**Updated on April 20th, 2012: Added some video explanations**

In **part (a)**, \(\overset{\longrightarrow}{OA} = \begin{pmatrix}-1 \\ 3\end{pmatrix}\), \(\overset{\longrightarrow}{OB} = \begin{pmatrix}5 \\ 1\end{pmatrix}\) and \(\overset{\longrightarrow}{BA} = \overset{\longrightarrow}{BO} + \overset{\longrightarrow}{OA} = \begin{pmatrix}-5 \\ -1\end{pmatrix} + \begin{pmatrix}-1 \\ 3\end{pmatrix} = \begin{pmatrix}-6 \\ 2\end{pmatrix}\).

Given that \(G\) is the mid-point of the line \(AB\), \(\overset{\longrightarrow}{BG} = \frac{1}{2} \overset{\longrightarrow}{BA} = \begin{pmatrix}-3 \\ 1\end{pmatrix}\) and \(\overset{\longrightarrow}{OG} = \overset{\longrightarrow}{OB} + \overset{\longrightarrow}{BG} = \begin{pmatrix}5 \\ 1\end{pmatrix} + \begin{pmatrix}-3 \\ 1\end{pmatrix} = \begin{pmatrix}2 \\ 2\end{pmatrix}\).

In **part (b)**, \(L = \begin{pmatrix}3 & 2 \\ 1 & 4\end{pmatrix}\) and \(M = \begin{pmatrix}-1 & 3 \\ 0 & 2\end{pmatrix}\).

$$

\begin{eqnarray}

L + 2M &=& \begin{pmatrix}3 & 2 \\ 1 & 4\end{pmatrix} + 2 \begin{pmatrix}-1 & 3 \\ 0 & 2\end{pmatrix} \\

&=& \begin{pmatrix}3-2 & 2+6 \\ 1+0 & 4+4\end{pmatrix} \\

&=& \begin{pmatrix}1 & 8 \\ 1 & 8\end{pmatrix}.

\end{eqnarray}

$$

$$

\begin{eqnarray}

LM &=& \begin{pmatrix}3 & 2 \\ 1 & 4\end{pmatrix} \begin{pmatrix}-1 & 3 \\ 0 & 2\end{pmatrix} \\

&=& \begin{pmatrix}3 \times -1 + 2 \times 0 & 3 \times 3 + 2 \times 2 \\ 1 \times -1 + 4 \times 0 & 1 \times 3 + 4 \times 2\end{pmatrix} \\

&=& \begin{pmatrix}-3 & 13 \\ -1 & 11\end{pmatrix}.

\end{eqnarray}

$$

Finally, in **part (c)** we are given a matrix \(Q\) such that \(Q = \begin{pmatrix}4 & 2 \\ 1 & 1\end{pmatrix}\).

$$

\begin{eqnarray}

Q^{-1} &=& \frac{1}{|Q|} \begin{pmatrix}1 & -2 \\ -1 & 4\end{pmatrix} \\

&=& \frac{1}{4 – 2} \begin{pmatrix}1 & -2 \\ -1 & 4\end{pmatrix} \\

&=& \frac{1}{2} \begin{pmatrix}1 & -2 \\ -1 & 4\end{pmatrix}.

\end{eqnarray}

$$

Using a matrix method, we need to find the values of \(x\) and \(y\) in the equation \(\begin{pmatrix}4 & 2 \\ 1 & 1\end{pmatrix} \begin{pmatrix}x \\ y\end{pmatrix} = \begin{pmatrix}8 \\ 3\end{pmatrix}\).

$$

\begin{eqnarray}

\begin{pmatrix}4 & 2 \\ 1 & 1\end{pmatrix} \begin{pmatrix}x \\ y\end{pmatrix} &=& \begin{pmatrix}8 \\ 3\end{pmatrix} \\

QX &=& B \\

X &=& Q^{-1}B \\

&=& \frac{1}{2} \begin{pmatrix}1 & -2 \\ -1 & 4\end{pmatrix} \begin{pmatrix}8 \\ 3\end{pmatrix} \\

&=& \frac{1}{2} \begin{pmatrix}2 \\ 4\end{pmatrix} \\

&=& \begin{pmatrix}1 \\ 2\end{pmatrix}.

\end{eqnarray}

$$

Therefore, \(x = 1\) and \(y = 2\).