Date | May 2016 | Marks available | 4 | Reference code | 16M.1.hl.TZ0.14 |
Level | HL only | Paper | 1 | Time zone | TZ0 |
Command term | Explain and Find | Question number | 14 | Adapted from | N/A |
Question
A matrix M is called idempotent if M2= M.
The idempotent matrix N has the form
N =(a−2aa−2a)
where a≠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.
Markscheme
(i) M2= 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
detM^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]
Examiners report
This was a more successful question for many candidates with a number of fully correct solutions being seen and a significant number of partially correct answers. Most candidates understood what was required from part (a)(i), but part (ii) often resulted in unnecessarily complex algebra which they were unable to manipulate. Part (b) resulted in many wholly successful answers, provided candidates realised the need for care in terms of the manipulation.
This was a more successful question for many candidates with a number of fully correct solutions being seen and a significant number of partially correct answers. Most candidates understood what was required from part (a)(i), but part (ii) often resulted in unnecessarily complex algebra which they were unable to manipulate. Part (b) resulted in many wholly successful answers, provided candidates realised the need for care in terms of the manipulation.