$z = 3 - \lambda i$ | B1 | si

$(3 + \lambda i)^2 + (3 - \lambda i)^2 = 2$ expansion attempt: $9 + 6\lambda i + i^2\lambda^2 + 9 - 6\lambda i + i^2\lambda^2 = 2$ | M1 | Attempt to expand

$9 + 6\lambda i - \lambda^2 + 9 - 6\lambda i - \lambda^2 = 2$ | A1 |

$2\lambda^2 = 16$, $\lambda = 2\sqrt{2}$ | A1 | oe (simplified); eg A0 for $\lambda = \sqrt{\frac{16}{2}}$

**Total: [4]**

# Question 2a

$\det A = -10$ | B1 | si

$A^{-1} = \frac{-1}{10}\begin{pmatrix} -7 & 1 \\ -4 & 2 \end{pmatrix}$ | B1 | FT their $\det A$

**[2]**

# Question 2b

$X = A^{-1}B$ | M1 | si FT their $A^{-1}$

$X = \frac{-1}{10}\begin{pmatrix} -4 & 2 \end{pmatrix}\begin{pmatrix} 4 & -20 & 13 \\ 0 & -40 & -10 \end{pmatrix}$ | m1 | Correct method multiplication

$X = \frac{-1}{10}\begin{pmatrix} -10 & -20 & -50 \\ 0 & -40 & -10 \end{pmatrix}$ | A1 | cao

**METHOD 2:**

$\begin{pmatrix} 2 & -1 \\ 4 & -7 \end{pmatrix}\begin{pmatrix} a & b & c \\ d & e & f \end{pmatrix} = \begin{pmatrix} 2 & 0 & 9 \\ 4 & -20 & 13 \end{pmatrix}$ | (M1) | Setting up and beginning to multiply matrices

Leading to: $2a - d = 2$, $2b - e = 0$, $2c - f = 9$, $4a - 7d = 4$, $4b - 7e = -20$, $4c - 7f = 13$ | | 

Solving at least 1 set of simultaneous equations, $X = \begin{pmatrix} 1 & 2 & 5 \\ 0 & 4 & 1 \end{pmatrix}$ | (m1), (A1) | cao

**[3]**

**Total: [5]**

# Question 3a

Another root is $5 + i$ | B1 | Accept 4 or -4

**[1]**

# Question 3b

**METHOD 1:**

$(x - 5 - i)(x - 5 + i)$, $x^2 - 10x + 26$ is the quadratic factor | M1 | Sum = 10, product = 26

$x^4 - 10x^3 + 10x^2 + 160x - 416 = 0$, $(x^2 - 10x + 26)(x^2 + ax - 16) = 0$ | A1 |

$\therefore x^2 - 16 = 0$, Solving, $x = \pm4$ | m1, A1 |

**METHOD 2:**

Let other two roots be $\alpha$ and $\beta$. $\therefore 5 + i + 5 - i + \alpha + \beta = 10$, $\alpha + \beta = 0$ | (M1) | 1 correct equation

$(5 - i)(5 + i)a\beta = -416$, $26a\beta = -416$, $a\beta = -16$ | (A1) | 2nd correct equation

Solving simultaneous equations, $x = \pm4$ | (m1), (A1) | Or $x^2 - 16 = 0$, convincing method

If $5 - i$ not considered, award SC1 for a use of Factor Theorem, SC2 for 1 correct root after FT, SC3 for 2 correct roots after FT | | 

**[5]**

**Total: [6]**

# Question 4a

Translation matrix: $\begin{pmatrix} 1 & 0 & 2 \\ 0 & 1 & -2 \\ 0 & 0 & 1 \end{pmatrix}$ | B1 |

Reflection matrix: $\begin{pmatrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{pmatrix}$ | B1 |

$T = \begin{pmatrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{pmatrix}\begin{pmatrix} 1 & 0 & 2 \\ 0 & 1 & -2 \\ 0 & 0 & 1 \end{pmatrix}$ | M1 | FT their translation and reflection matrix

$T = \begin{pmatrix} 0 & 1 & -2 \\ 1 & 0 & 2 \\ 0 & 0 & 1 \end{pmatrix}$ | A1 | cao

M0A0 For multiplying the wrong way, which gives $T = \begin{pmatrix} 0 & 1 & 2 \\ 1 & 0 & -2 \\ 0 & 0 & 1 \end{pmatrix}$ | | 

**[4]**

# Question 4b

Invariant points given by $\begin{pmatrix} 0 & 1 & -2 \\ 1 & 0 & 2 \\ 0 & 0 & 1 \end{pmatrix}\begin{pmatrix} x \\ y \\ 1 \end{pmatrix} = \begin{pmatrix} x \\ y \\ 1 \end{pmatrix}$ | M1 | FT their $T$ from (a)

Giving, $y - 2 = x$ and $x + 2 = y$. | A1 |

As these are equivalent there is an infinite number of invariant points. | A1 |

**[3]**

**Total: [7]**

# Question 5a

$\vec{AB} = (-2i + 7k) - (3i + 4j - 2k) = -5i - 4j + 9k$ | B1 | si

Therefore, $\vec{r} = 3i + 4j - 2k + \lambda(-5i - 4j + 9k)$, $\vec{r} = (3 - 5\lambda)i + (4 - 4\lambda)j + (-2 + 9\lambda)k$ | M1 | Accept equivalent convincing

**[3]**

# Question 5b

Substituting into plane equation: $2(3 - 5\lambda) + 3(4 - 4\lambda) + 3(-2 + 9\lambda) = 27$, $5\lambda = 15$, $\lambda = 3$ | M1 |

$\lambda = 3$ | A1 |

Therefore, point of intersection: $(-12, -8, 25)$ | A1 | FT their $\lambda$

**[3]**

**Total: [6]**