# Question 7:

## Part (a)

| Working | Mark | Guidance |
|---------|------|----------|
| $|\mathbf{M}|=2(-k-8)+1(-3-12)+1(6-3k)=0 \Rightarrow k=\ldots$ | M1 | 1.1b - Sets determinant to zero and solves |
| $k \neq -5$ | A1 | 2.4 - Correct value with rejection of $k=-5$ implied |

---

## Part (b) Way 1:

| Working | Mark | Guidance |
|---------|------|----------|
| $\mathbf{M}=\begin{pmatrix}2&-1&1\\3&-6&4\\3&2&-1\end{pmatrix}\Rightarrow\begin{pmatrix}x\\y\\z\end{pmatrix}=\mathbf{M}^{-1}\begin{pmatrix}p\\1\\0\end{pmatrix}$ | M1 | 3.1a - Sets up matrix equation |
| $\mathbf{M}^{-1}=\frac{1}{5}\begin{pmatrix}-2&1&2\\15&-5&-5\\24&-7&-9\end{pmatrix}$ | B1 | 1.1b - Correct inverse |
| $\begin{pmatrix}x\\y\\z\end{pmatrix}=\frac{1}{5}\begin{pmatrix}-2&1&2\\15&-5&-5\\24&-7&-9\end{pmatrix}\begin{pmatrix}p\\1\\0\end{pmatrix}\Rightarrow\begin{pmatrix}x\\y\\z\end{pmatrix}=\ldots$ | M1 | 2.1 - Multiplies correctly |
| $\begin{pmatrix}x\\y\\z\end{pmatrix}=\frac{1}{5}\begin{pmatrix}-2p+1\\15p-5\\24p-7\end{pmatrix}$ | A1 | 1.1b |
| $\left(\frac{-2p+1}{5},\ 3p-1,\ \frac{24p-7}{5}\right)$ | A1ft | 2.5 - Correct coordinates as functions of $p$ |

---

## Part (b) Way 2:

| Working | Mark | Guidance |
|---------|------|----------|
| $2x-y+z=p$, $3x-6y+4z=1$, $3x+2y-z=0 \Rightarrow$ e.g. $8y-5z=-1$, $9y-5z=3p-2 \Rightarrow y=\ldots$ | M1 | 3.1a - Eliminates to reach solvable equations |
| $y=3p-1$ (or $x=\frac{-2p+1}{5}$ or $z=\frac{24p-7}{5}$) | B1 | 1.1b |
| $8(3p-1)-5z=-1\Rightarrow z=\ldots\Rightarrow x=\ldots$ | M1 | 2.1 |
| $z=\frac{24p-7}{5}$, $x=\frac{-2p+1}{5}$ | A1 | 1.1b |
| $\left(\frac{-2p+1}{5},\ 3p-1,\ \frac{24p-7}{5}\right)$ | A1ft | 2.5 |

## Question 7(c)(i):

| Answer/Working | Mark | Guidance |
|---|---|---|
| For consistency: E.g. $5x + y = 4 - q$ and $15x + 3y = q$ | M1 | 3.1a — Uses a correct strategy that will lead to establishing a value for $q$, e.g. eliminating one of $x$, $y$ or $z$ |
| $4 - q = \dfrac{q}{3} \Rightarrow q = \ldots$ | M1 | 2.1 — Solves a suitable equation to obtain a value for $q$ |
| $q = 3$ | A1 | 1.1b — Correct value |

**Alternative for (c)(i):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $x=1 \Rightarrow 2-y+z=1,\ 3+2y-z=0 \Rightarrow y=\ldots, z=\ldots$ | M1 | Allocating a number to one variable and solving for the other 2: $x=1, y=-4, z=-5 \Rightarrow 3+20-20=q$ |
| Substitutes into second equation and solves for $q$; $q=3$ | M1, A1 | |

## Question 7(c)(ii):

| Answer/Working | Mark | Guidance |
|---|---|---|
| Three **planes** that intersect in a **line**, OR three **planes** that form a **sheaf** (allow "sheath") | B1 | 2.4 — Must include the two ideas of **planes** and meeting in a **line** or forming a **sheaf** with no contradictory statements |

---