IB Mathematics AHL 1.14 Definition of a matrix AI HL Paper 2- Exam Style Questions- New Syllabus
▶️ Answer/Explanation
a(i)
Let \( A = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \).
Transpose: \( A^T = \begin{pmatrix} a & c \\ b & d \end{pmatrix} \).
Inverse: \( (A^T)^{-1} = \frac{1}{ad – bc} \begin{pmatrix} d & -c \\ -b & a \end{pmatrix} \), where \( \det(A^T) = ad – bc \neq 0 \).
Inverse of \( A \): \( A^{-1} = \frac{1}{ad – bc} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix} \).
Transpose: \( (A^{-1})^T = \frac{1}{ad – bc} \begin{pmatrix} d & -c \\ -b & a \end{pmatrix} \).
Compare: \( (A^T)^{-1} = (A^{-1})^T \).
Explanation:
Using the formula for the inverse of a \( 2 \times 2 \) matrix and properties of transposes, both sides yield the same matrix.
Result: \( (A^T)^{-1} = (A^{-1})^T \).
a(ii)
Let \( A = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \), \( B = \begin{pmatrix} e & f \\ g & h \end{pmatrix} \).
Product: \( AB = \begin{pmatrix} ae + bg & af + bh \\ ce + dg & cf + dh \end{pmatrix} \).
Transpose: \( (AB)^T = \begin{pmatrix} ae + bg & ce + dg \\ af + bh & cf + dh \end{pmatrix} \).
Transposes: \( B^T = \begin{pmatrix} e & g \\ f & h \end{pmatrix} \), \( A^T = \begin{pmatrix} a & c \\ b & d \end{pmatrix} \).
Product: \( B^T A^T = \begin{pmatrix} ae + bg & ce + dg \\ af + bh & cf + dh \end{pmatrix} \).
Compare: \( (AB)^T = B^T A^T \).
Explanation:
Matrix multiplication and transposition properties show that the transpose of the product equals the product of the transposes in reverse order.
Result: \( (AB)^T = B^T A^T \).
b
Reflexive: For \( I \in S \), \( IA I^T = A \), so \( A R A \).
Symmetric: If \( A R B \), then \( XA X^T = B \).
Rewrite: \( A = X^{-1} B (X^T)^{-1} \).
Use a(i): \( (X^T)^{-1} = (X^{-1})^T \), so \( A = X^{-1} B (X^{-1})^T \).
Let \( Y = X^{-1} \): \( A = Y B Y^T \), so \( B R A \).
Transitive: If \( A R B \) (\( XA X^T = B \)) and \( B R C \) (\( Y B Y^T = C \)), then:
Substitute: \( Y (XA X^T) Y^T = C \).
Rewrite: \( (YX) A (YX)^T = C \), using a(ii).
Since \( YX \in S \), \( A R C \).
Explanation:
The relation is reflexive (using the identity matrix), symmetric (using the inverse and part a(i)), and transitive (using part a(ii) and matrix properties).
Result: \( R \) is an equivalence relation.
▶️ Answer/Explanation
a(i)
Given the transformation \( r’ = P r + q \), for \( r = \begin{pmatrix} 0 \\ 0 \end{pmatrix} \), the image is \( (0, 1) \):
Substitute:
\( \begin{pmatrix} 0 \\ 1 \end{pmatrix} = P \begin{pmatrix} 0 \\ 0 \end{pmatrix} + q \)
Compute:
\( P \begin{pmatrix} 0 \\ 0 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix} \)
Solve:
\( q = \begin{pmatrix} 0 \\ 1 \end{pmatrix} \)
Explanation:
The transformation maps the origin \( (0, 0) \) to \( (0, 1) \). Since the linear part \( P r \) is zero for the origin, the translation vector \( q \) equals the image, indicating a shift upward by 1 unit along the y-axis.
Result: \( q = \begin{pmatrix} 0 \\ 1 \end{pmatrix} \).
a(ii)
For \( r = \begin{pmatrix} 1 \\ 0 \end{pmatrix} \), image is \( \left( \frac{\sqrt{3}}{4}, \frac{3}{4} \right) \):
Substitute:
\( P \begin{pmatrix} 1 \\ 0 \end{pmatrix} + \begin{pmatrix} 0 \\ 1 \end{pmatrix} = \begin{pmatrix} \frac{\sqrt{3}}{4} \\ \frac{3}{4} \end{pmatrix} \)
Solve:
\( \begin{pmatrix} p_{11} \\ p_{21} \end{pmatrix} = \begin{pmatrix} \frac{\sqrt{3}}{4} \\ -\frac{1}{4} \end{pmatrix} \)
For \( r = \begin{pmatrix} 0 \\ 1 \end{pmatrix} \), image is \( \left( \frac{1}{4}, 1 + \frac{\sqrt{3}}{4} \right) \):
Substitute:
\( P \begin{pmatrix} 0 \\ 1 \end{pmatrix} + \begin{pmatrix} 0 \\ 1 \end{pmatrix} = \begin{pmatrix} \frac{1}{4} \\ 1 + \frac{\sqrt{3}}{4} \end{pmatrix} \)
Solve:
\( \begin{pmatrix} p_{12} \\ p_{22} \end{pmatrix} = \begin{pmatrix} \frac{1}{4} \\ \frac{\sqrt{3}}{4} \end{pmatrix} \)
Combine:
\( P = \begin{pmatrix} \frac{\sqrt{3}}{4} & \frac{1}{4} \\ -\frac{1}{4} & \frac{\sqrt{3}}{4} \end{pmatrix} \)
Explanation:
The matrix \( P \) is constructed from the images of the basis vectors \( (1, 0) \) and \( (0, 1) \), adjusted for the translation \( q \).
Result: \( P = \begin{pmatrix} \frac{\sqrt{3}}{4} & \frac{1}{4} \\ -\frac{1}{4} & \frac{\sqrt{3}}{4} \end{pmatrix} \).
b
For an enlargement with scale factor 0.5, centre \( (0, 0) \):
Compute:
\( S = 0.5 \times I \)
Solve:
\( S = \begin{pmatrix} 0.5 & 0 \\ 0 & 0.5 \end{pmatrix} \)
Explanation:
The enlargement scales coordinates by 0.5 relative to the origin, represented by the identity matrix scaled by 0.5.
Result: \( S = \begin{pmatrix} 0.5 & 0 \\ 0 & 0.5 \end{pmatrix} \).
c(i)
Given \( P = RS \), solve for \( R \):
Compute:
\( S^{-1} = \begin{pmatrix} 2 & 0 \\ 0 & 2 \end{pmatrix} \)
Multiply:
\( R = \begin{pmatrix} \frac{\sqrt{3}}{4} & \frac{1}{4} \\ -\frac{1}{4} & \frac{\sqrt{3}}{4} \end{pmatrix} \times \begin{pmatrix} 2 & 0 \\ 0 & 2 \end{pmatrix} \)
Solve:
\( R = \begin{pmatrix} \frac{\sqrt{3}}{2} & \frac{1}{2} \\ -\frac{1}{2} & \frac{\sqrt{3}}{2} \end{pmatrix} \)
Explanation:
The rotation matrix \( R \) is obtained by multiplying \( P \) by the inverse of the enlargement matrix \( S \).
Result: \( R = \begin{pmatrix} \frac{\sqrt{3}}{2} & \frac{1}{2} \\ -\frac{1}{2} & \frac{\sqrt{3}}{2} \end{pmatrix} \).
c(ii)
Compare \( R \) to the standard rotation matrix \( \begin{pmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{pmatrix} \):
Identify:
\( \cos \theta = \frac{\sqrt{3}}{2} \implies \theta = 30^\circ \)
Note:
Negative \( \sin \theta = -\frac{1}{2} \) indicates clockwise rotation.
Explanation:
The matrix elements correspond to a 30° rotation, with the negative sign of \( \sin \theta \) indicating clockwise direction.
Result: 30° clockwise.
d(i)
For the fixed point \( (a, b) \):
Set:
\( \begin{pmatrix} a \\ b \end{pmatrix} = P \begin{pmatrix} a \\ b \end{pmatrix} + q \)
Explanation:
The centre \( (a, b) \) is invariant under \( T \), defining a fixed point equation.
Result: \( \begin{pmatrix} a \\ b \end{pmatrix} = P \begin{pmatrix} a \\ b \end{pmatrix} + q \).
d(ii)
From the fixed point equation:
Solve:
\( a = \frac{\sqrt{3}}{4} a + \frac{1}{4} b \)
\( b = -\frac{1}{4} a + \frac{\sqrt{3}}{4} b + 1 \)
From first:
\( b = a (4 – \sqrt{3}) \)
Substitute into second:
\( a (4 – \sqrt{3}) = -\frac{1}{4} a + \frac{\sqrt{3}}{4} a (4 – \sqrt{3}) + 1 \)
Simplify:
\( a (14 – 8\sqrt{3}) = 4 \)
Solve:
\( a = \frac{5 + 2\sqrt{3}}{13} \)
Then:
\( b = \frac{5 + 2\sqrt{3}}{13} (4 – \sqrt{3}) = \frac{14 + 3\sqrt{3}}{13} \)
Explanation:
The system of equations from the fixed point condition yields the coordinates of the centre.
Result: \( a = \frac{5 + 2\sqrt{3}}{13} \), \( b = \frac{14 + 3\sqrt{3}}{13} \).