Given vectors a = [1, -3], b = [2, 3], and c = [8, 3], find m and n such that c = ma + nb. Then find |d| if d = c − 2a.

$$\begin{pmatrix} 8 \\ 3 \end{pmatrix} = m \begin{pmatrix} 1 \\ -3 \end{pmatrix} + n \begin{pmatrix} 2 \\ 3 \end{pmatrix}$$ $$\begin{pmatrix} 8 \\ 3 \end{pmatrix} = \begin{pmatrix} m \\ -3m \end{pmatrix} + \begin{pmatrix} 2n \\ 3n \end{pmatrix}$$ $$\begin{pmatrix} 8 \\ 3 \end{pmatrix} = \begin{pmatrix} m + 2n \\ -3m + 3n \end{pmatrix}$$

$$m + 2n = 8 \quad (1)$$ $$-3m + 3n = 3 \quad (2)$$

From equation (2), divide by 3:

$$-m + n = 1 \implies n = m + 1$$

Substitute $n = m + 1$ into equation (1):

$$m + 2(m + 1) = 8$$ $$m + 2m + 2 = 8$$ $$3m = 6$$ $$m = 2$$

Substitute $m = 2$ back into $n = m + 1$:

$$n = 2 + 1$$ $$n = 3$$

The values are $m=2$ and $n=3$.

$$\boxed{m=2, n=3}$$

12. (a) (ii) Find $|\mathbf{d}|$, if $\mathbf{d} = \mathbf{c} - 2\mathbf{a}$

$$\mathbf{d} = \begin{pmatrix} 8 \\ 3 \end{pmatrix} - 2 \begin{pmatrix} 1 \\ -3 \end{pmatrix}$$ $$\mathbf{d} = \begin{pmatrix} 8 \\ 3 \end{pmatrix} - \begin{pmatrix} 2 \\ -6 \end{pmatrix}$$ $$\mathbf{d} = \begin{pmatrix} 8 - 2 \\ 3 - (-6) \end{pmatrix}$$ $$\mathbf{d} = \begin{pmatrix} 6 \\ 9 \end{pmatrix}$$

$$|\mathbf{d}| = \sqrt{6^2 + 9^2}$$ $$|\mathbf{d}| = \sqrt{36 + 81}$$ $$|\mathbf{d}| = \sqrt{117}$$

$$|\mathbf{d}| = \sqrt{9 \times 13}$$ $$|\mathbf{d}| = 3\sqrt{13}$$

The magnitude of $\mathbf{d}$ is $3\sqrt{13}$.

$$\boxed{|\mathbf{d}| = 3\sqrt{13}}$$

12. (b) If $\begin{pmatrix} x+3 \\ 2 \end{pmatrix} - \begin{pmatrix} y \\ x+y \end{pmatrix} = \begin{pmatrix} 2 \\ 1 \end{pmatrix}$, find the values for $x$ and $y$.

$$\begin{pmatrix} (x+3) - y \\ 2 - (x+y) \end{pmatrix} = \begin{pmatrix} 2 \\ 1 \end{pmatrix}$$ $$\begin{pmatrix} x+3-y \\ 2-x-y \end{pmatrix} = \begin{pmatrix} 2 \\ 1 \end{pmatrix}$$

$$x + 3 - y = 2 \implies x - y = -1 \quad (1)$$ $$2 - x - y = 1 \implies -x - y = -1 \quad (2)$$

Add equation (1) and equation (2):

$$(x - y) + (-x - y) = -1 + (-1)$$ $$-2y = -2$$ $$y = 1$$

Substitute $y = 1$ into equation (1):

$$x - 1 = -1$$ $$x = 0$$

The values are $x=0$ and $y=1$.

$$\boxed{x=0, y=1}$$