AS October 2020 Q3
3.
| Scheme | Marks | AO |
|---|---|---|
| \(\begin{vmatrix} 1-\lambda & -2 \\ 1 & 4-\lambda \end{vmatrix} = (1-\lambda)(4-\lambda) + 2 = 0\) | M1 | 1.1b |
| \(\Rightarrow 4 - 5\lambda + \lambda^2 + 2 = 0 \Rightarrow \lambda^2 - 5\lambda + 6 = 0\) * | A1* | 1.1b |
| (2) |
Notes
M1: Attempts the determinant of \(\mathbf{A} - \lambda\mathbf{I}\)
A1*: Fully correct proof
| Scheme | Marks | AO |
|---|---|---|
| \(\mathbf{A}^2 - 5\mathbf{A} + 6\mathbf{I} = 0\) | M1 | 1.1b |
| \(\mathbf{A}^3 - 5\mathbf{A}^2 + 6\mathbf{A} = 0 \Rightarrow \mathbf{A}^3 = 5(5\mathbf{A} - 6\mathbf{I}) - 6\mathbf{A}\) | M1 | 3.1a |
| \(\mathbf{A}^3 = 19\mathbf{A} - 30\mathbf{I}\) | A1 | 1.1b |
| (3) |
Notes
M1: Applies the Cayley-Hamilton theorem to the equation given in (a)(i)
M1: A full method leading to \(\mathbf{A}^3\) by multiplying by \(\mathbf{A}\) and substituting for \(\mathbf{A}^2\)
A1: Deduces the correct expression or correct values for \(p\) and \(q\)
Alternative to part (b)
| Scheme | Marks | AO |
|---|---|---|
| \(\lambda^2 - 5\lambda + 6 = 0 \Rightarrow \lambda^3 - 5\lambda^2 + 6\lambda = 0 \Rightarrow \lambda^3 = 5(5\lambda - 6) - 6\lambda\) | M1 | 3.1a |
| \(\mathbf{A}^3 = 5(5\mathbf{A} - 6\mathbf{I}) - 6\mathbf{A}\) | M1 | 1.1b |
| \(\mathbf{A}^3 = 19\mathbf{A} - 30\mathbf{I}\) | A1 | 1.1b |
| (3) |
M1: A full method leading to \(\lambda^3\) in terms of \(\lambda\)
M1: Applies the Cayley-Hamilton theorem
A1: Deduces the correct expression or correct values for \(p\) and \(q\)
| Scheme | Marks | AO |
|---|---|---|
| \(\mathbf{M} = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \Rightarrow \begin{pmatrix} a & b \\ c & d \end{pmatrix}\begin{pmatrix} 1 \\ 2-\mathrm{i} \end{pmatrix} = (-1+\mathrm{i})\begin{pmatrix} 1 \\ 2-\mathrm{i} \end{pmatrix}\) or \(\mathbf{M} = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \Rightarrow \begin{pmatrix} a & b \\ c & d \end{pmatrix}\begin{pmatrix} 1 \\ 2+\mathrm{i} \end{pmatrix} = (-1-\mathrm{i})\begin{pmatrix} 1 \\ 2+\mathrm{i} \end{pmatrix}\) | M1 | 1.1b |
| \(a + b(2-\mathrm{i}) = -1+\mathrm{i} \quad a + b(2+\mathrm{i}) = -1-\mathrm{i}\) \(c + d(2-\mathrm{i}) = -1+3\mathrm{i} \quad c + d(2+\mathrm{i}) = -1-3\mathrm{i}\) | A1 | 1.1b |
| \(a + b(2-\mathrm{i}) = -1+\mathrm{i},\ a + b(2+\mathrm{i}) = -1-\mathrm{i} \Rightarrow a = 1,\ b = -1\) or \(c + d(2-\mathrm{i}) = -1+3\mathrm{i},\ c + d(2+\mathrm{i}) = -1-3\mathrm{i} \Rightarrow c = 5,\ d = -3\) | M1 A1 | 3.1a 1.1b |
| \(\mathbf{M} = \begin{pmatrix} 1 & -1 \\ 5 & -3 \end{pmatrix}\) | A1 | 2.2a |
| (5) | ||
| (10 marks) |
Notes
M1: Uses a general matrix and sets up at least one matrix equation using the information given in the question
A1: Correct equations in terms of \(a\), \(b\), \(c\) and \(d\)
M1: Solves simultaneously to find values for all of \(a\), \(b\), \(c\) and \(d\)
A1: One correct pair of values
A1: Deduces the correct matrix \(\mathbf{M}\)
Alternative to part (ii)
| Scheme | Marks | AO |
|---|---|---|
| \(\mathbf{P} = \begin{pmatrix} 1 & 1 \\ 2-\mathrm{i} & 2+\mathrm{i} \end{pmatrix} \Rightarrow \mathbf{P}^{-1} = \dfrac{1}{2\mathrm{i}}\begin{pmatrix} 2+\mathrm{i} & -1 \\ \mathrm{i}-2 & 1 \end{pmatrix}\) | M1 A1 | 1.1b 1.1b |
| \(\mathbf{D} = \mathbf{P}^{-1}\mathbf{M}\mathbf{P} \Rightarrow \mathbf{M} = \mathbf{P}\mathbf{D}\mathbf{P}^{-1}\) \(\mathbf{M} = \dfrac{1}{2\mathrm{i}}\begin{pmatrix} 1 & 1 \\ 2-\mathrm{i} & 2+\mathrm{i} \end{pmatrix}\begin{pmatrix} -1+\mathrm{i} & 0 \\ 0 & -1-\mathrm{i} \end{pmatrix}\begin{pmatrix} 2+\mathrm{i} & -1 \\ \mathrm{i}-2 & 1 \end{pmatrix} = \ldots\) | M1 | 3.1a |
| \(\mathbf{M} = \begin{pmatrix} 1 & -1 \\ 5 & -3 \end{pmatrix}\) | A1 A1 | 1.1b 2.2a |
| (5) |
M1: Attempts to find the inverse of the matrix of eigenvectors
A1: Correct matrix
M1: Attempts \(\mathbf{PDP}^{-1}\) where \(\mathbf{D}\) is the diagonal matrix of eigenvalues
A1: At least 2 elements correct
A1: Deduces the correct matrix \(\mathbf{M}\)