{A} = \boldsymbol{BX}\) and \(\boldsymbol{B} = \boldsymbol{CY}\) where \(\det (\boldsymbol{X}) = \det (\boldsymbol{Y}) = 1\) M1
it follows that \(\boldsymbol{A} = \boldsymbol{CYX}\) A1
\(\det (\boldsymbol{YX}) = \det (\boldsymbol{Y})\det (\boldsymbol{X}) = 1\) A1
\(R\) is transitive
hence \(R\) is an equivalence relation AG
[8 marks]
for reflexivity, we require \(\boldsymbol{ARA}\) so that \(\boldsymbol{A} = \boldsymbol{AI}\) (for all \(\boldsymbol{A} \in S\) ) M1
since \(\det (\boldsymbol{I}) = 1\) and we require \(\boldsymbol{I} \in S\) the only possibility is \(n = 1\) A1
[2 marks]
Question
The matrix A is given by A = \(\left( {\begin{array}{*{20}{c}}1&2&1\\1&1&2\\2&3&1\end{array}} \right)\).
(a) Given that A\(^3\) can be expressed in the form A\(^3 = a\)A\(^2 = b\)A \( + c\)I, determine the values of the constants \(a\), \(b\), \(c\).
(b) (i) Hence express A\(^{ – 1}\) in the form A\(^{ – 1} = d\)A\(^2 = e\)A \( + f\)I where \(d,{\text{ }}e,{\text{ }}f \in \mathbb{Q}\).
(ii) Use this result to determine A\(^{ – 1}\).
Answer/Explanation
Markscheme
(a) successive powers of A are given by
A\(^2 = \) \(\left( {\begin{array}{*{20}{c}}5&7&6\\6&9&5\\7&{10}&9\end{array}} \right)\) (M1)A1
A\(^3 = \) \(\left( {\begin{array}{*{20}{c}}{24}&{35}&{25}\\{25}&{36}&{29}\\{35}&{51}&{36}\end{array}} \right)\) A1
it follows, considering elements in the first rows, that
\(5a + b + c = 24\)
\(7a + 2b = 35\)
\(6a + b = 25\) M1A1
solving, (M1)
\((a,{\text{ }}b,{\text{ }}c) = (3,{\text{ }}7,{\text{ }}2)\) A1
Note: Accept any other three correct equations.
Note: Accept the use of the Cayley–Hamilton Theorem.
[7 marks]
(b) (i) it has been shown that
A\(^3 = 3\)A\(^2 + 7\)A\( + 2\)I
multiplying by A\(^{ – 1}\), M1
A\(^2 = 3\)A\( + 7\)I\( + 2\)A\(^{ – 1}\) A1
whence
A\(^{ – 1} = 0.5\)A\(^2 – 1.5\)A \( – 3.5\)I A1
(ii) substituting powers of A,
A\(^{ – 1} = 0.5\)\(\left( {\begin{array}{*{20}{c}}5&7&6\\6&9&5\\7&{10}&9\end{array}} \right) – 1.5\left( {\begin{array}{*{20}{c}}1&2&1\\1&1&2\\2&3&1\end{array}} \right) – 3.5\left( {\begin{array}{*{20}{c}}1&0&0\\0&1&0\\0&0&1\end{array}} \right)\) M1
=\(\left( {\begin{array}{*{20}{c}}{ – 2.5}&{0.5}&{1.5}\\{1.5}&{ – 0.5}&{ – 0.5}\\{0.5}&{0.5}&{ – 0.5}\end{array}} \right)\) A1
Note: Follow through their equation in (b)(i).
Note: Line (ii) of (ii) must be seen.
[5 marks]
Question
A matrix M is called idempotent if M\(^2 = \) M.
The idempotent matrix N has the form
N \( = \left( {\begin{array}{*{20}{c}} a&{ – 2a} \\ a&{ – 2a} \end{array}} \right)\)
where \(a \ne 0\).
(i) Explain why M is a square matrix.
(ii) Find the set of possible values of det(M).
(i) Find the value of \(a\).
(ii) Find the eigenvalues of N.
(iii) Find corresponding eigenvectors.
Answer/Explanation
Markscheme
(i) M\(^2 = \) MM only exists if the number of columns of M equals the number of rows of M R1
hence M is square AG
(ii) apply the determinant function to both sides M1
\(\det (\)M\(^2) = \det (\)M\()\)
use the multiplicative property of the determinant
\(\det (\)M\(^2) = \det (\)M\(){\text{ }}\det (\)M\() = \det (\)M\()\) (M1)
hence \(\det (\)M\() = 0\) or 1 A1
[4 marks]
(i) attempt to calculate N\(^2\) M1
obtain \(\left( {\begin{array}{*{20}{c}} { – {a^2}}&{2{a^2}} \\ { – {a^2}}&{2{a^2}} \end{array}} \right)\) A1
equating to N M1
to obtain \(a = – 1\) A1
(ii) N \( = \left( {\begin{array}{*{20}{c}} { – 1}&2 \\ { – 1}&2 \end{array}} \right)\)
N \( – \lambda \)I \( = \left( {\begin{array}{*{20}{c}} { – 1 – \lambda }&2 \\ { – 1}&{2 – \lambda } \end{array}} \right)\) M1
\(( – 1 – \lambda )(2 – \lambda ) + 2 = 0\) (A1)
\({\lambda ^2} – \lambda = 0\) (A1)
\(\lambda \) is 1 or 0 A1
(iii) let \(\lambda = 1\)
to obtain \(\left( {\begin{array}{*{20}{c}} { – 1}&2 \\ { – 1}&2 \end{array}} \right)\left( {\begin{array}{*{20}{c}} x \\ y \end{array}} \right) = \left( {\begin{array}{*{20}{c}} x \\ y \end{array}} \right){\text{ or }}\left( {\begin{array}{*{20}{c}} { – 2}&2 \\ { – 1}&1 \end{array}} \right)\left( {\begin{array}{*{20}{c}} x \\ y \end{array}} \right) = \left( {\begin{array}{*{20}{c}} 0 \\ 0 \end{array}} \right)\) M1
hence eigenvector is \(\left( {\begin{array}{*{20}{c}} x \\ x \end{array}} \right)\) A1
let \(\lambda = 0\)
to obtain \(\left( {\begin{array}{*{20}{c}} { – 1}&2 \\ { – 1}&2 \end{array}} \right)\left( {\begin{array}{*{20}{c}} x \\ y \end{array}} \right) = \left( {\begin{array}{*{20}{c}} 0 \\ 0 \end{array}} \right)\) M1
hence eigenvector is \(\left( {\begin{array}{*{20}{c}} {2y} \\ y \end{array}} \right)\) A1
Note: Accept specific eigenvectors.
[12 marks]
Question
The set \(S\) contains the eight matrices of the form\[\left( {\begin{array}{*{20}{c}}
a&0&0\\
0&b&0\\
0&0&c
\end{array}} \right)\]where \(a\), \(b\), \(c\) can each take one of the values \( + 1\) or \( – 1\) .
Show that any matrix of this form is its own inverse.
Show that \(S\) forms an Abelian group under matrix multiplication.
Giving a reason, state whether or not this group is cyclic.
Answer/Explanation
Markscheme
\(\left( {\begin{array}{*{20}{c}}
a&0&0\\
0&b&0\\
0&0&c
\end{array}} \right)\left( {\begin{array}{*{20}{c}}
a&0&0\\
0&b&0\\
0&0&c
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
{{a^2}}&0&0\\
0&{{b^2}}&0\\
0&0&{{c^2}}
\end{array}} \right)\) A1M1
\( = \left( {\begin{array}{*{20}{c}}
1&0&0\\
0&1&0\\
0&0&1
\end{array}} \right)\) A1
this shows that each matrix is self-inverse
[3 marks]
closure:
\(\left( {\begin{array}{*{20}{c}}
{{a_1}}&0&0\\
0&{{b_1}}&0\\
0&0&{{c_1}}
\end{array}} \right)\left( {\begin{array}{*{20}{c}}
{{a_2}}&0&0\\
0&{{b_2}}&0\\
0&0&{{c_2}}
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
{{a_1}{a_2}}&0&0\\
0&{{b_1}{b_2}}&0\\
0&0&{{c_1}{c_2}}
\end{array}} \right)\) M1A1
\( = \left( {\begin{array}{*{20}{c}}
{{a_3}}&0&0\\
0&{{b_3}}&0\\
0&0&{{c_3}}
\end{array}} \right)\)
where each of \({a_3}\), \({b_3}\), \({c_3}\) can only be \( \pm 1\) A1
this proves closure
identity: the identity matrix is the group identity A1
inverse: as shown above, every element is self-inverse A1
associativity: this follows because matrix multiplication is associative A1
\(S\) is therefore a group AG
Abelian:
\(\left( {\begin{array}{*{20}{c}}
{{a_2}}&0&0\\
0&{{b_2}}&0\\
0&0&{{c_2}}
\end{array}} \right)\left( {\begin{array}{*{20}{c}}
{{a_1}}&0&0\\
0&{{b_1}}&0\\
0&0&{{c_1}}
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
{{a_2}{a_1}}&0&0\\
0&{{b_2}{b_1}}&0\\
0&0&{{c_2}{c_1}}
\end{array}} \right)\) A1
\(\left( {\begin{array}{*{20}{c}}
{{a_1}}&0&0\\
0&{{b_1}}&0\\
0&0&{{c_1}}
\end{array}} \right)\left( {\begin{array}{*{20}{c}}
{{a_2}}&0&0\\
0&{{b_2}}&0\\
0&0&{{c_2}}
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
{{a_1}{a_2}}&0&0\\
0&{{b_1}{b_2}}&0\\
0&0&{{c_1}{c_2}}
\end{array}} \right)\) A1
Note: Second line may have been shown whilst proving closure, however a reference to it must be made here.
we see that the same result is obtained either way which proves commutativity so that the group is Abelian R1
[9 marks]
since all elements (except the identity) are of order \(2\), the group is not cyclic (since \(S\) contains \(8\) elements) R1
[1 mark]