Detailed Answer

a) We need to find such a vector that satisfies \[ \left(\begin{array}{c} 1 \\ 2 \end{array}\right) + \left(\begin{array}{c} a \\ b \end{array}\right) = \left(\begin{array}{c} 7 \\ -3 \end{array}\right) \] It is clear to see that \( a = 6 \) and \( b = -5 \).

b) We need to find such a vector that satisfies \[ \left(\begin{array}{c} 8 \\ 5 \end{array}\right) + \left(\begin{array}{c} a \\ b \end{array}\right) = \left(\begin{array}{c} 2 \\ 0 \end{array}\right) \] It is clear to see that \( a = -6 \) and \( b = -5 \).

c) We need to find such a vector that satisfies \[ \left(\begin{array}{c} -4 \\ 1 \end{array}\right) + \left(\begin{array}{c} a \\ b \end{array}\right) = \left(\begin{array}{c} 0 \\ 0 \end{array}\right) \] It is clear to see that \( a = 4 \) and \( b = -1 \).

Detailed Answer

\[ \tan{\theta} = 2 \]

\[ 2\theta = 2\arctan{2} \]

So:

\[ \cos{2\theta} = \cos{(2\arctan{2})} = -\frac{3}{5} \]

And:

\[ \sin{2\theta} = \sin{(2\arctan{2})} = \frac{4}{5} \]

So the reflection matrix is:

\[ \mathbf{R} = \left(\begin{array}{cc} -\frac{3}{5} & \frac{4}{5} \\ \frac{4}{5} & \frac{3}{5} \end{array}\right) \]

So the reflected point will be at:

\[ \left(\begin{array}{cc} -\frac{3}{5} & \frac{4}{5} \\ \frac{4}{5} & \frac{3}{5} \end{array}\right)\cdot \left(\begin{array}{c} 5 \\ -2 \end{array}\right) = \left(\begin{array}{c} -\frac{23}{5} \\ \frac{14}{5} \end{array}\right) \]

Detailed Answer

Our matrix \( \textbf{A} \) has a form:

\[ \left(\begin{array}{cc} a & b \\ c & d \end{array}\right) \]

We can use the fact that by linear transformation we mean:

\[ \mathbf{x}'=\mathbf{A}\mathbf{x} \]

So plugging in our values we get:

\[ \left(\begin{array}{cc} -4 \\ 7 \end{array}\right) = \left(\begin{array}{cc} a & b \\ c & d \end{array}\right) \cdot \left(\begin{array}{cc} 3 \\ 1 \end{array}\right) \]

\[ \left(\begin{array}{cc} -4 \\ 7 \end{array}\right) = \left(\begin{array}{cc} 3a + b \\ 3c + d \end{array}\right) \]

We can do the same for the other point transformation:

\[ \left(\begin{array}{cc} 26 \\ 0 \end{array}\right) = \left(\begin{array}{cc} -2a + 4b \\ -2c + 4d \end{array}\right) \]

As such, we get a system of 4 equations, which we can solve either algebraically or with our GDC:

\[ -4 = 3a + b \]

\[ 7 = 3c + d \]

\[ 26 = -2a + 4b \]

\[ 0 = -2c + 4d \]

Solving this system gives:

\[ a = -3 \]

\[ b = 5 \]

\[ c = 2 \]

\[ d = 1 \]

Detailed Answer

a) \( \mathbf{A} = \left(\begin{array}{cc} \frac{5}{4} & 0 \\ 0 & \frac{3}{2} \end{array}\right) \)

b) First, we need to find the determinant of our matrix:

\[ \text{det} \, \mathbf{A} = \left|\begin{array}{cc} \frac{5}{4} & 0 \\ 0 & \frac{3}{2} \end{array}\right| = \frac{15}{8} \]

The area of the original circle is:

\[ \pi r^2 = 4\pi \]

So the area of the new circle is:

\[ \frac{15}{8} \cdot 4\pi = \frac{15}{2}\pi \]

c)

Horizontally, the new radius is:

\[ 2 \cdot \frac{5}{4} = 2.5 \]

And vertically:

\[ 2 \cdot \frac{3}{2} = 3 \]

Detailed Answer

a)

\[ \tan{\theta} = \sqrt{3} \]

\[ \theta = \frac{\pi}{3} \]

\[ 2\theta = \frac{2\pi}{3} \]

Thus:

\[ \cos{2\theta} = -\frac{1}{2} \]

\[ \sin{2\theta} = \frac{\sqrt{3}}{2} \]

So the reflection matrix is:

\[ \mathbf{X} = \left(\begin{array}{cc} -\frac{1}{2} & \frac{\sqrt{3}}{2} \\ \frac{\sqrt{3}}{2} & \frac{1}{2} \end{array}\right) \]

Now for the rotation:

\[ \theta = \frac{\pi}{4} \]

\[ \cos{\frac{\pi}{4}} = \sin{\frac{\pi}{4}} = \frac{\sqrt{2}}{2} \]

So the rotation matrix is:

\[ \mathbf{R} = \left(\begin{array}{cc} \frac{\sqrt{2}}{2} & -\frac{\sqrt{2}}{2} \\ \frac{\sqrt{2}}{2} & \frac{\sqrt{2}}{2} \end{array}\right) \]

So the final matrix is:

\[ \mathbf{XR} = \left(\begin{array}{cc} -\frac{1}{2} & \frac{\sqrt{3}}{2} \\ \frac{\sqrt{3}}{2} & \frac{1}{2} \end{array}\right) \cdot \left(\begin{array}{cc} \frac{\sqrt{2}}{2} & -\frac{\sqrt{2}}{2} \\ \frac{\sqrt{2}}{2} & \frac{\sqrt{2}}{2} \end{array}\right) = \left(\begin{array}{cc} \frac{-\sqrt{2}+\sqrt{6}}{4} & \frac{\sqrt{2}+\sqrt{6}}{4} \\ \frac{\sqrt{2}+\sqrt{6}}{4} & \frac{\sqrt{2}-\sqrt{6}}{4} \end{array}\right) \]

b) Reflection on the y-axis:

\[ \left(\begin{array}{cc} -1 & 0 \\ 0 & 1 \end{array}\right) \]

Reflection on the x-axis:

\[ \left(\begin{array}{cc} 1 & 0 \\ 0 & -1 \end{array}\right) \]

So the final matrix is:

\[ \left(\begin{array}{cc} -1 & 0 \\ 0 & 1 \end{array}\right) \cdot \left(\begin{array}{cc} 1 & 0 \\ 0 & -1 \end{array}\right) = \left(\begin{array}{cc} -1 & 0 \\ 0 & -1 \end{array}\right) \]

Detailed Answer

a) Enlargement by factor 2 has matrix:

\[ \mathbf{E} = \left(\begin{array}{cc} 2 & 0 \\ 0 & 2 \end{array}\right) \]

So the equation is:

\[ \mathbf{x}' = \left(\begin{array}{cc} 2 & 0 \\ 0 & 2 \end{array}\right) \mathbf{x} + \left(\begin{array}{c} -3 \\ 5 \end{array}\right) \]

b) Here we need to be very careful with the order we do things in, due to the brackets involved. The transformation is as follows:

\[ \mathbf{x}' = \left(\begin{array}{cc} 1 & 0 \\ 0 & 3 \end{array}\right) \left[\mathbf{x} + \left(\begin{array}{c} 1 \\ 2 \end{array}\right) \right] \]

Which expands to:

\[ = \left(\begin{array}{cc} 1 & 0 \\ 0 & 3 \end{array}\right) \mathbf{x} + \left(\begin{array}{cc} 1 & 0 \\ 0 & 3 \end{array}\right) \left(\begin{array}{c} 1 \\ 2 \end{array}\right) \]

Thus:

\[ = \left(\begin{array}{cc} 1 & 0 \\ 0 & 3 \end{array}\right) \mathbf{x} + \left(\begin{array}{c} 1 \\ 6 \end{array}\right) \]

Detailed Answer

a) Horizontally stretched by a factor of 5, vertically stretched by a factor of 4.

b) (i) Sides AB and CD are parallel, and have the same magnitude, and so are AD and BC.

b) (ii) \( \text{Area} = |AB|^2 = (-2 + 6)^2 + (5 - 2)^2 = 25 \)

c) (i) The coordinates after transformation are:

\[ A' = (-36, 8), \quad B' = \left(\begin{array}{cc} 6 & 0 \\ 0 & 4 \end{array}\right) \cdot \left(\begin{array}{c} -2 \\ 5 \end{array}\right) = \left(\begin{array}{c} -12 \\ 20 \end{array}\right) \]

Similarly:

\[ C' = (-30, 36), \quad D' = (-54, 24) \]

c) (ii) We need to check whether the sides are parallel to one another. The gradient of \(A'B'\) is:

\[ \frac{20 - 8}{-12 + 36} = \frac{12}{24} = 0.5 \]

The gradient of \(C'D'\) is:

\[ \frac{36 - 24}{-30 + 54} = 0.5 \]

The gradient of \(B'C'\) is:

\[ \frac{36 - 20}{-30 + 12} = -\frac{16}{18} \]

The gradient of \(A'D'\) is:

\[ \frac{24 - 8}{-54 + 36} = -\frac{16}{18} \]

Thus, these two pairs of sides are parallel and not perpendicular to one another (dot product not 0), so we have a parallelogram.

c) (iii) \( \text{Area} = \det(\textbf{Y}) \cdot \text{area of square} = 24 \cdot 25 = 600 \)

Detailed Answer

a) Horizontal stretch with scale factor 2, and vertical stretch with scale factor 3.

a) (i)

\[ A' = \mathbf{X} \cdot A = \left(\begin{array}{cc} -12 \\ 0 \end{array}\right), \quad B' = \mathbf{X} \cdot B = \left(\begin{array}{cc} 0 \\ 12 \end{array}\right), \quad O' = \mathbf{X} \cdot O = \left(\begin{array}{cc} 0 \\ 0 \end{array}\right) \]

b) The area of triangle AOB is calculated as:

\[ A = \frac{1}{2} \cdot 6 \cdot 4 = 12 \]

b) (i)

\[ \det \mathbf{X} = 6 \quad \Rightarrow \quad A_{\text{new}} = 6 \cdot 12 = 72 \quad \text{or} \quad A_{\text{new}} = \frac{1}{2} \cdot 12 \cdot 12 = 72 \]

c) \( \mathbf{R} = \left(\begin{array}{cc} 0 & 1 \\ -1 & 0 \end{array}\right) \)

c) (i) After rotation, the vertices are:

\[ O''(0,0), \quad A''(0,6), \quad B''(4,0) \]

d) (i) \( \mathbf{P} = \left(\begin{array}{cc} 0 & 2 \\ -3 & 0 \end{array}\right) \)

d) (ii) First, it applies translations from matrix \(\mathbf{X}\), and then it rotates the figure clockwise by 90 degrees.