A2 June 2022 Paper 1 Q5
5.
\[\mathbf{M} = \begin{pmatrix}a & 2 & -3\\ 2 & 3 & 0\\ 4 & a & 2\end{pmatrix} \qquad \text{where } a \text{ is a constant}\]| Scheme | Marks | AO |
|---|---|---|
| \(\det(\mathbf{M}) = a(6) - 2(4) - 3(2a - 12)\) | M1 | 1.1b |
| \(\det(\mathbf{M}) = 28 \neq 0\) therefore, non-singular for all values of \(a\) | A1 | 2.4 |
| (2) |
Notes
M1: Finds the determinant of the matrix \(\mathbf{M}\). Must be seen in part (a). Allow one slip if no method shown.
A1: Correct value for determinant, states doesn’t equal 0 (accept \(\gt 0\)) and draws the conclusion that the matrix is non-singular. If non-singular meaning determinant is non-zero is given in a preamble then accept a minimal conclusion (e.g. “hence shown”), but there must be a conclusion.
| Scheme | Marks | AO |
|---|---|---|
| Finds the matrix of minors \(\begin{pmatrix}6 & 4 & 2a - 12\\ 4 + 3a & 2a + 12 & a^2 - 8\\ 9 & 6 & 3a - 4\end{pmatrix}\) | M1 | 1.1b |
| Finds the matrix of cofactors and transposes. \(\begin{pmatrix}6 & -4 - 3a & 9\\ -4 & 2a + 12 & -6\\ 2a - 12 & 8 - a^2 & 3a - 4\end{pmatrix}\) | M1 | 1.1b |
| \(\dfrac{1}{28}\begin{pmatrix}6 & -4 - 3a & 9\\ -4 & 2a + 12 & -6\\ 2a - 12 & 8 - a^2 & 3a - 4\end{pmatrix}\) | M1 A1 | 1.1b 2.1 |
| (4) | ||
| (6 marks) |
Notes
M1: Finds the matrix of minors, at least 5 correct values.
M1: Finds the matrix of cofactors and transposes (in either order). Note: some will do all these steps in one go, which is fine as long as it is clear what they have done. Allow minor slips if the process is clearly correct.
M1: Completes the process to find the inverse matrix, divides by the determinant.
A1: Correct matrix.