AS June 2024 Q3
3.
In this question you must show all stages of your working.
Solutions relying on calculator technology are not acceptable.
\[\mathbf{A} = \begin{pmatrix} 3 & k \\ -5 & 2 \end{pmatrix}\]where \(k\) is a constant.
Given that there exists a matrix \(\mathbf{P}\) such that \(\mathbf{P}^{-1}\mathbf{A}\mathbf{P}\) is a diagonal matrix where
\[\mathbf{P}^{-1}\mathbf{A}\mathbf{P} = \begin{pmatrix} 8 & 0 \\ 0 & -3 \end{pmatrix}\]| Scheme | Marks | AO |
|---|---|---|
| Uses entries on the diagonal matrix so eigenvalues are \(-3\) and 8 | B1 | 1.2 |
| \(\begin{vmatrix} 3-\lambda & k \\ -5 & 2-\lambda \end{vmatrix} = 0 \Rightarrow (3-\lambda)(2-\lambda) + 5k = 0\) \(\Rightarrow \lambda^2 - 5\lambda + 6 + 5k = 0\) \(\Rightarrow 6 + 5k = -24 \Rightarrow k = \ldots\) | M1 | 3.1a |
| \(k = -6\) | A1 | 1.1b |
| (3) |
Notes
B1: Recalls that the eigenvalues form the entries of the diagonal matrix in the diagonalisation process. May be implied by working or stated, for work in (a) or (b). Implied by an attempt at equating the determinants of the two matrices.
M1: For a full method to find \(k\), e.g. by finding the characteristic equation and using (one of) the eigenvalues to find \(k\). The factor theorem can be used with either eigenvalue, for example, after finding the characteristic equation.
A1: Correct value.
Note Using \(k = -6\)
B1: Recalls that the eigenvalues form the entries of the diagonal matrix in the diagonalisation process.
M1: Uses \(k = -6\), finds the characteristic equation and solves to find the eigenvalues.
A1: Draws a conclusion, same eigenvalues therefore \(k = -6\)
| Scheme | Marks | AO |
|---|---|---|
| For \(\lambda = -3\) eigenvector equations are \(\left\{\begin{matrix} 6x - \text{“}6\text{”}y = 0 \\ -5x + 5y = 0 \end{matrix}\right. \Rightarrow x, y = \ldots\) OR For \(\lambda = 8\) eigenvector equations are \(\left\{\begin{matrix} -5x - \text{“}6\text{”}y = 0 \\ -5x - 6y = 0 \end{matrix}\right. \Rightarrow x, y = \ldots\) | M1 | 2.1 |
| (For \(\lambda = -3\), \(x = y\), for \(\lambda = 8\), \(5x = -6y\) so eigenvectors are) One of \(\begin{pmatrix} 1 \\ 1 \end{pmatrix}\) o.e and \(\begin{pmatrix} 6 \\ -5 \end{pmatrix}\) o.e. | A1 | 1.1b |
| Both of \(\begin{pmatrix} 1 \\ 1 \end{pmatrix}\) o.e. and \(\begin{pmatrix} 6 \\ -5 \end{pmatrix}\) o.e. | A1 | 1.1b |
| \(\mathbf{P} = \begin{pmatrix} 6 & 1 \\ -5 & 1 \end{pmatrix}\) o.e. | B1ft | 2.2a |
| (4) | ||
| (7 marks) |
Notes
M1: Correct method to find an eigenvector for either of their eigenvalues.
A1: One correct eigenvector – accept any non-zero multiples of their eigenvalues.
A1: A correct eigenvector for each of their eigenvalues.
B1ft: Gives \(\mathbf{P}\) as a matrix with their columns in the correct order for the given diagonal matrix.