A2 June 2024 Paper 1 Q3
3 A transformation T is represented by the matrix \(\mathbf{N} = \begin{pmatrix} a & 4 & 2 \\ 5 & 1 & 0 \\ 3 & 6 & 3 \end{pmatrix}\), where \(a\) is a constant.
The value of \(a\) is 13 to the nearest integer.
A shape \(S_1\) has volume 11.6 to 1 decimal place. Shape \(S_1\) is mapped to shape \(S_2\) by the transformation T.
A student claims that the volume of \(S_2\) is less than 400.
| Scheme | Marks | AO |
|---|---|---|
| For reference: \(\mathbf{N} = \begin{pmatrix} a & 4 & 2 \\ 5 & 1 & 0 \\ 3 & 6 & 3 \end{pmatrix}\) | ||
| \(\mathbf{N}^2 = \begin{pmatrix} a^2 + 26 & 4a + 16 & 2a + 6 \\ 5a + 5 & 21 & 10 \\ 3a + 39 & 36 & 15 \end{pmatrix}\) | M1 M1 A1 | 1.1 1.1 1.1 |
| [3] |
Notes
M1: A 3 by 3 matrix with at least three correct (simplified) entries.
M1: A 3 by 3 matrix with at least two correctly simplified rows or at least two correctly simplified columns.
A1: cao
| Scheme | Marks | AO |
|---|---|---|
| \(\det\mathbf{N} = a\begin{vmatrix} 1 & 0 \\ 6 & 3 \end{vmatrix} - 4\begin{vmatrix} 5 & 0 \\ 3 & 3 \end{vmatrix} + 2\begin{vmatrix} 5 & 1 \\ 3 & 6 \end{vmatrix}\) | M1 | 1.1 |
| \(= 3a - 6\) | A1 | 1.1 |
| [2] |
Notes
M1: Correct method. May see \(a(3 - 0) - 4(15 - 0) + 2(30 - 3)\). Ignore sign errors in \(2 \times 2\) determinant calculations, but cofactors must have correct signs. Do look out for expanding by other rows and columns.
If using Sarrus’ method, must see correct 66 and \(3a + 60\).
Or for \(\begin{pmatrix} a \\ 5 \\ 3 \end{pmatrix} \bullet \left(\begin{pmatrix} 4 \\ 1 \\ 6 \end{pmatrix} \times \begin{pmatrix} 2 \\ 0 \\ 3 \end{pmatrix}\right) = \begin{pmatrix} a \\ 5 \\ 3 \end{pmatrix} \bullet \begin{pmatrix} k_1 \\ k_2 \\ k_3 \end{pmatrix}\) (oe) with at least one of \(k_1 = 3, k_2 = 0, k_3 = -2\) correct.
A1: cao
| Scheme | Marks | AO |
|---|---|---|
| \(11.6 \times \det\mathbf{N}\) | M1 | 1.1 |
| For example, Upper bound is \(11.65 \times (3 \times 13.5 - 6) = 401.925\) Lower bound is \(11.55 \times (3 \times 12.5 - 6) = 363.825\) | A1 | 3.1a |
| So, the student’s claim (that the volume is less than 400) is not necessarily true. | A1 | 2.2b |
| [3] |
Notes
M1: Or \(k \times \det\mathbf{N}\) where \(11.55 \leqslant k \leqslant 11.65\) and \(12.5 \leqslant a \leqslant 13.5\) in their \(\det\mathbf{N}\) from part (b). If not calculating the correct UB and LB (from a correct determinant) then we must see either a correct calculation (for their determinant from part (b)) or they must state, the values they’ve used in their calculation for both \(a\) and the volume of \(S_1\) to award any marks.
A1: For either correct upper or lower bound, or any correct \(V_2\) for any value of \(a\) between 12.5 and 13.5, and \(11.55 \lt V_1 \lt 11.65\). Allow answers rounded or truncated to the nearest integer or greater degree of accuracy.
A1: Demonstrates that there are values of \(12.5 \leqslant a \lt 13.5\) and \(11.55 \leqslant k \lt 11.65\) such that both \(V \gt 400\), and \(V \lt 400\) and concludes that the volume may be greater or less than 400 (depending on a more accurately determined value of \(a\) and the volume of \(S_1\)). Conclusion must indicate that the claim could be correct but not necessarily so (oe).
If M0 then SC1 for LB = 144.375 and UB = 157.275 (using 13 for \(\det\mathbf{N}\)).