Let $\sqrt{24+10i} = a + bi$ where $a$ and $b$ are real numbers. | M1 | 3.1a |

Then $24 + 10i = (a+bi)^2 = a^2 - b^2 + 2abi$ | A1 | 1.1 |

$\Rightarrow a^2 - b^2 = 24$ and $2ab = 10$ | M1 | 2.2a |

$\Rightarrow b = \frac{5}{a} \Rightarrow a^2 - \frac{25}{a^2} = 24$ | A1 | 1.1 |

$\Rightarrow a^4 - 24a^2 - 25 = 0$ | M1 | 2.2a |

$\Rightarrow (a^2+1)(a^2-25) = 0 \Rightarrow a = \pm 5, b = \pm 1$ | A1 | 1.1 |

$\Rightarrow (5+i)$ and $-(5+i)$ | A1 | 2.2a |

**OR**

Square root of $(r,\theta)$ is $\left(\sqrt{r}, \frac{\theta}{2}\right)$ | M1, A1 | |

where $r = 26$, $\theta = \tan^{-1}\frac{10}{24} = 22.62°$ | M1, A1 | |

$\Rightarrow \pm\left(\sqrt{26}, 11.31\right) = \pm\left(\sqrt{26}\cos 11.31 + \sqrt{26}\sin 11.31 i\right)$ | M1, A1 | |

$= \pm(5+i)$ | A1 | |

| [5] | |

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# Question 2 (i):

$AB = \begin{pmatrix} 1 & a \\ 3 & 0 \end{pmatrix}\begin{pmatrix} 4 & 2 \\ 3 & 3 \end{pmatrix} = \begin{pmatrix} 4+3a & 2+3a \\ 12 & 6 \end{pmatrix}$ | M1 | 2.1 |

$BA = \begin{pmatrix} 4 & 2 \\ 3 & 3 \end{pmatrix}\begin{pmatrix} 1 & a \\ 3 & 0 \end{pmatrix} = \begin{pmatrix} 10 & 4a \\ 12 & 3a \end{pmatrix}$ | A1 | 1.1 |

Equate any element give $a = 2$ | A1 | 1.1 |

| [3] | |

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# Question 2 (ii):

Choose any two matrices that are not commutative | M1 | 2.1 |

Demonstrate $BC \neq CB$ | A1 | 2.1 |

| [2] | |

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# Question 2 (iii):

Det $B = 6$ gives area $= 24$ | B1 | 1.1 |

| [1] | |

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# Question 2 (iv):

$\begin{pmatrix} 4 & 2 \\ 3 & 3 \end{pmatrix}\begin{pmatrix} 1 \\ m \end{pmatrix} = \begin{pmatrix} \lambda \\ \lambda m \end{pmatrix}$ | M1 | 2.1 |

$\Rightarrow 4 + 2m = \lambda, 3 + 3m = \lambda m$ | A1 | 1.1 |

$\Rightarrow 3 + 3m = (4 + 2m)m$ | A1 | 1.1 |

$\Rightarrow 2m^2 + m - 3 = 0 \Rightarrow (m-1)(2m+3) = 0$ | A1 | 1.1 |

$\Rightarrow m = 1$ and $m = -\frac{3}{2}$

i.e. $2y + 3x = 0$ and $y = x$ | A1 | 2.2a |

| [4] | |

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