Challenging +1.2 This question requires finding eigenvalues by solving the characteristic equation det(M - λI) = 0, which gives λ² - (2k+1)λ + (k²+k+1) = 0, then using the discriminant to show no real eigenvalues exist for any k. While it involves multiple steps (characteristic equation, discriminant, completing the square or showing discriminant is always negative), the technique is standard for Further Maths students and the algebraic manipulation is straightforward. It's moderately harder than average due to the parameter k and proof requirement, but not exceptionally challenging.
9 A transformation T of the plane is represented by the matrix \(\mathbf { M } = \left( \begin{array} { c c } k + 1 & - 1 \\ 1 & k \end{array} \right)\), where \(k\) is a
constant. constant.
Show that, for all values of \(k , \mathrm {~T}\) has no invariant lines through the origin.
If a specific value for k used, allow max of 3 M marks (SC)
or x + ky = m(kx + x − y) + c
x + k(mx + c) = m(kx + x − mx − c) + c
1 + km = km + m – m2 and kc = c − mc
soi
1 3
oe or by solving to get m = i
2 2
without wrong working
If invariant point (instead of line) only first M1 is available
Question 9:
9 | k+1 −1x kx+x−y
=
1 k y x+ky
y = mx x + ky = m(kx + x – y)
x + kmx = m(kx + x – mx)
1 + km = km + m – m2
m2 – m + 1 = 0
discriminant = (−1)2 – 4 = −3 < 0
so no real roots, and no invariant lines | M1
M1
A1
A1
M1
A1
[6] | 1.1
3.1a
2.1
1.1
3.1a
3.2a | If a specific value for k used, allow max of 3 M marks (SC)
or x + ky = m(kx + x − y) + c
x + k(mx + c) = m(kx + x − mx − c) + c
1 + km = km + m – m2 and kc = c − mc
soi
1 3
oe or by solving to get m = i
2 2
without wrong working
If invariant point (instead of line) only first M1 is available
9 A transformation T of the plane is represented by the matrix $\mathbf { M } = \left( \begin{array} { c c } k + 1 & - 1 \\ 1 & k \end{array} \right)$, where $k$ is a\\
constant. constant.
Show that, for all values of $k , \mathrm {~T}$ has no invariant lines through the origin.
\hfill \mbox{\textit{OCR MEI Further Pure Core AS 2023 Q9 [6]}}