## Question 6(i)(a):

| Answer/Working | Mark | Guidance |
|---|---|---|
| Multiplies **A** by itself and sets equal to **I**: $\begin{pmatrix}2 & a\\a-4 & b\end{pmatrix}\begin{pmatrix}2 & a\\a-4 & b\end{pmatrix}=\begin{pmatrix}1 & 0\\0 & 1\end{pmatrix} \Rightarrow 4+a(a-4)=1$ and either $2a+ab=0$ or $2(a-4)+b(a-4)=0$ or $a(a-4)+b^2=1$ | M1 | 3.1a — Forming two equations, one involving $a$ only and one involving $a$ and $b$ |
| $a^2-4a+3=0 \Rightarrow (a-3)(a-1)=0 \Rightarrow a=\ldots$ | dM1 | 1.1b — Dependent on previous mark, solves 3TQ involving $a$ only |
| $a=1,\ a=3$ | A1 | 1.1b — Correct values for $a$ |
| Substitutes value of $a$ into equation involving both $a$ and $b$, solves for $b$. e.g. $2(1)+(1)b \Rightarrow b=\ldots$; $2(1-4)b+(1-4)=0 \Rightarrow b=\ldots$; $(1)(1-4)+b^2=1 \Rightarrow b=\ldots$ Alternatively: $2a+ab=0$, $a(2+b)=0$, as $a\neq 0$, $2+b=0 \Rightarrow b=\ldots$ | dM1 | 1.1b — Dependent on first M1; substitutes one value of $a$ to find $b$ |
| $b=-2$ | A1 | 1.1b — Correct value for $b$ |

**Alternative (i)(a):**

| Answer/Working | Mark | Guidance |
|---|---|---|
| Finds $\mathbf{A}^{-1}$ in terms of $a$ and $b$, sets equal to **A**: $\frac{1}{2b-a(a-4)}\begin{pmatrix}b & -a\\-(a-4) & 2\end{pmatrix}=\begin{pmatrix}2 & a\\a-4 & b\end{pmatrix}$. One equation from $\frac{b}{2b-a(a-4)}=2$, $\frac{2}{2b-a(a-4)}=b$. One equation from $\frac{-a}{2b-a(a-4)}=a$, $\frac{-(a-4)}{2b-a(a-4)}=a-4$ | M1 | 3.1a — Finds $\mathbf{A}^{-1}$ and sets equal to **A**, forms two different equations |
| $a^2-4a+3=0 \Rightarrow (a-3)(a-1)=0 \Rightarrow a=\ldots$ | dM1 | 1.1b — Eliminates $b$ and solves 3TQ involving only $a$ |
| $a=1,\ a=3$ | A1 | 1.1b |
| $\frac{-a}{2b-a(a-4)}=a \Rightarrow 2b-a(a-4)=-1 \Rightarrow \frac{b}{-1}=2$ or $\frac{-(a-4)}{2b-a(a-4)}=a-4 \Rightarrow 2b-a(a-4)=-1 \Rightarrow \frac{2}{-1}=b$ | dM1 | 1.1b — Substitutes value of $a$ to find $b$ |
| $b=-2$ | A1 | 1.1b |

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## Question 6(i)(b):

| Answer/Working | Mark | Guidance |
|---|---|---|
| Uses smallest value of $a$ and value of $b$ to form two equations: $\begin{pmatrix}2 & 1\\-3 & -2\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}=\begin{pmatrix}x\\y\end{pmatrix} \Rightarrow 2x+y=x$ and $-3x-2y=y$ | M1 | 3.1a — Extracts simultaneous equations using matrix **A** with smallest value of $a$ |
| $2x+y=x \Rightarrow x+y=0$ o.e. and $-3x-2y=y \Rightarrow x+y=0$ o.e. | M1 | 1.1b — Gathers terms from both equations |
| $x+y=0$ o.e. | A1 | 2.1 — Achieves correct equation and deduces correct line; accept equivalent equations as long as both shown to be the same |

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## Question 6(ii)(a):

| Answer/Working | Mark | Guidance |
|---|---|---|
| Area of triangle $T=3$ | B1 | 1.1b |
| Determinant $3p\times p-(-1)\times 2p = \frac{15}{\text{their area}} \Rightarrow p=\ldots$. Resulting 3TQ solved to find $p$ | M1 | 3.1a — Full method: finds determinant, sets equal to $15\div$their area, solves 3TQ |
| $3p^2+2p-5\ (=0)$ | A1 | 1.1b — Correct quadratic |
| $p=1$ only; must reject $p=-\frac{5}{3}$ | A1 | 1.1b — $p=1$ only |

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## Question 6(ii)(b):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\begin{pmatrix}3 & 0\\0 & -2\end{pmatrix}$ | B1 | 1.1b — One correct row or column |
| | B1 | 1.1b — All correct |

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## Question 6(ii)(c):

| Answer/Working | Mark | Guidance |
|---|---|---|
| (their matrix from (b))$\begin{pmatrix}p' & 2p'\\-1 & 3p'\end{pmatrix}=\begin{pmatrix}\ldots & \ldots\\\ldots & \ldots\end{pmatrix}$, i.e. $\begin{pmatrix}3 & 0\\0 & -2\end{pmatrix}\begin{pmatrix}1 & 2\\-1 & 3\end{pmatrix}=\begin{pmatrix}\ldots & \ldots\\\ldots & \ldots\end{pmatrix}$ | M1 | 1.1b — Multiplies matrices **QP** in correct order |
| $\begin{pmatrix}3 & 6\\2 & -6\end{pmatrix}$ | A1ft | 1.1b — Correct matrix; follow through on part (b) and value of $p$, as long as positive constant |

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