Year: 2019
Paper: 2
Question Number: 8
Course: LFM Pure
Section: Matrices
The Pure questions were again the most popular of the paper, with only one of the questions attempted by fewer than half of the candidates (of the remaining four questions, only question 9 was attempted by more than a quarter of the candidates). In many of the questions candidates were often unable to make good use of the results shown in the earlier parts of the question in order to solve the more complex later parts. Nevertheless, some good solutions were seen to all of the questions. For many of the questions, solutions were seen in which the results were reached, but without sufficient justification of some of the steps.
Difficulty Rating: 1500.0
Difficulty Comparisons: 0
Banger Rating: 1638.7
Banger Comparisons: 14
The domain of the function f is the set of all $2 \times 2$ matrices and its range is the set of real numbers. Thus, if $M$ is a $2 \times 2$ matrix, then $f(M) \in \mathbb{R}$.
The function f has the property that $f(MN) = f(M)f(N)$ for any $2 \times 2$ matrices $M$ and $N$.
\begin{questionparts}
\item You are given that there is a matrix $M$ such that $f(M) \neq 0$. Let $I$ be the $2 \times 2$ identity matrix. By considering $f(MI)$, show that $f(I) = 1$.
\item Let $J = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$. You are given that $f(J) \neq 1$. By considering $J^2$, evaluate $f(J)$.
Using $J$, show that, for any real numbers $a$, $b$, $c$ and $d$,
$$.f\left(\begin{pmatrix} a & b \\ c & d \end{pmatrix}\right) = -f\left(\begin{pmatrix} c & d \\ a & b \end{pmatrix}\right) = f\left(\begin{pmatrix} d & c \\ b & a \end{pmatrix}\right)$$
\item Let $K = \begin{pmatrix} 1 & 0 \\ 0 & k \end{pmatrix}$ where $k \in \mathbb{R}$. Use $K$ to show that, if the second row of the matrix $A$ is a multiple of the first row, then $f(A) = 0$.
\item Let $P = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}$. By considering the matrices $P^2$, $P^{-1}$, and $K^{-1}PK$ for suitable values of $k$, evaluate $f(P)$.
\end{questionparts}\begin{questionparts}
\item Consider $f(M) = f(MI) = f(M)f(I)$. Since $f(M) \neq 0$ we can divide by $f(M)$ to obtain $f(I) = 1$
\item Let $J = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$, then $J^2 = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = I$. Therefore $1 = f(I) = f(J^2) = f(J)f(J) \Rightarrow f(J) = \pm 1 \Rightarrow f(J) = -1$ since $f(J) \neq 1$.
\begin{align*}
\begin{pmatrix} a & b \\ c & d \end{pmatrix}J &= \begin{pmatrix} a & b \\ c & d \end{pmatrix}\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} \\
&= \begin{pmatrix} b & a \\ d & c \end{pmatrix} \\
J\begin{pmatrix} a & b \\ c & d \end{pmatrix} &= \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}\begin{pmatrix} a & b \\ c & d \end{pmatrix} \\
&= \begin{pmatrix} c & d \\ a & b \end{pmatrix} \\
J\begin{pmatrix} a & b \\ c & d \end{pmatrix}J &=\begin{pmatrix} d & c \\ b & a \end{pmatrix}
\end{align*}
Therefore $f\left(\begin{pmatrix} c & d \\ a & b \end{pmatrix}\right) = f \left (J\begin{pmatrix} a & b \\ c & d \end{pmatrix} \right) = f(J) f\left(\begin{pmatrix} a & b \\ c & d \end{pmatrix}\right) = -f\left(\begin{pmatrix} a & b \\ c & d \end{pmatrix}\right)$ and
$f\left(\begin{pmatrix} d & c \\ b & a \end{pmatrix}\right) = f\left(J\begin{pmatrix} a & b \\ c & d \end{pmatrix}J \right) = f(J)f\left(\begin{pmatrix} a & b \\ c & d \end{pmatrix}\right)f(J) = f\left(\begin{pmatrix} a & b \\ c & d \end{pmatrix}\right)$ as required.
\item First consider $O$ the matrix of $0$, then
\begin{align*}
&& JO &= O \\
\Rightarrow && f(JO) &= f(O) \\
\Rightarrow && f(J)f(O) &= f(O) \\
\Rightarrow && -f(O) &= f(O) \\
\Rightarrow && f(O) &= 0
\end{align*}
Now consider $K_{k} = \begin{pmatrix} 1 & 0 \\ 0 & k \end{pmatrix}$. Suppose $A = \begin{pmatrix} a & b \\ ka & kb \end{pmatrix}$ then
\begin{align*}
K_{\frac1k}A &= \begin{pmatrix} 1 & 0 \\ 0 & \frac1k \end{pmatrix} \begin{pmatrix} a & b \\ ka & kb \end{pmatrix} \\
&= \begin{pmatrix} a & b \\ a & b \end{pmatrix}
\end{align*}
And so $f(K_{\frac1k}A) = f\left ( \begin{pmatrix} a & b \\ a & b \end{pmatrix} \right) = - f \left ( \begin{pmatrix} a & b \\ a & b \end{pmatrix} \right) = 0$, therefore either $f(K_{\frac1k}) = 0$ or $f(A) = 0$, but we know that $f(I) \neq 0$ therefore $f(A) = 0$.
\item Let $P = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}$, then $P^2 = \begin{pmatrix} 1 &2 \\ 0 & 1 \end{pmatrix}$, $P^{-1} = \begin{pmatrix} 1 & -1 \\ 0 & 1 \end{pmatrix}$, $K_k^{-1}PK_k = K_k^{-1}\begin{pmatrix} 1 & k \\ 0 & k \end{pmatrix} = \begin{pmatrix} 1 & k \\ 0 & 1 \end{pmatrix}$.
If $A$ has an inverse then $f(A) \neq 0$ since $1 = f(I) = f(A)f(A^{-1})$, in particular, $f(A)f(A^{-1}) = 1$. Using this for $K_2$ we have:
$f(P)^2 = f(P^2) = f(K_2^{-1}PK_2) = f(P)$ therefore $f(P) = 0, 1$, but since $f(P)$ has an inverse, $f(P) \neq 0$ so $f(P) = 1$
\end{questionparts}
Many good solutions were seen to this question, but solutions often lacked clear enough justification to be awarded full marks. However, there were also a surprising number of candidates who did not manage to invert the 2x2 matrices successfully. Candidates who claimed that the function f was the determinant of the matrix were not able to score high marks as the solutions did not then demonstrate that the results were true of any function satisfying the property given. The first two parts of this question were largely done well. The third part was found more difficult, with few candidates realising that the matrix with rows (a, b) and (ka, kb) could be factored as a product involving K. Those who did were then often able to provide a full solution, although often these were not fully justified. Several candidates instead used an alternative factorisation to produce a solution which covered all cases apart from the one where k = 0. In some cases, candidates did not appear to consider the matrix with rows (a, b) and (0, 0) to be an example of a matrix in which the second row was a multiple of the first. In part (iv) many candidates made use of the fact that f(P) ≠ 0 without showing that this must be the case.