# Question 2:

## Part (a):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\det(\mathbf{M} - \lambda\mathbf{I}) = \begin{vmatrix} 2-\lambda & 0 & 3 \\ 0 & -4-\lambda & -3 \\ 0 & -4 & -\lambda \end{vmatrix}$ $= (2-\lambda)[(-4-\lambda)(-\lambda)-(-4)(-3)] - 0 + 3(0)$ or $(2-\lambda)[(-4-\lambda)(-\lambda)-(-4)(-3)] - 0 + 0$ Sarrus $\Rightarrow (2-\lambda)(-4-\lambda)(-\lambda) - (2-\lambda)(-3)(-4)$ | M1 | Obtains an unsimplified cubic expression for $\det(\mathbf{M}-\lambda\mathbf{I})$ condoning sign/copying slips only. Allow poor bracketing if intention clear. |
| Note: possible to use $\mathbf{M}x = \lambda x$ e.g., $-4y = \lambda z \Rightarrow y = -\frac{\lambda z}{4}$ and $-4y - 3z = \lambda y \Rightarrow \lambda z - 3z = -\frac{\lambda^2 z}{4} \Rightarrow \lambda^2 + 4\lambda - 12 = 0 \Rightarrow \ldots$ Score M1 for achieving a 3TQ in $\lambda$ from appropriate work condoning copying/sign slips only | | |
| $(2-\lambda)(\lambda^2 + 4\lambda - 12) = 0$ or $\lambda^3 + 2\lambda^2 - 20\lambda + 24 = 0$ or $-\lambda^3 - 2\lambda^2 + 20\lambda - 24 = 0$ $(2-\lambda)(\lambda-2)(\lambda+6) = 0$ or $(\lambda+6)(\lambda-2)(\lambda-2) = 0$ $\lambda_1 = -6\;\;(\lambda_2 = 2)$ | dM1 A1 | dM1: Solves $\det(\mathbf{M}-\lambda\mathbf{I}) = 0$ to obtain any value for $\lambda$ including 2. Not usual rules – award for any value seen that is consistent with their equation. The "$= 0$" can be implied by a solution. Note that they may disregard the $(2-\lambda)$ and solve a quadratic. A1: $-6$ from a correct equation. Accept both solutions e.g., "$-6, 2$" and allow if mislabelled and/or $-6$ rejected. No incorrect solutions. |
| $\mathbf{M}x = -6x \Rightarrow \begin{cases} 2x+3z = -6x \\ -4y-3z = -6y \\ -4y = -6z \end{cases} \Rightarrow x = \ldots, y = \ldots, z = \ldots$ $(\mathbf{M}+6\mathbf{I})x = 0 \Rightarrow \begin{cases} 8x+3z = 0 \\ 2y-3z = 0 \end{cases} \Rightarrow x = \ldots, y = \ldots, z = \ldots$ $-4y+6z = 0$ | M1 | Uses $\mathbf{M}x = \lambda x$ or $(\mathbf{M}-\lambda\mathbf{I})x = 0$ with any of their non-zero eigenvalues (however obtained) to form simultaneous equations and solves. No requirement for a vector for this mark. There is no need to check their values but award M0 for a zero solution. Note: Could find vector product of first 2 rows of $\mathbf{M}-\lambda\mathbf{I}$ i.e., $(8\mathbf{i}+3\mathbf{k})\times(2\mathbf{j}-3\mathbf{k}) = (-6\mathbf{i}+24\mathbf{j}+16\mathbf{k})$ (two correct components) |
| $\Rightarrow \begin{pmatrix}x\\y\\z\end{pmatrix} = \begin{pmatrix}-3\\12\\8\end{pmatrix} \Rightarrow \frac{1}{\sqrt{3^2+12^2+8^2}}\begin{pmatrix}-3\\12\\8\end{pmatrix}$ | M1 | Correct method to normalise their eigenvector no matter how this vector is obtained provided it has at least 2 non-zero components. Only allow slips if there is working. |
| e.g., $\frac{1}{\sqrt{217}}\begin{pmatrix}-3\\12\\8\end{pmatrix}$ or $\begin{pmatrix}-\frac{3\sqrt{217}}{217}\\ \frac{12\sqrt{217}}{217}\\ \frac{8\sqrt{217}}{217}\end{pmatrix}$ or $\begin{pmatrix}-\frac{3}{\sqrt{217}}\\ \frac{12}{\sqrt{217}}\\ \frac{8}{\sqrt{217}}\end{pmatrix}$ or $\frac{1}{2\sqrt{217}}\begin{pmatrix}-6\\24\\16\end{pmatrix}$ | A1 | A correct normalised eigenvector in any form. Note direction may be reversed. May use $\mathbf{i}, \mathbf{j}, \mathbf{k}$ notation |

## Part (b):

| Answer/Working | Mark | Guidance |
|---|---|---|
| Multiplies position and direction by $\mathbf{M}$ (not e.g., $\mathbf{M}-\lambda\mathbf{I}$). In parametric form: $\begin{pmatrix}2&0&3\\0&-4&-3\\0&-4&0\end{pmatrix}\begin{pmatrix}4+2\mu\\-1\\-\mu\end{pmatrix} = \ldots \left\{\begin{pmatrix}8+4\mu-3\mu\\4+3\mu\\4\end{pmatrix} = \begin{pmatrix}8\\4\\4\end{pmatrix} + \mu\begin{pmatrix}1\\3\\0\end{pmatrix}\right\}$ Allow without a parameter: $\begin{pmatrix}2&0&3\\0&-4&-3\\0&-4&0\end{pmatrix}\begin{pmatrix}4\\-1\\0\end{pmatrix} = \ldots\left\{\begin{pmatrix}8\\4\\4\end{pmatrix}\right\}$ and $\begin{pmatrix}2&0&3\\0&-4&-3\\0&-4&0\end{pmatrix}\begin{pmatrix}2\\0\\-1\end{pmatrix} = \ldots\left\{\begin{pmatrix}1\\3\\0\end{pmatrix}\right\}$ or $\begin{pmatrix}2&0&3\\0&-4&-3\\0&-4&0\end{pmatrix}\begin{pmatrix}4&2\\-1&0\\0&-1\end{pmatrix} = \ldots\left\{\begin{pmatrix}8&1\\4&3\\4&0\end{pmatrix}\right\}$ Alternatively: Could find 2 points on $l_1$, transform them both and **subtract** to find direction. Allow slips and condone the matrix product written the wrong way round provided they have attempted to multiply the elements appropriately and they obtain a vector (or 3×2 matrix) with the resulting values correctly placed. Condone if they proceed to confuse which is the position and which is the direction. | M1 | |
| $\mathbf{r}\times\begin{pmatrix}1\\3\\0\end{pmatrix} = \begin{pmatrix}8\\4\\4\end{pmatrix}\times\begin{pmatrix}1\\3\\0\end{pmatrix}$ | dM1 | Forms: $\mathbf{r}\times$ direction $=$ position $\times$ direction. Must not clearly confuse their vectors. Allow if RHS $=$ direction $\times$ position. **Requires previous M mark.** No requirement to calculate vector product but the RHS could be implied by 2 correct components (or the negative version if the product is reversed) |
| $\mathbf{r}\times\begin{pmatrix}1\\3\\0\end{pmatrix} = \begin{pmatrix}-12\\4\\20\end{pmatrix}$ | A1 | Any correct equation in the correct form. Not $\mathbf{b} = \ldots, \mathbf{c} = \ldots$ unless $\mathbf{r}\times\mathbf{b} = \mathbf{c}$ seen. Is w once a correct answer is seen. |

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