A2 June 2022 Q2
2. Matrix \(\mathbf{M}\) is given by
\[\mathbf{M} = \begin{pmatrix} 1 & 0 & a \\ -3 & b & 1 \\ 0 & 1 & a \end{pmatrix}\]where \(a\) and \(b\) are integers, such that \(a \lt b\)
Given that the characteristic equation for \(\mathbf{M}\) is
\[\lambda^3 - 7\lambda^2 + 13\lambda + c = 0\]where \(c\) is a constant,
| Scheme | Marks | AO |
|---|---|---|
| \(\begin{vmatrix} 1-\lambda & 0 & a \\ -3 & b-\lambda & 1 \\ 0 & 1 & a-\lambda \end{vmatrix}\) \(= (1-\lambda)[(b-\lambda)(a-\lambda) - 1] + a(-3) (= 0)\) | M1 | 1.1b |
| \(\lambda^3 - (a+b+1)\lambda^2 + (a+b+ab-1)\lambda + (3a+1-ab) (= 0)\) o.e. \(-\lambda^3 + (a+b+1)\lambda^2 - (a+b+ab-1)\lambda + (ab-3a-1) (= 0)\) o.e. | A1 | 1.1b |
| \(\lambda^2 \Rightarrow\ a+b+1 = 7\) and \(\lambda \Rightarrow\ a+b+ab-1 = 13\) Solves simultaneously e.g. \(a+b = 6,\ ab = 8\) For example: leading to \(a^2 - 6a + 8 = 0 \Rightarrow a = \ldots\) | M1 | 3.1a |
| \(a = 2,\ b = 4\) | A1 | 1.1b |
| \(c = -1\) | A1 | 2.2a |
| (5) |
Notes
M1: Correct method to find the characteristic equation for \(\mathbf{M}\), condone missing = 0, and one slip as long as the intention is clear
A1: Multiplies out to achieve a correct characteristic equation, condone missing = 0
M1: A complete method to find the values of the constants \(a\) or \(b\). Equates their coefficients for \(\lambda^2\) and \(\lambda\) and solves simultaneously to find values for \(a\) or \(b\).
A1: Deduces the correct values for \(a\) and \(b\). \((a \lt b)\) following correct simultaneous equations
A1: Deduces the correct value for \(c\).
| Scheme | Marks | AO |
|---|---|---|
| \(\mathbf{M}^3 - 7\mathbf{M}^2 + 13\mathbf{M} + \text{‘}\text{their } c\text{’}\,\mathbf{I} = 0\) | B1ft | 1.1b |
| \(I = M^3 - 7M^2 + 13M \Rightarrow M^{-1} = M^2 - 7M + 13I\) \(\Rightarrow M^{-1} = \begin{pmatrix} 1 & 0 & 2 \\ -3 & 4 & 1 \\ 0 & 1 & 2 \end{pmatrix}^2 - 7\begin{pmatrix} 1 & 0 & 2 \\ -3 & 4 & 1 \\ 0 & 1 & 2 \end{pmatrix} + \begin{pmatrix} 13 & 0 & 0 \\ 0 & 13 & 0 \\ 0 & 0 & 13 \end{pmatrix} = \ldots\) \(= \begin{pmatrix} 1 & 2 & 6 \\ -15 & 17 & 0 \\ -3 & 6 & 5 \end{pmatrix} - 7\begin{pmatrix} 1 & 0 & 2 \\ -3 & 4 & 1 \\ 0 & 1 & 2 \end{pmatrix} + \begin{pmatrix} 13 & 0 & 0 \\ 0 & 13 & 0 \\ 0 & 0 & 13 \end{pmatrix} = \ldots\) | M1 | 3.1a |
| \(M^{-1} = \begin{pmatrix} 7 & 2 & -8 \\ 6 & 2 & -7 \\ -3 & -1 & 4 \end{pmatrix}\) | A1 | 1.1b |
| (3) | ||
| (8 marks) |
Notes
B1ft: Uses Cayley-Hamilton theorem to produce equation replacing \(\lambda\) with \(\mathbf{M}\) and constant term with constant multiple of the identity matrix \(\mathbf{I}\). Follow through on their value for \(c\). This mark may be implied by the M mark.
M1: A complete method to find \(M^{-1}\) using the Cayley-Hamilton theorem. The minimum is for writing an expression for \(M^{-1}\) from their characteristic equation, for example \(M^{-1} = M^2 - 7M + 13I\) and then stating an answer for \(M^{-1}\), they may have used their calculator, there is no need to check.
A1: Correct \(M^{-1}\)