Stretching Question 2

Friday 13 October 2006 at 3:14 pm | In Articles | 3 Comments

In Stretching Question I wondered how students would cope with a type of question they wouldn’t have seen before:

    If M=\begin{pmatrix}
\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\ \vspace{-2mm} \\
-\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}}
\end{pmatrix} calculate M^2,\,M^4 and M^{2006}

The examiners’ report is now out and it says (edited to show context)

    Attempts at calculating M^{2006} were rarely completely successful. Many candidates found that M^8 was the identity matrix but then went on to say that M^{2006} must also be the identity matrix. Others gave the value of M^2 or gave the correct answer without any indication as to their reason for doing so, or with incomplete reasoning.

How disappointing that so few managed to complete the question successfully. The lack of reasoning doesn’t tell us whether students guessed or found it too easy to bother putting down the working. It might have helped if students were taught that if a and b are integers then there exist integers q and r with 0 \leq r<b such that a=qb+r but these fundamental ideas seemed to have been lost to A level.

In the first AS exam the examiners comment that

    When asked to use the Factor Theorem or Remainder Theorem, no marks can be earned for using long division.

which means

    British students haven’t a clue how to do long division of polynomials so we’ve made it easier for you, so don’t try and pretend you know best. (See A Level Exam Questions).

3 Comments »

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  1. the matrix looks like the rotational matrix for rotating a point 45 degrees. so since M is a rotation by 45 degress then M^n is a rotation by (45)(n) degrees.

    Comment by nick james — Sunday 22 October 2006 8:45 am #

  2. Good point especially as students were asked to identify the transformation that M represents. I wonder how many, if any, students used that when they found that M^8=I.

    Whatever way is used, students still need to be somehow aware of the fact that 2006\equiv 6 \pmod{8}

    Comment by Steve — Sunday 22 October 2006 12:37 pm #

  3. it is easy to see some cosinuses or sinuses there,so the problem is solved whereas
    there are many trigonometric relations known

    Comment by stefan petrea — Tuesday 31 October 2006 1:29 am #

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