# Question 6:

## Part (a):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $|\mathbf{M}| = k(-1-1) - 5(-1-2) + 7(1-2) = \{8-2k\}$ | M1 | Correct method to find the determinant. Condone one sign slip |
| Minors: $\begin{pmatrix} -2 & -3 & -1 \\ -12 & -k-14 & k-10 \\ -2 & k-7 & k-5 \end{pmatrix}$ | M1 | A correct first step in obtaining the inverse. Could be matrix of minors or cofactors. Condone sign slips as long as intention is clear |
| Cofactors: $\begin{pmatrix} -2 & 3 & -1 \\ 12 & -k-14 & 10-k \\ -2 & 7-k & k-5 \end{pmatrix}$ | | |
| $\mathbf{M}^{-1} = \frac{1}{8-2k}\begin{pmatrix} -2 & 12 & -2 \\ 3 & -k-14 & 7-k \\ -1 & 10-k & k-5 \end{pmatrix}$ | M1, A1 | Fully correct method: minors, cofactors, transposes and $\frac{1}{\det}$. Correct matrix |

## Part (b):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\mathbf{M}^{-1} = \frac{1}{4}\begin{pmatrix} -2 & 12 & -2 \\ 3 & -16 & 5 \\ -1 & 8 & -3 \end{pmatrix} \Rightarrow \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \mathbf{M}^{-1}\begin{pmatrix} 1 \\ p \\ 2 \end{pmatrix}$ | M1 | Complete strategy for solving: multiplies coordinates by inverse or solves simultaneously to achieve values for $x$, $y$ and $z$ |
| $\begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} -\frac{1}{2}+3p-1 \\ \frac{3}{4}-4p+\frac{5}{2} \\ -\frac{1}{4}+2p-\frac{3}{2} \end{pmatrix}$ | A1ft | Correct calculation on their inverse matrix (unsimplified) or at least on correct value if solving simultaneously |
| $\left(\frac{12p-6}{4}, \frac{13-16p}{4}, \frac{8p-7}{4}\right)$ or $\left(3p-\frac{3}{2}, \frac{13}{4}-4p, 2p-\frac{7}{4}\right)$ | A1 | Correct coordinates |

## Part (c)(i):

| Answer/Working | Mark | Guidance |
|---|---|---|
| E.g. eliminates $z$: $3x+2y = q+2$, $3x+2y = 7q-1$, $18x+12y=15$ | M1 | Uses a correct strategy leading to establishing a value for $q$, e.g. eliminating one of $x$, $y$ or $z$ |
| E.g. $q+2 = 7q-1 \Rightarrow q = \ldots$ or $-3 = 3(1-4q) \Rightarrow q = \ldots$ or $-9 = 6(1-5q) \Rightarrow q = \ldots$ | M1 | Solves a suitable equation to obtain a value for $q$ |
| $q = \frac{1}{2}$ | A1 | Correct value |

**Alternative 1:**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $4a+b=2,\ 5a+b=1,\ 7a+b=-1 \Rightarrow \{a=-1, b=6\}$ | M1 | Equating coefficients leading to two out of three equations, solves to find $a$ and $b$ |
| Forms $a+bq=2$ and substitutes values of $a$ and $b$ | M1 | Solves suitable equation to obtain value of $q$ |
| $q = \frac{1}{2}$ | A1 | Correct value |

**Alternative 2:**
| Answer/Working | Mark | Guidance |
|---|---|---|
| Finds coordinate of intersection of $4x+5y+7z=1$ and $2x+y-z=2$, e.g. let $z=0$: $y=-1, x=1.5$ | M1 | Finds a coordinate of intersection of the two planes |
| Substitutes into $x+y+z=q$ | M1 | Substitutes $x$, $y$, $z$ into $x+y+z=q$ to reach value for $q$ |
| $q = \frac{1}{2}$ | A1 | Correct value |

## Part (c)(ii):

| Answer/Working | Mark | Guidance |
|---|---|---|
| E.g. Let $x=\lambda$: $3\lambda+2y=\frac{5}{2}$, $\lambda-2z=\frac{3}{2} \Rightarrow y=f(\lambda), z=f(\lambda)$ | M1 | Uses a correct strategy to obtain the Cartesian equation of the line or general coordinates |
| $\lambda = \dfrac{y-\frac{5}{4}}{-\frac{3}{2}} = \dfrac{z+\frac{3}{4}}{\frac{1}{2}}$ or $y=\frac{5}{4}-\frac{3}{2}\lambda,\ z=-\frac{3}{4}+\frac{1}{2}\lambda$ | A1 | Correct Cartesian equation or coordinates in terms of parameter |
| Uses Cartesian equation to extract position and direction to form vector equation | M1 | Correctly extracts position and direction vector |
| $\mathbf{r} = \frac{5}{4}\mathbf{j} - \frac{3}{4}\mathbf{k} + t(2\mathbf{i}-3\mathbf{j}+\mathbf{k})$ o.e. | A1 | Correct equation (o.e.), look out for multiples of direction vector. Must have $\mathbf{r}=\ldots$ |

**Alternative (ii):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| Finds two different coordinates on the line, e.g. $x=0$: $\left(0, \frac{5}{4}, -\frac{3}{4}\right)$; $y=0$: $\left(\frac{5}{6}, 0, -\frac{1}{3}\right)$; $z=0$: $\left(\frac{3}{2}, -1, 0\right)$ | M1, A1 | Finds two different coordinates that lie on the line. Correct coordinates |
| Finds vector equation of line passing through the two points | M1 | Uses their coordinates to find vector equation |
| $\mathbf{r} = \frac{5}{4}\mathbf{j} - \frac{3}{4}\mathbf{k} + t(2\mathbf{i}-3\mathbf{j}+\mathbf{k})$ o.e. | A1 | Correct equation (o.e.), look out for multiples of direction vector. Must have $\mathbf{r}=\ldots$ |

**Alternative (ii) outside spec:**
| Answer/Working | Mark | Guidance |
|---|---|---|
| Finds cross product of two normal vectors and a coordinate lying on all three planes | M1 | |
| $\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 4 & 5 & 7 \\ 1 & 1 & 1 \end{vmatrix} = \begin{pmatrix} 2 \\ 3 \\ -1 \end{pmatrix}$ or multiple thereof | A1 | Correct cross product |
| Uses point and direction vector to find equation of line | M1 | |
| $\mathbf{r} = \frac{5}{4}\mathbf{j} - \frac{3}{4}\mathbf{k} + t(2\mathbf{i}-3\mathbf{j}+\mathbf{k})$ o.e. | A1 | Correct equation (o.e.) |