14 exam-style questions. Tap Show answer for full working.
Given \( \mathbf{A} = \begin{pmatrix} 2 & 1 \\ 0 & 3 \end{pmatrix} \) and \( \mathbf{B} = \begin{pmatrix} 1 & -1 \\ 4 & 2 \end{pmatrix} \), find \( \mathbf{AB} \).
\( \mathbf{AB} = \begin{pmatrix} 2\cdot1 + 1\cdot4 & 2\cdot(-1) + 1\cdot2 \\ 0\cdot1 + 3\cdot4 & 0\cdot(-1) + 3\cdot2 \end{pmatrix} = \begin{pmatrix} 6 & 0 \\ 12 & 6 \end{pmatrix} \)
Find the determinant and inverse of \( \mathbf{M} = \begin{pmatrix} 4 & 1 \\ 2 & 3 \end{pmatrix} \).
\( \det\mathbf{M} = 4\cdot3 - 1\cdot2 = 12 - 2 = 10 \)
\( \mathbf{M}^{-1} = \dfrac{1}{10}\begin{pmatrix} 3 & -1 \\ -2 & 4 \end{pmatrix} \)
The matrix \( \mathbf{T} = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} \) represents a transformation.
Describe the transformation fully.
Rotation of \( 90^\circ \) anticlockwise about the origin.
(Check: \( (1,0) \mapsto (0,1) \), \( (0,1) \mapsto (-1,0) \).)
Find the invariant points of the transformation given by \( \mathbf{A} = \begin{pmatrix} 2 & 1 \\ 0 & 1 \end{pmatrix} \).
Invariant points satisfy \( \mathbf{A}\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} x \\ y \end{pmatrix} \)
\( \begin{pmatrix} 2x + y \\ y \end{pmatrix} = \begin{pmatrix} x \\ y \end{pmatrix} \Rightarrow 2x + y = x \) and \( y = y \)
\( x + y = 0 \Rightarrow y = -x \)
All points on the line \( y = -x \) are invariant.
Given \( \mathbf{P} = \begin{pmatrix} 1 & 2 & 0 \\ 0 & 1 & 3 \\ 0 & 0 & 1 \end{pmatrix} \), find \( \mathbf{P}^{-1} \).
This is upper triangular. Using row reduction or formula for this form:
\( \mathbf{P}^{-1} = \begin{pmatrix} 1 & -2 & 6 \\ 0 & 1 & -3 \\ 0 & 0 & 1 \end{pmatrix} \)
(Check: \( \mathbf{P}\mathbf{P}^{-1} = \mathbf{I} \).)
The matrix \( \mathbf{R} = \begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix} \) represents a rotation.
(a) State the angle and sense of the rotation.
(b) Find \( \mathbf{R}^2 \) and interpret the result.
(a) Rotation by angle \( \theta \) anticlockwise about the origin.
(b) \( \mathbf{R}^2 = \begin{pmatrix} \cos 2\theta & -\sin 2\theta \\ \sin 2\theta & \cos 2\theta \end{pmatrix} \) — rotation by \( 2\theta \) anticlockwise.
Find the eigenvalues of \( \mathbf{A} = \begin{pmatrix} 3 & 1 \\ 0 & 2 \end{pmatrix} \).
\( \det(\mathbf{A} - \lambda\mathbf{I}) = 0 \)
\( \begin{vmatrix} 3-\lambda & 1 \\ 0 & 2-\lambda \end{vmatrix} = (3-\lambda)(2-\lambda) = 0 \)
\( \lambda = 3 \) or \( \lambda = 2 \)
The transformation \( T \) is represented by \( \mathbf{A} = \begin{pmatrix} 2 & 1 \\ 1 & 2 \end{pmatrix} \).
(a) Find the invariant lines through the origin.
(b) Describe the effect of \( T \) on each invariant line.
(a) Invariant lines: \( \mathbf{A}\begin{pmatrix} x \\ y \end{pmatrix} = \lambda\begin{pmatrix} x \\ y \end{pmatrix} \) for some scalar \( \lambda \).
Eigenvalues: \( \det\begin{pmatrix} 2-\lambda & 1 \\ 1 & 2-\lambda \end{pmatrix} = (2-\lambda)^2 - 1 = 0 \Rightarrow 2-\lambda = \pm 1 \)
\( \lambda = 1 \) or \( \lambda = 3 \)
For \( \lambda = 1 \): \( (2-1)x + y = 0 \Rightarrow x + y = 0 \Rightarrow y = -x \)
For \( \lambda = 3 \): \( (2-3)x + y = 0 \Rightarrow -x + y = 0 \Rightarrow y = x \)
Invariant lines: \( y = x \) and \( y = -x \).
(b) On \( y = x \), points are scaled by factor 3 from the origin. On \( y = -x \), scaled by factor 1 (fixed).
Solve the simultaneous equations using matrices:
\( 2x + y - z = 1 \)
\( x - y + 2z = 3 \)
\( 3x + 2y + z = 4 \)
\( \begin{pmatrix} 2 & 1 & -1 \\ 1 & -1 & 2 \\ 3 & 2 & 1 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 1 \\ 3 \\ 4 \end{pmatrix} \)
Using inverse or elimination: one solution is \( x = 1 \), \( y = 0 \), \( z = 1 \).
Check: \( 2+0-1=1 \), \( 1-0+2=3 \), \( 3+0+1=4 \). ✓
Show that the matrix \( \mathbf{A} = \begin{pmatrix} 1 & 2 \\ 3 & 6 \end{pmatrix} \) is singular, and find the equation of the image of the plane under the transformation represented by \( \mathbf{A} \) (interpreting as a mapping of \( \mathbb{R}^2 \)).
\( \det\mathbf{A} = 1\cdot6 - 2\cdot3 = 0 \), so singular (not invertible).
Image of \( \begin{pmatrix} x \\ y \end{pmatrix} \) is \( \begin{pmatrix} x+2y \\ 3x+6y \end{pmatrix} = (x+2y)\begin{pmatrix} 1 \\ 3 \end{pmatrix} \).
All image points lie on the line \( Y = 3X \) (through the origin).
Given \( \mathbf{A} = \begin{pmatrix} 3 & -1 \\ 2 & 4 \end{pmatrix} \), find \( \mathbf{A}^2 \) and \( \operatorname{tr}(\mathbf{A}) \).
\( \mathbf{A}^2 = \begin{pmatrix} 3\cdot3 + (-1)\cdot2 & 3\cdot(-1) + (-1)\cdot4 \\ 2\cdot3 + 4\cdot2 & 2\cdot(-1) + 4\cdot4 \end{pmatrix} = \begin{pmatrix} 7 & -7 \\ 14 & 14 \end{pmatrix} \)
\( \operatorname{tr}(\mathbf{A}) = 3 + 4 = 7 \)
The matrix \( \mathbf{M} = \begin{pmatrix} 1 & 2 & 0 \\ 0 & 1 & 3 \\ 0 & 0 & 1 \end{pmatrix} \) represents a linear transformation.
(a) Find \( \mathbf{M}^{-1} \).
(b) Describe the geometric effect of the transformation represented by \( \mathbf{M} \).
(a) This is an upper-triangular matrix with 1s on the diagonal; inverse is
\( \mathbf{M}^{-1} = \begin{pmatrix} 1 & -2 & 6 \\ 0 & 1 & -3 \\ 0 & 0 & 1 \end{pmatrix} \)
(Verify by multiplication or row reduction.)
(b) Shear / translation-like in the plane: it leaves the \( z \)-direction (or third basis vector) fixed and applies a shear that mixes the first two coordinates. Explicitly, \( (x,y,z) \mapsto (x+2y, y+3z, z) \).
Find the eigenvalues and corresponding eigenvectors of \( \mathbf{A} = \begin{pmatrix} 5 & 4 \\ 1 & 2 \end{pmatrix} \).
Characteristic equation: \( \det(\mathbf{A}-\lambda I) = (5-\lambda)(2-\lambda) - 4 = \lambda^2 - 7\lambda + 6 = 0 \)
\( (\lambda-6)(\lambda-1) = 0 \), so \( \lambda = 6 \) or \( \lambda = 1 \).
For \( \lambda=6 \): \( (A-6I)\mathbf{v} = \begin{pmatrix} -1 & 4 \\ 1 & -4 \end{pmatrix}\mathbf{v} = 0 \) ⇒ \( v_1 = 4v_2 \). Eigenvector \( \begin{pmatrix} 4 \\ 1 \end{pmatrix} \).
For \( \lambda=1 \): \( (A-I)\mathbf{v} = \begin{pmatrix} 4 & 4 \\ 1 & 1 \end{pmatrix}\mathbf{v} = 0 \) ⇒ \( v_1 = -v_2 \). Eigenvector \( \begin{pmatrix} 1 \\ -1 \end{pmatrix} \).
The matrix \( \mathbf{A} = \begin{pmatrix} 2 & 1 & 0 \\ 0 & 2 & 1 \\ 0 & 0 & 2 \end{pmatrix} \) has a single eigenvalue.
(a) State the eigenvalue and find a basis for the corresponding eigenspace.
(b) Show that \( \mathbf{A} \) is not diagonalisable.
(c) Find the Jordan form of \( \mathbf{A} \) (or describe its structure).
(a) Eigenvalue \( \lambda = 2 \) (only, algebraic multiplicity 3). \( (A-2I) = \begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{pmatrix} \). Rank 2, nullity 1. Eigenspace spanned by \( \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} \).
(b) Geometric multiplicity 1 < algebraic multiplicity 3, so not diagonalisable.
(c) Jordan form consists of a single 3×3 block: \( \begin{pmatrix} 2 & 1 & 0 \\ 0 & 2 & 1 \\ 0 & 0 & 2 \end{pmatrix} \) (already in Jordan form, or similar).