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Matrices provide a compact algebraic framework for representing and manipulating systems of linear Equations, geometric transformations, and, at a more advanced level, quantum mechanical states and Data structures. This topic develops the algebra of matrices and their interpretation as linear Transformations of the plane and space.

BoardPaperNotes
AQAPaper 13D transformations, eigenvalues and eigenvectors
EdexcelFP12D transformations, 3×33\times3 matrices, determinants
OCR (A)Paper 12D transformations, 3×33\times3 matrices
CIEP12D and basic 3D transformations, inverses, determinants

Definition. An m×nm \times n matrix is a rectangular array of numbers arranged in mm rows and nn columns. We write A=(aij)\mathbf{A} = (a_{ij}) where aija_{ij} is the entry in row ii Column jj. The set of all m×nm \times n matrices with real entries is denoted Mm×n(R)M_{m\times n}(\mathbb{R}).

A matrix with a single row is a row vector, and a matrix with a single column is a column Vector.

1.1 Matrix Addition and Scalar Multiplication

Section titled “1.1 Matrix Addition and Scalar Multiplication”

For matrices A,B∈Mm×n(R)\mathbf{A}, \mathbf{B} \in M_{m \times n}(\mathbb{R}) and scalar k∈Rk \in \mathbb{R}:

(A+B)ij=aij+bij(kA)ij=k⋅aij\begin{aligned} (\mathbf{A} + \mathbf{B})_{ij} &= a_{ij} + b_{ij} \\ (k\mathbf{A})_{ij} &= k \cdot a_{ij} \end{aligned}

Matrix addition is commutative (A+B=B+A\mathbf{A} + \mathbf{B} = \mathbf{B} + \mathbf{A}) and Associative.

Definition. If A∈Mm×p(R)\mathbf{A} \in M_{m \times p}(\mathbb{R}) and B∈Mp×n(R)\mathbf{B} \in M_{p \times n}(\mathbb{R}) The product AB∈Mm×n(R)\mathbf{AB} \in M_{m \times n}(\mathbb{R}) is defined by:

(AB)ij=∑k=1paik bkj\boxed{(\mathbf{AB})_{ij} = \sum_{k=1}^{p} a_{ik}\, b_{kj}}

Matrix multiplication is associative but not commutative : AB≠BA\mathbf{AB} \neq \mathbf{BA}.

The n×nn \times n identity matrix In\mathbf{I}_n satisfies AIn=InA=A\mathbf{A}\mathbf{I}_n = \mathbf{I}_n\mathbf{A} = \mathbf{A} for any A∈Mm×n(R)\mathbf{A} \in M_{m \times n}(\mathbb{R}).


Definition. For A=(abcd)\mathbf{A} = \begin{pmatrix} a & b \\ c & d \end{pmatrix}:

det⁡(A)=ad−bc\boxed{\det(\mathbf{A}) = ad - bc}

Definition. For A=(a11a12a13a21a22a23a31a32a33)\mathbf{A} = \begin{pmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{pmatrix} the determinant is computed by cofactor expansion along any row or column:

det⁡(A)=a11∣a22a23a32a33∣−a12∣a21a23a31a33∣+a13∣a21a22a31a32∣\boxed{\det(\mathbf{A}) = a_{11}\begin{vmatrix} a_{22} & a_{23} \\ a_{32} & a_{33} \end{vmatrix} - a_{12}\begin{vmatrix} a_{21} & a_{23} \\ a_{31} & a_{33} \end{vmatrix} + a_{13}\begin{vmatrix} a_{21} & a_{22} \\ a_{31} & a_{32} \end{vmatrix}}

The signs alternate ++, −-, ++ along the first row (following the checkerboard pattern).

For n×nn \times n matrices A,B\mathbf{A}, \mathbf{B} and scalar kk:

  1. det⁡(In)=1\det(\mathbf{I}_n) = 1
  2. det⁡(AT)=det⁡(A)\det(\mathbf{A}^T) = \det(\mathbf{A})
  3. det⁡(kA)=kndet⁡(A)\det(k\mathbf{A}) = k^n \det(\mathbf{A})
  4. Swapping two rows (or columns) multiplies the determinant by −1-1
  5. A matrix with a zero row (or column) has det⁡=0\det = 0
  6. det⁡(A)=0\det(\mathbf{A}) = 0 if and only if A\mathbf{A} is singular (non-invertible)

Proof of det⁡(AB)=det⁡(A)det⁡(B)\det(\mathbf{AB}) = \det(\mathbf{A})\det(\mathbf{B}) (for 2×22\times2 matrices)

Section titled “Proof of det⁡(AB)=det⁡(A)det⁡(B)\det(\mathbf{AB}) = \det(\mathbf{A})\det(\mathbf{B})det(AB)=det(A)det(B) (for 2×22\times22×2 matrices)”

Let A=(abcd)\mathbf{A} = \begin{pmatrix} a & b \\ c & d \end{pmatrix} and B=(efgh)\mathbf{B} = \begin{pmatrix} e & f \\ g & h \end{pmatrix}.

AB=(ae+bgaf+bhce+dgcf+dh)\mathbf{AB} = \begin{pmatrix} ae + bg & af + bh \\ ce + dg & cf + dh \end{pmatrix} det⁡(AB)=(ae+bg)(cf+dh)−(af+bh)(ce+dg)=acef+adeh+bcfg+bdgh−acef−adfg−bceh−bdgh=adeh+bcfg−adfg−bceh=ad(eh−fg)−bc(eh−fg)=(ad−bc)(eh−fg)=det⁡(A)⋅det⁡(B)□\begin{aligned} \det(\mathbf{AB}) &= (ae + bg)(cf + dh) - (af + bh)(ce + dg) \\ &= acef + adeh + bcfg + bdgh - acef - adfg - bceh - bdgh \\ &= adeh + bcfg - adfg - bceh \\ &= ad(eh - fg) - bc(eh - fg) \\ &= (ad - bc)(eh - fg) \\ &= \det(\mathbf{A}) \cdot \det(\mathbf{B}) \quad \square \end{aligned}

Intuition. The determinant measures how a matrix scales area (in 2D) or volume (in 3D). Composing two transformations multiplies their area/volume scaling factors, which is why Determinants multiply.


Definition. The inverse of a square matrix A\mathbf{A} Written A−1\mathbf{A}^{-1} Is the Unique matrix satisfying:

AA−1=A−1A=I\mathbf{A}\mathbf{A}^{-1} = \mathbf{A}^{-1}\mathbf{A} = \mathbf{I}

An inverse exists if and only if det⁡(A)≠0\det(\mathbf{A}) \neq 0. Such a matrix is called non-singular.

3.1 Inverse of a 2×22 \times 2 Matrix

Section titled “3.1 Inverse of a 2×22 \times 22×2 Matrix”

For A=(abcd)\mathbf{A} = \begin{pmatrix} a & b \\ c & d \end{pmatrix} with det⁡(A)=ad−bc≠0\det(\mathbf{A}) = ad - bc \neq 0:

A−1=1ad−bc(d−b−ca)\boxed{\mathbf{A}^{-1} = \frac{1}{ad - bc}\begin{pmatrix} d & -b \\ -c & a \end{pmatrix}}

3.2 The Adjugate Method (3×33 \times 3)

Section titled “3.2 The Adjugate Method (3×33 \times 33×3)”

For a 3×33 \times 3 matrix A\mathbf{A}:

A−1=1det⁡(A)adj⁡(A)\boxed{\mathbf{A}^{-1} = \frac{1}{\det(\mathbf{A})}\operatorname{adj}(\mathbf{A})}

Where the adjugate (or adjoint) matrix adj⁡(A)\operatorname{adj}(\mathbf{A}) is the transpose of the cofactor matrix.

Definition. The cofactor CijC_{ij} of entry aija_{ij} is (−1)i+j(-1)^{i+j} times the determinant of The submatrix obtained by deleting row ii and column jj. The cofactor matrix has entries CijC_{ij} And adj⁡(A)=(Cij)T\operatorname{adj}(\mathbf{A}) = (C_{ij})^T.

Suppose B\mathbf{B} and C\mathbf{C} are both inverses of A\mathbf{A}. Then:

B=BI=B(AC)=(BA)C=IC=C□\mathbf{B} = \mathbf{B}\mathbf{I} = \mathbf{B}(\mathbf{AC}) = (\mathbf{BA})\mathbf{C} = \mathbf{IC} = \mathbf{C} \quad \square
Worked Example: $3\times3$ inverse

Find A−1\mathbf{A}^{-1} where A=(120013101)\mathbf{A} = \begin{pmatrix} 1 & 2 & 0 \\ 0 & 1 & 3 \\ 1 & 0 & 1 \end{pmatrix}.

det⁡(A)=1∣1301∣−2∣0311∣+0=1(1)−2(−3)=7\det(\mathbf{A}) = 1\begin{vmatrix} 1 & 3 \\ 0 & 1 \end{vmatrix} - 2\begin{vmatrix} 0 & 3 \\ 1 & 1 \end{vmatrix} + 0 = 1(1) - 2(-3) = 7.

Cofactors:

C11=+∣1301∣=1,C12=−∣0311∣=3,C13=+∣0110∣=−1C21=−∣2001∣=−2,C22=+∣1011∣=1,C23=−∣1210∣=2C31=+∣2013∣=6,C32=−∣1003∣=−3,C33=+∣1201∣=1\begin{aligned} C_{11} &= +\begin{vmatrix} 1 & 3 \\ 0 & 1 \end{vmatrix} = 1, \quad C_{12} &= -\begin{vmatrix} 0 & 3 \\ 1 & 1 \end{vmatrix} = 3, \quad C_{13} &= +\begin{vmatrix} 0 & 1 \\ 1 & 0 \end{vmatrix} = -1 \\ C_{21} &= -\begin{vmatrix} 2 & 0 \\ 0 & 1 \end{vmatrix} = -2, \quad C_{22} &= +\begin{vmatrix} 1 & 0 \\ 1 & 1 \end{vmatrix} = 1, \quad C_{23} &= -\begin{vmatrix} 1 & 2 \\ 1 & 0 \end{vmatrix} = 2 \\ C_{31} &= +\begin{vmatrix} 2 & 0 \\ 1 & 3 \end{vmatrix} = 6, \quad C_{32} &= -\begin{vmatrix} 1 & 0 \\ 0 & 3 \end{vmatrix} = -3, \quad C_{33} &= +\begin{vmatrix} 1 & 2 \\ 0 & 1 \end{vmatrix} = 1 \end{aligned} adj⁡(A)=(1−2631−3−121)\operatorname{adj}(\mathbf{A}) = \begin{pmatrix} 1 & -2 & 6 \\ 3 & 1 & -3 \\ -1 & 2 & 1 \end{pmatrix} A−1=17(1−2631−3−121)\mathbf{A}^{-1} = \frac{1}{7}\begin{pmatrix} 1 & -2 & 6 \\ 3 & 1 & -3 \\ -1 & 2 & 1 \end{pmatrix}

A system of nn linear equations in nn unknowns can be written as Ax=b\mathbf{A}\mathbf{x} = \mathbf{b} where A\mathbf{A} is the coefficient matrix, x\mathbf{x} is the column vector of unknowns, and b\mathbf{b} is the column vector of constants.

If A\mathbf{A} is non-singular, the unique solution is:

x=A−1b\boxed{\mathbf{x} = \mathbf{A}^{-1}\mathbf{b}}

For a 2×22 \times 2 system:

  • det⁡(A)≠0\det(\mathbf{A}) \neq 0: the two lines intersect at a unique point.
  • det⁡(A)=0\det(\mathbf{A}) = 0 and the equations are consistent: the lines are coincident (infinitely many solutions).
  • det⁡(A)=0\det(\mathbf{A}) = 0 and the equations are inconsistent: the lines are parallel (no solutions).

For a system Ax=b\mathbf{A}\mathbf{x} = \mathbf{b} where det⁡(A)≠0\det(\mathbf{A}) \neq 0:

xi=det⁡(Ai)det⁡(A)x_i = \frac{\det(\mathbf{A}_i)}{\det(\mathbf{A})}

Where Ai\mathbf{A}_i is A\mathbf{A} with column ii replaced by b\mathbf{b}.

Worked Example: Solving a $3\times3$ system

Solve: {x+2y=4y+3z=5x+z=2\begin{cases} x + 2y = 4 \\ y + 3z = 5 \\ x + z = 2 \end{cases}

In matrix form: (120013101)(xyz)=(452)\begin{pmatrix} 1 & 2 & 0 \\ 0 & 1 & 3 \\ 1 & 0 & 1 \end{pmatrix}\begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 4 \\ 5 \\ 2 \end{pmatrix}.

Using the inverse from the previous example:

(xyz)=17(1−2631−3−121)(452)=17(4−10+1212+5−6−4+10+2)=17(6118)=(6/711/78/7)\begin{pmatrix} x \\ y \\ z \end{pmatrix} = \frac{1}{7}\begin{pmatrix} 1 & -2 & 6 \\ 3 & 1 & -3 \\ -1 & 2 & 1 \end{pmatrix}\begin{pmatrix} 4 \\ 5 \\ 2 \end{pmatrix} = \frac{1}{7}\begin{pmatrix} 4 - 10 + 12 \\ 12 + 5 - 6 \\ -4 + 10 + 2 \end{pmatrix} = \frac{1}{7}\begin{pmatrix} 6 \\ 11 \\ 8 \end{pmatrix} = \begin{pmatrix} 6/7 \\ 11/7 \\ 8/7 \end{pmatrix}

A 2×22 \times 2 matrix T\mathbf{T} represents a linear transformation of R2\mathbb{R}^2: the point (x,y)(x, y) is mapped to (x′,y′)(x', y') where (x′y′)=T(xy)\begin{pmatrix} x' \\ y' \end{pmatrix} = \mathbf{T}\begin{pmatrix} x \\ y \end{pmatrix}.

Reflection in the xx-axis: (100−1)\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}

Reflection in the yy-axis: (−1001)\begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix}

Reflection in the line y=xy = x: (0110)\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}

Reflection in the line y=tan⁡θ xy = \tan\theta\, x:

R=(cos⁡2θsin⁡2θsin⁡2θ−cos⁡2θ)\boxed{\mathbf{R} = \begin{pmatrix} \cos 2\theta & \sin 2\theta \\ \sin 2\theta & -\cos 2\theta \end{pmatrix}}

The reflection of a vector in a line through the origin making angle θ\theta with the xx-axis can be decomposed: first rotate by −θ-\theta to align the mirror with the xx-axis, reflect in the xx-axis, then rotate back by θ\theta.

R=(cos⁡θ−sin⁡θsin⁡θcos⁡θ)(100−1)(cos⁡θsin⁡θ−sin⁡θcos⁡θ)\mathbf{R} = \begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix}\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}\begin{pmatrix} \cos\theta & \sin\theta \\ -\sin\theta & \cos\theta \end{pmatrix} =(cos⁡θsin⁡θsin⁡θ−cos⁡θ)(cos⁡θsin⁡θ−sin⁡θcos⁡θ)= \begin{pmatrix} \cos\theta & \sin\theta \\ \sin\theta & -\cos\theta \end{pmatrix}\begin{pmatrix} \cos\theta & \sin\theta \\ -\sin\theta & \cos\theta \end{pmatrix} =(cos⁡2θ−sin⁡2θcos⁡θsin⁡θ+sin⁡θcos⁡θsin⁡θcos⁡θ+cos⁡θsin⁡θsin⁡2θ−cos⁡2θ)=(cos⁡2θsin⁡2θsin⁡2θ−cos⁡2θ)□= \begin{pmatrix} \cos^2\theta - \sin^2\theta & \cos\theta\sin\theta + \sin\theta\cos\theta \\ \sin\theta\cos\theta + \cos\theta\sin\theta & \sin^2\theta - \cos^2\theta \end{pmatrix} = \begin{pmatrix} \cos 2\theta & \sin 2\theta \\ \sin 2\theta & -\cos 2\theta \end{pmatrix} \quad \square

Rotation anticlockwise by angle θ\theta about the origin:

Rθ=(cos⁡θ−sin⁡θsin⁡θcos⁡θ)\boxed{\mathbf{R}_\theta = \begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix}}

Note: det⁡(Rθ)=cos⁡2θ+sin⁡2θ=1\det(\mathbf{R}_\theta) = \cos^2\theta + \sin^2\theta = 1.

Enlargement by scale factor kk about the origin: (k00k)\begin{pmatrix} k & 0 \\ 0 & k \end{pmatrix}

det⁡=k2\det = k^2 Confirming the area is scaled by k2k^2.

Shear parallel to the xx-axis by factor kk: (1k01)\begin{pmatrix} 1 & k \\ 0 & 1 \end{pmatrix}

Shear parallel to the yy-axis by factor kk: (10k1)\begin{pmatrix} 1 & 0 \\ k & 1 \end{pmatrix}

Note: det⁡=1\det = 1 for shears, so area is preserved.

If transformation A\mathbf{A} is followed by transformation B\mathbf{B} The combined Transformation is represented by BA\mathbf{BA} (rightmost applied first).

Worked Example: Combined transformation

Find the matrix representing a rotation of 90∘90^\circ anticlockwise about the origin followed by a Reflection in the line y=xy = x.

Rotation by 90∘90^\circ: R=(0−110)\mathbf{R} = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}

Reflection in y=xy = x: S=(0110)\mathbf{S} = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}

Combined (reflection applied after rotation): SR=(0110)(0−110)=(100−1)\mathbf{S}\mathbf{R} = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}

This is a reflection in the xx-axis. det⁡=−1\det = -1 Consistent with an orientation-reversing Transformation.


Rotation about the xx-axis by angle θ\theta:

Rx=(1000cos⁡θ−sin⁡θ0sin⁡θcos⁡θ)\mathbf{R}_x = \begin{pmatrix} 1 & 0 & 0 \\ 0 & \cos\theta & -\sin\theta \\ 0 & \sin\theta & \cos\theta \end{pmatrix}

Rotation about the yy-axis by angle θ\theta:

Ry=(cos⁡θ0sin⁡θ010−sin⁡θ0cos⁡θ)\mathbf{R}_y = \begin{pmatrix} \cos\theta & 0 & \sin\theta \\ 0 & 1 & 0 \\ -\sin\theta & 0 & \cos\theta \end{pmatrix}

Rotation about the zz-axis by angle θ\theta:

Rz=(cos⁡θ−sin⁡θ0sin⁡θcos⁡θ0001)\mathbf{R}_z = \begin{pmatrix} \cos\theta & -\sin\theta & 0 \\ \sin\theta & \cos\theta & 0 \\ 0 & 0 & 1 \end{pmatrix}

Each has det⁡=1\det = 1 and represents a rigid motion preserving distances and orientation.

Reflection in the plane x=0x = 0 (the yzyz-plane): (−100010001)\begin{pmatrix} -1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}

Reflection in the plane y=0y = 0 (the xzxz-plane): (1000−10001)\begin{pmatrix} 1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 1 \end{pmatrix}

Reflection in the plane z=0z = 0 (the xyxy-plane): (10001000−1)\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -1 \end{pmatrix}

Each has det⁡=−1\det = -1 Confirming orientation reversal.


Info: info require this topic at A Level. Definition. Let A\mathbf{A} be an n×nn \times n matrix. A scalar λ\lambda is an eigenvalue of A\mathbf{A} if there exists a non-zero vector v\mathbf{v} such that:

Av=λv\boxed{\mathbf{A}\mathbf{v} = \lambda\mathbf{v}}

The vector v\mathbf{v} is called an eigenvector corresponding to λ\lambda.

Av=λv  ⟺  (A−λI)v=0\mathbf{A}\mathbf{v} = \lambda\mathbf{v} \iff (\mathbf{A} - \lambda\mathbf{I})\mathbf{v} = \mathbf{0}.

For a non-trivial solution (v≠0\mathbf{v} \neq \mathbf{0}), we require det⁡(A−λI)=0\det(\mathbf{A} - \lambda\mathbf{I}) = 0.

det⁡(A−λI)=0\boxed{\det(\mathbf{A} - \lambda\mathbf{I}) = 0}

This is the characteristic equation of A\mathbf{A}. Its roots are the eigenvalues.

For each eigenvalue λ\lambda Solve (A−λI)v=0(\mathbf{A} - \lambda\mathbf{I})\mathbf{v} = \mathbf{0} by Row reduction.

Definition. A matrix A\mathbf{A} is diagonalisable if there exists an invertible matrix P\mathbf{P} and a diagonal matrix D\mathbf{D} such that:

A=PDP−1\boxed{\mathbf{A} = \mathbf{P}\mathbf{D}\mathbf{P}^{-1}}

The columns of P\mathbf{P} are the eigenvectors of A\mathbf{A} And the diagonal entries of D\mathbf{D} are the corresponding eigenvalues.

A matrix is diagonalisable if and only if it has nn linearly independent eigenvectors (always true for nn distinct eigenvalues).

Intuition. Diagonalisation changes to a coordinate system where the transformation acts Independently on each axis (stretching by eigenvalues). In this basis, the matrix takes its simplest Possible form.

Worked Example: Eigenvalues and eigenvectors

Find the eigenvalues and eigenvectors of A=(4123)\mathbf{A} = \begin{pmatrix} 4 & 1 \\ 2 & 3 \end{pmatrix}.

Characteristic equation: det⁡ ⁣(4−λ123−λ)=0\det\!\begin{pmatrix} 4 - \lambda & 1 \\ 2 & 3 - \lambda \end{pmatrix} = 0

(4−λ)(3−λ)−2=λ2−7λ+10=(λ−5)(λ−2)=0(4 - \lambda)(3 - \lambda) - 2 = \lambda^2 - 7\lambda + 10 = (\lambda - 5)(\lambda - 2) = 0

Eigenvalues: λ1=5\lambda_1 = 5, λ2=2\lambda_2 = 2.

For λ1=5\lambda_1 = 5: (−112−2)(xy)=(00)  ⟹  −x+y=0  ⟹  y=x\begin{pmatrix} -1 & 1 \\ 2 & -2 \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix} \implies -x + y = 0 \implies y = x.

Eigenvector: v1=(11)\mathbf{v}_1 = \begin{pmatrix} 1 \\ 1 \end{pmatrix} (or any non-zero scalar Multiple).

For λ2=2\lambda_2 = 2: (2121)(xy)=(00)  ⟹  2x+y=0  ⟹  y=−2x\begin{pmatrix} 2 & 1 \\ 2 & 1 \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix} \implies 2x + y = 0 \implies y = -2x.

Eigenvector: v2=(1−2)\mathbf{v}_2 = \begin{pmatrix} 1 \\ -2 \end{pmatrix}.

Diagonalisation: P=(111−2)\mathbf{P} = \begin{pmatrix} 1 & 1 \\ 1 & -2 \end{pmatrix} D=(5002)\mathbf{D} = \begin{pmatrix} 5 & 0 \\ 0 & 2 \end{pmatrix} Giving A=PDP−1\mathbf{A} = \mathbf{P}\mathbf{D}\mathbf{P}^{-1}.


det⁡(AB)=det⁡(A)det⁡(B)\boxed{\det(\mathbf{AB}) = \det(\mathbf{A})\det(\mathbf{B})} A−1=1det⁡(A)adj⁡(A)whendet⁡(A)≠0\boxed{\mathbf{A}^{-1} = \frac{1}{\det(\mathbf{A})}\operatorname{adj}(\mathbf{A}) \quad \mathrm{when } \det(\mathbf{A}) \neq 0} Rotationbyθ:(cos⁡θ−sin⁡θsin⁡θcos⁡θ)\boxed{\mathrm{Rotation by } \theta: \begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix}} Reflectioniny=(tan⁡θ)x:(cos⁡2θsin⁡2θsin⁡2θ−cos⁡2θ)\boxed{\mathrm{Reflection in } y = (\tan\theta)x: \begin{pmatrix} \cos 2\theta & \sin 2\theta \\ \sin 2\theta & -\cos 2\theta \end{pmatrix}} det⁡(A−λI)=0  ⟹  eigenvaluesofA\boxed{\det(\mathbf{A} - \lambda\mathbf{I}) = 0 \implies \mathrm{eigenvalues of } \mathbf{A}}

Problem 1. Given A=(3−124)\mathbf{A} = \begin{pmatrix} 3 & -1 \\ 2 & 4 \end{pmatrix} and B=(15−20)\mathbf{B} = \begin{pmatrix} 1 & 5 \\ -2 & 0 \end{pmatrix} Compute AB−BA\mathbf{AB} - \mathbf{BA}.

Hint

Compute both products separately and subtract. They will not be equal.

AnswerAB=(3−124)(15−20)=(515−610)\mathbf{AB} = \begin{pmatrix} 3 & -1 \\ 2 & 4 \end{pmatrix}\begin{pmatrix} 1 & 5 \\ -2 & 0 \end{pmatrix} = \begin{pmatrix} 5 & 15 \\ -6 & 10 \end{pmatrix} BA=(15−20)(3−124)=(1319−62)\mathbf{BA} = \begin{pmatrix} 1 & 5 \\ -2 & 0 \end{pmatrix}\begin{pmatrix} 3 & -1 \\ 2 & 4 \end{pmatrix} = \begin{pmatrix} 13 & 19 \\ -6 & 2 \end{pmatrix} AB−BA=(515−610)−(1319−62)=(−8−408)\mathbf{AB} - \mathbf{BA} = \begin{pmatrix} 5 & 15 \\ -6 & 10 \end{pmatrix} - \begin{pmatrix} 13 & 19 \\ -6 & 2 \end{pmatrix} = \begin{pmatrix} -8 & -4 \\ 0 & 8 \end{pmatrix}

This confirms AB≠BA\mathbf{AB} \neq \mathbf{BA}.


Problem 2. Find the determinant and inverse of A=(2130−12101)\mathbf{A} = \begin{pmatrix} 2 & 1 & 3 \\ 0 & -1 & 2 \\ 1 & 0 & 1 \end{pmatrix}.

Hint

Expand the determinant along the first row. Then compute cofactors for the adjugate.

Answer

det⁡(A)=2∣−1201∣−1∣0211∣+3∣0−110∣\det(\mathbf{A}) = 2\begin{vmatrix} -1 & 2 \\ 0 & 1 \end{vmatrix} - 1\begin{vmatrix} 0 & 2 \\ 1 & 1 \end{vmatrix} + 3\begin{vmatrix} 0 & -1 \\ 1 & 0 \end{vmatrix}

=2(−1)−1(−2)+3(1)=−2+2+3=3= 2(-1) - 1(-2) + 3(1) = -2 + 2 + 3 = 3

Cofactors:

C11=+(−1)=−1,C12=−(0⋅1−2⋅1)=2,C13=+(0⋅0−(−1)⋅1)=1C21=−(1⋅1−3⋅0)=−1,C22=+(2⋅1−3⋅1)=−1,C23=−(2⋅0−1⋅1)=1C31=+(2+2)=4,C32=−(4−0)=−4,C33=+(−2−0)=−2\begin{aligned} C_{11} = +(-1) = -1, \quad C_{12} = -(0 \cdot 1 - 2 \cdot 1) = 2, \quad C_{13} = +(0 \cdot 0 - (-1) \cdot 1) = 1 \\ C_{21} = -(1 \cdot 1 - 3 \cdot 0) = -1, \quad C_{22} = +(2 \cdot 1 - 3 \cdot 1) = -1, \quad C_{23} = -(2 \cdot 0 - 1 \cdot 1) = 1 \\ C_{31} = +(2 + 2) = 4, \quad C_{32} = -(4 - 0) = -4, \quad C_{33} = +(-2 - 0) = -2 \end{aligned} adj⁡(A)=(−1−142−1−411−2)\operatorname{adj}(\mathbf{A}) = \begin{pmatrix} -1 & -1 & 4 \\ 2 & -1 & -4 \\ 1 & 1 & -2 \end{pmatrix} A−1=13(−1−142−1−411−2)\mathbf{A}^{-1} = \frac{1}{3}\begin{pmatrix} -1 & -1 & 4 \\ 2 & -1 & -4 \\ 1 & 1 & -2 \end{pmatrix}

Problem 3. Find the matrix representing a reflection in the line y=3 xy = \sqrt{3}\,x And verify That M2=I\mathbf{M}^2 = \mathbf{I}.

Hint

The line makes angle θ\theta with the xx-axis where tan⁡θ=3\tan\theta = \sqrt{3}. A reflection applied Twice is the identity.

Answer

tan⁡θ=3  ⟹  θ=π3\tan\theta = \sqrt{3} \implies \theta = \dfrac{\pi}{3}.

M=(cos⁡2π3sin⁡2π3sin⁡2π3−cos⁡2π3)=(−12323212)\mathbf{M} = \begin{pmatrix} \cos\frac{2\pi}{3} & \sin\frac{2\pi}{3} \\ \sin\frac{2\pi}{3} & -\cos\frac{2\pi}{3} \end{pmatrix} = \begin{pmatrix} -\frac{1}{2} & \frac{\sqrt{3}}{2} \\ \frac{\sqrt{3}}{2} & \frac{1}{2} \end{pmatrix}

Verification:

M2=(−12323212)(−12323212)=(14+34−34+34−34+3434+14)=(1001)=I□\mathbf{M}^2 = \begin{pmatrix} -\frac{1}{2} & \frac{\sqrt{3}}{2} \\ \frac{\sqrt{3}}{2} & \frac{1}{2} \end{pmatrix}\begin{pmatrix} -\frac{1}{2} & \frac{\sqrt{3}}{2} \\ \frac{\sqrt{3}}{2} & \frac{1}{2} \end{pmatrix} = \begin{pmatrix} \frac{1}{4} + \frac{3}{4} & -\frac{\sqrt{3}}{4} + \frac{\sqrt{3}}{4} \\ -\frac{\sqrt{3}}{4} + \frac{\sqrt{3}}{4} & \frac{3}{4} + \frac{1}{4} \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = \mathbf{I} \quad \square

Problem 4. The triangle with vertices (0, 0)$$(2, 0)$$(0, 1) is transformed by the matrix T=(3102)\mathbf{T} = \begin{pmatrix} 3 & 1 \\ 0 & 2 \end{pmatrix}. Find the coordinates of the vertices of the image, and verify that the area scales by ∣det⁡(T)∣|\det(\mathbf{T})|.

Hint

Apply T\mathbf{T} to each vertex. The original triangle has area 1.

Answer

(0, 0) \mapsto (0, 0)$$(2, 0) \mapsto (6, 0)$$(0, 1) \mapsto (1, 2).

Image vertices: (0, 0)$$(6, 0)$$(1, 2).

Original area: 12×2×1=1\dfrac{1}{2} \times 2 \times 1 = 1.

Image area using the determinant formula: 12∣6⋅2−0⋅1∣=12×12=6\dfrac{1}{2}\left|6 \cdot 2 - 0 \cdot 1\right| = \dfrac{1}{2} \times 12 = 6.

det⁡(T)=6\det(\mathbf{T}) = 6 And ∣det⁡(T)∣×originalarea=6×1=6|\det(\mathbf{T})| \times \mathrm{original area} = 6 \times 1 = 6. ✓


Problem 5. Solve the system of equations using matrices:

{2x+y−z=3x−y+2z=13x+2y+z=10\begin{cases} 2x + y - z = 3 \\ x - y + 2z = 1 \\ 3x + 2y + z = 10 \end{cases}
Hint

Write as Ax=b\mathbf{A}\mathbf{x} = \mathbf{b} and compute x=A−1b\mathbf{x} = \mathbf{A}^{-1}\mathbf{b}.

Answer

A=(21−11−12321)\mathbf{A} = \begin{pmatrix} 2 & 1 & -1 \\ 1 & -1 & 2 \\ 3 & 2 & 1 \end{pmatrix} b=(3110)\mathbf{b} = \begin{pmatrix} 3 \\ 1 \\ 10 \end{pmatrix}.

det⁡(A)=2∣−1221∣−1∣1231∣+(−1)∣1−132∣\det(\mathbf{A}) = 2\begin{vmatrix} -1 & 2 \\ 2 & 1 \end{vmatrix} - 1\begin{vmatrix} 1 & 2 \\ 3 & 1 \end{vmatrix} + (-1)\begin{vmatrix} 1 & -1 \\ 3 & 2 \end{vmatrix}

=2(−5)−1(−5)−1(5)=−10+5−5=−10= 2(-5) - 1(-5) - 1(5) = -10 + 5 - 5 = -10

Cofactors: C_{11} = -5$$C_{12} = 5$$C_{13} = 5$$C_{21} = -3$$C_{22} = 5$$C_{23} = -1 C_{31} = 1$$C_{32} = -5$$C_{33} = -3.

A−1=−110(−5−3155−55−1−3)\mathbf{A}^{-1} = -\frac{1}{10}\begin{pmatrix} -5 & -3 & 1 \\ 5 & 5 & -5 \\ 5 & -1 & -3 \end{pmatrix} x=−110(−5−3155−55−1−3)(3110)=−110(−15−3+1015+5−5015−1−30)=−110(−8−30−16)=(4/538/5)\mathbf{x} = -\frac{1}{10}\begin{pmatrix} -5 & -3 & 1 \\ 5 & 5 & -5 \\ 5 & -1 & -3 \end{pmatrix}\begin{pmatrix} 3 \\ 1 \\ 10 \end{pmatrix} = -\frac{1}{10}\begin{pmatrix} -15 - 3 + 10 \\ 15 + 5 - 50 \\ 15 - 1 - 30 \end{pmatrix} = -\frac{1}{10}\begin{pmatrix} -8 \\ -30 \\ -16 \end{pmatrix} = \begin{pmatrix} 4/5 \\ 3 \\ 8/5 \end{pmatrix}

So x = \dfrac{4}{5}$$y = 3$$z = \dfrac{8}{5}.


Problem 6. Find the single 2×22 \times 2 matrix that represents an enlargement by scale factor 2 About the origin followed by a rotation of 90∘90^\circ anticlockwise. Show that this is equivalent to A single rotation of 90∘90^\circ combined with an enlargement by factor 2.

Hint

The enlargement matrix is 2I2\mathbf{I} and the rotation is Rπ/2\mathbf{R}_{\pi/2}. Since 2I2\mathbf{I} commutes with all matrices, the order doesn’t matter.

Answer

Enlargement by 2: E=(2002)\mathbf{E} = \begin{pmatrix} 2 & 0 \\ 0 & 2 \end{pmatrix}.

Rotation by 90∘90^\circ: R=(0−110)\mathbf{R} = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}.

Enlargement then rotation: RE=(0−110)(2002)=(0−220)\mathbf{R}\mathbf{E} = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}\begin{pmatrix} 2 & 0 \\ 0 & 2 \end{pmatrix} = \begin{pmatrix} 0 & -2 \\ 2 & 0 \end{pmatrix}.

Rotation then enlargement: ER=(2002)(0−110)=(0−220)\mathbf{E}\mathbf{R} = \begin{pmatrix} 2 & 0 \\ 0 & 2 \end{pmatrix}\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} = \begin{pmatrix} 0 & -2 \\ 2 & 0 \end{pmatrix}.

Both give the same result: (0−220)=2(0−110)\begin{pmatrix} 0 & -2 \\ 2 & 0 \end{pmatrix} = 2\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}.

This is a rotation by 90∘90^\circ combined with an enlargement by factor 2, and the order is Irrelevant because scalar multiples of the identity commute with all matrices.


Problem 7. AQA only. Find the eigenvalues and corresponding eigenvectors of A=(5412)\mathbf{A} = \begin{pmatrix} 5 & 4 \\ 1 & 2 \end{pmatrix} And write down a matrix P\mathbf{P} And diagonal matrix D\mathbf{D} such that A=PDP−1\mathbf{A} = \mathbf{P}\mathbf{D}\mathbf{P}^{-1}.

Hint

Solve det⁡(A−λI)=0\det(\mathbf{A} - \lambda\mathbf{I}) = 0 for λ\lambda Then find eigenvectors.

Answer

Characteristic equation: det⁡(5−λ412−λ)=(5−λ)(2−λ)−4=λ2−7λ+6=(λ−1)(λ−6)=0\det\begin{pmatrix} 5 - \lambda & 4 \\ 1 & 2 - \lambda \end{pmatrix} = (5 - \lambda)(2 - \lambda) - 4 = \lambda^2 - 7\lambda + 6 = (\lambda - 1)(\lambda - 6) = 0.

Eigenvalues: \lambda_1 = 6$$\lambda_2 = 1.

For λ1=6\lambda_1 = 6: (−141−4)(xy)=0  ⟹  −x+4y=0  ⟹  x=4y\begin{pmatrix} -1 & 4 \\ 1 & -4 \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix} = \mathbf{0} \implies -x + 4y = 0 \implies x = 4y.

Eigenvector: v1=(41)\mathbf{v}_1 = \begin{pmatrix} 4 \\ 1 \end{pmatrix}.

For λ2=1\lambda_2 = 1: (4411)(xy)=0  ⟹  x+y=0  ⟹  x=−y\begin{pmatrix} 4 & 4 \\ 1 & 1 \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix} = \mathbf{0} \implies x + y = 0 \implies x = -y.

Eigenvector: v2=(1−1)\mathbf{v}_2 = \begin{pmatrix} 1 \\ -1 \end{pmatrix}.

P=(411−1),D=(6001)\mathbf{P} = \begin{pmatrix} 4 & 1 \\ 1 & -1 \end{pmatrix}, \quad \mathbf{D} = \begin{pmatrix} 6 & 0 \\ 0 & 1 \end{pmatrix}

Problem 8. AQA only. The matrix A=(3−210)\mathbf{A} = \begin{pmatrix} 3 & -2 \\ 1 & 0 \end{pmatrix} has Eigenvalues 1 and 2. Use this to compute A5\mathbf{A}^5 without multiplying matrices five times.

Hint

Diagonalise A=PDP−1\mathbf{A} = \mathbf{P}\mathbf{D}\mathbf{P}^{-1} Then A5=PD5P−1\mathbf{A}^5 = \mathbf{P}\mathbf{D}^5\mathbf{P}^{-1}.

Answer

For λ=1\lambda = 1: (2−21−1)v=0  ⟹  x=y\begin{pmatrix} 2 & -2 \\ 1 & -1 \end{pmatrix}\mathbf{v} = \mathbf{0} \implies x = y. Eigenvector (11)\begin{pmatrix} 1 \\ 1 \end{pmatrix}.

For λ=2\lambda = 2: (1−21−2)v=0  ⟹  x=2y\begin{pmatrix} 1 & -2 \\ 1 & -2 \end{pmatrix}\mathbf{v} = \mathbf{0} \implies x = 2y. Eigenvector (21)\begin{pmatrix} 2 \\ 1 \end{pmatrix}.

P=(1211)\mathbf{P} = \begin{pmatrix} 1 & 2 \\ 1 & 1 \end{pmatrix} D=(1002)\mathbf{D} = \begin{pmatrix} 1 & 0 \\ 0 & 2 \end{pmatrix}.

det⁡(P)=1−2=−1\det(\mathbf{P}) = 1 - 2 = -1 So P−1=(−121−1)\mathbf{P}^{-1} = \begin{pmatrix} -1 & 2 \\ 1 & -1 \end{pmatrix}.

D5=(10032)\mathbf{D}^5 = \begin{pmatrix} 1 & 0 \\ 0 & 32 \end{pmatrix}.

A5=PD5P−1=(1211)(10032)(−121−1)=(164132)(−121−1)=(63−6231−30)\mathbf{A}^5 = \mathbf{P}\mathbf{D}^5\mathbf{P}^{-1} = \begin{pmatrix} 1 & 2 \\ 1 & 1 \end{pmatrix}\begin{pmatrix} 1 & 0 \\ 0 & 32 \end{pmatrix}\begin{pmatrix} -1 & 2 \\ 1 & -1 \end{pmatrix} = \begin{pmatrix} 1 & 64 \\ 1 & 32 \end{pmatrix}\begin{pmatrix} -1 & 2 \\ 1 & -1 \end{pmatrix} = \begin{pmatrix} 63 & -62 \\ 31 & -30 \end{pmatrix}

Problem 9. Find the 3×33 \times 3 matrix representing a rotation of 90∘90^\circ anticlockwise About the zz-axis. Verify that this matrix has determinant 1 and that it maps (100)\begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} to (010)\begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix}.

Hint

Use the standard formula for Rz\mathbf{R}_z with θ=π/2\theta = \pi/2.

AnswerRz=(cos⁡π2−sin⁡π20sin⁡π2cos⁡π20001)=(0−10100001)\mathbf{R}_z = \begin{pmatrix} \cos\frac{\pi}{2} & -\sin\frac{\pi}{2} & 0 \\ \sin\frac{\pi}{2} & \cos\frac{\pi}{2} & 0 \\ 0 & 0 & 1 \end{pmatrix} = \begin{pmatrix} 0 & -1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{pmatrix}

det⁡(Rz)=0⋅(0−0)−(−1)(1−0)+0=1\det(\mathbf{R}_z) = 0 \cdot (0 - 0) - (-1)(1 - 0) + 0 = 1. ✓

Rz(100)=(010)✓\mathbf{R}_z\begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} = \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix} \quad \checkmark

The xx-axis is correctly rotated to the yy-axis by a 90∘90^\circ anticlockwise rotation about zz.


Problem 10. Prove that if A\mathbf{A} and B\mathbf{B} are non-singular n×nn \times n matrices, Then (AB)−1=B−1A−1(\mathbf{AB})^{-1} = \mathbf{B}^{-1}\mathbf{A}^{-1}.

Hint

Show that B−1A−1\mathbf{B}^{-1}\mathbf{A}^{-1} satisfies the definition of the inverse of AB\mathbf{AB} and invoke uniqueness.

Answer

We need to show that (AB)(B−1A−1)=I(\mathbf{AB})(\mathbf{B}^{-1}\mathbf{A}^{-1}) = \mathbf{I} and (B−1A−1)(AB)=I(\mathbf{B}^{-1}\mathbf{A}^{-1})(\mathbf{AB}) = \mathbf{I}.

(AB)(B−1A−1)=A(BB−1)A−1=AIA−1=AA−1=I(\mathbf{AB})(\mathbf{B}^{-1}\mathbf{A}^{-1}) = \mathbf{A}(\mathbf{B}\mathbf{B}^{-1})\mathbf{A}^{-1} = \mathbf{A}\mathbf{I}\mathbf{A}^{-1} = \mathbf{A}\mathbf{A}^{-1} = \mathbf{I} (B−1A−1)(AB)=B−1(A−1A)B=B−1IB=B−1B=I(\mathbf{B}^{-1}\mathbf{A}^{-1})(\mathbf{AB}) = \mathbf{B}^{-1}(\mathbf{A}^{-1}\mathbf{A})\mathbf{B} = \mathbf{B}^{-1}\mathbf{I}\mathbf{B} = \mathbf{B}^{-1}\mathbf{B} = \mathbf{I}

Since the inverse is unique, (AB)−1=B−1A−1(\mathbf{AB})^{-1} = \mathbf{B}^{-1}\mathbf{A}^{-1}. □\square

Intuition. The order reverses, just like putting on and taking off socks and shoes. To undo “A Then B,” you must undo B first, then undo A.


Example 8.1: Diagonalisation of a 3×3 matrix

Section titled “Example 8.1: Diagonalisation of a 3×3 matrix”

Problem. Diagonalise A=(210020013)\mathbf{A} = \begin{pmatrix}2&1&0\\0&2&0\\0&1&3\end{pmatrix}.

Solution. Find eigenvalues: det⁡(A−λI)=0\det(\mathbf{A}-\lambda\mathbf{I}) = 0.

det⁡(2−λ1002−λ0013−λ)=(2−λ)2(3−λ)=0\det\begin{pmatrix}2-\lambda&1&0\\0&2-\lambda&0\\0&1&3-\lambda\end{pmatrix} = (2-\lambda)^2(3-\lambda) = 0

λ1=2\lambda_1 = 2 (repeated), λ2=3\lambda_2 = 3.

For λ=2\lambda = 2: (A−2I)v=0(\mathbf{A}-2\mathbf{I})\mathbf{v} = \mathbf{0} gives (010000011)v=0\begin{pmatrix}0&1&0\\0&0&0\\0&1&1\end{pmatrix}\mathbf{v} = \mathbf{0} So v2=0v_2 = 0 and v3=0v_3 = 0 With v1v_1 free. Only one eigenvector: (1,0,0)(1,0,0). Since the geometric multiplicity (1) is less than the algebraic multiplicity (2), A\mathbf{A} is not diagonalisable.

Example 8.2: Finding An\mathbf{A}^n using Cayley—Hamilton

Section titled “Example 8.2: Finding An\mathbf{A}^nAn using Cayley—Hamilton”

Problem. For A=(1203)\mathbf{A} = \begin{pmatrix}1&2\\0&3\end{pmatrix} Find A5\mathbf{A}^5.

Solution. By Cayley—Hamilton, A2−4A+3I=O\mathbf{A}^2 - 4\mathbf{A} + 3\mathbf{I} = \mathbf{O} So A2=4A−3I\mathbf{A}^2 = 4\mathbf{A} - 3\mathbf{I}.

A3=A(4A−3I)=4(4A−3I)−3A=13A−12I\mathbf{A}^3 = \mathbf{A}(4\mathbf{A}-3\mathbf{I}) = 4(4\mathbf{A}-3\mathbf{I}) - 3\mathbf{A} = 13\mathbf{A} - 12\mathbf{I}.

A4=13(4A−3I)−12A=40A−39I\mathbf{A}^4 = 13(4\mathbf{A}-3\mathbf{I}) - 12\mathbf{A} = 40\mathbf{A} - 39\mathbf{I}.

A5=40(4A−3I)−39A=121A−120I\mathbf{A}^5 = 40(4\mathbf{A}-3\mathbf{I}) - 39\mathbf{A} = 121\mathbf{A} - 120\mathbf{I}.

=121(1203)−120(1001)=(1212420363)−(12000120)=(12420243)= 121\begin{pmatrix}1&2\\0&3\end{pmatrix} - 120\begin{pmatrix}1&0\\0&1\end{pmatrix} = \begin{pmatrix}121&242\\0&363\end{pmatrix} - \begin{pmatrix}120&0\\0&120\end{pmatrix} = \boxed{\begin{pmatrix}1&242\\0&243\end{pmatrix}}

Example 8.3: Invariant points and invariant lines

Section titled “Example 8.3: Invariant points and invariant lines”

Problem. A=(3102)\mathbf{A} = \begin{pmatrix}3&1\\0&2\end{pmatrix}. Find all invariant points and Invariant lines of the transformation x↦Ax\mathbf{x} \mapsto \mathbf{Ax}.

Solution. Invariant points: Ax=x  ⟹  (A−I)x=0\mathbf{Ax} = \mathbf{x} \implies (\mathbf{A}-\mathbf{I})\mathbf{x} = \mathbf{0}.

(2101)x=0  ⟹  x2=0,  2x1=0\begin{pmatrix}2&1\\0&1\end{pmatrix}\mathbf{x} = \mathbf{0} \implies x_2 = 0, \; 2x_1 = 0

Only the origin (0,0)(0,0) is an invariant point.

Invariant lines through the origin: These are the eigenspaces. Eigenvalues: (3−λ)(2−λ)=0(3-\lambda)(2-\lambda) = 0 So λ=3\lambda = 3 and λ=2\lambda = 2.

For λ=3\lambda = 3: (010−1)v=0  ⟹  v2=0\begin{pmatrix}0&1\\0&-1\end{pmatrix}\mathbf{v} = \mathbf{0} \implies v_2 = 0. Line: y=0y = 0 (the xx-axis).

For λ=2\lambda = 2: (1100)v=0  ⟹  v1+v2=0\begin{pmatrix}1&1\\0&0\end{pmatrix}\mathbf{v} = \mathbf{0} \implies v_1 + v_2 = 0. Line: y=−xy = -x.

Example 8.4: Determinant as a scaling factor

Section titled “Example 8.4: Determinant as a scaling factor”

Problem. The matrix T=(21−13)\mathbf{T} = \begin{pmatrix}2&1\\-1&3\end{pmatrix} represents a Transformation. A triangle has vertices (0,0)(0,0), (1,0)(1,0), (0,1)(0,1). Find the area of its image.

Solution. det⁡(T)=6−(−1)=7\det(\mathbf{T}) = 6 - (-1) = 7.

Original area =12= \dfrac{1}{2}. Image area =∣det⁡(T)∣×original area=7×12=3.5= |\det(\mathbf{T})| \times \text{original area} = 7 \times \dfrac{1}{2} = \boxed{3.5}.

Example 8.5: Commutator and non-commuting matrices

Section titled “Example 8.5: Commutator and non-commuting matrices”

Problem. For A=(0100)\mathbf{A} = \begin{pmatrix}0&1\\0&0\end{pmatrix} and B=(0010)\mathbf{B} = \begin{pmatrix}0&0\\1&0\end{pmatrix} Compute the commutator [A,B]=AB−BA[\mathbf{A}, \mathbf{B}] = \mathbf{AB} - \mathbf{BA}.

Solution. AB=(0100)(0010)=(1000)\mathbf{AB} = \begin{pmatrix}0&1\\0&0\end{pmatrix}\begin{pmatrix}0&0\\1&0\end{pmatrix} = \begin{pmatrix}1&0\\0&0\end{pmatrix}.

BA=(0010)(0100)=(0001)\mathbf{BA} = \begin{pmatrix}0&0\\1&0\end{pmatrix}\begin{pmatrix}0&1\\0&0\end{pmatrix} = \begin{pmatrix}0&0\\0&1\end{pmatrix}.

[A,B]=(100−1)\boxed{[\mathbf{A},\mathbf{B}] = \begin{pmatrix}1&0\\0&-1\end{pmatrix}}

Since [A,B]≠O[\mathbf{A},\mathbf{B}] \neq \mathbf{O}, A\mathbf{A} and B\mathbf{B} do not commute.

Example 8.6: Matrix representing successive transformations

Section titled “Example 8.6: Matrix representing successive transformations”

Problem. Transformation RR is a reflection in the line y=x3y = x\sqrt{3}. Transformation SS is a Rotation by 90°90° anticlockwise about the origin. Find the matrix representing RSRS.

Solution. The line y=x3y = x\sqrt{3} makes angle 60°60° with the xx-axis.

R=(cos⁡120°sin⁡120°sin⁡120°−cos⁡120°)=(−12323212)R = \begin{pmatrix}\cos 120°&\sin 120°\\\sin 120°&-\cos 120°\end{pmatrix} = \begin{pmatrix}-\frac{1}{2}&\frac{\sqrt{3}}{2}\\\frac{\sqrt{3}}{2}&\frac{1}{2}\end{pmatrix}.

S=(0−110)S = \begin{pmatrix}0&-1\\1&0\end{pmatrix}.

RS=(−12323212)(0−110)=(321212−32)RS = \begin{pmatrix}-\frac{1}{2}&\frac{\sqrt{3}}{2}\\\frac{\sqrt{3}}{2}&\frac{1}{2}\end{pmatrix}\begin{pmatrix}0&-1\\1&0\end{pmatrix} = \begin{pmatrix}\frac{\sqrt{3}}{2}&\frac{1}{2}\\\frac{1}{2}&-\frac{\sqrt{3}}{2}\end{pmatrix}.

det⁡(RS)=−34−14=−1\det(RS) = -\dfrac{3}{4} - \dfrac{1}{4} = -1 and tr(RS)=0\text{tr}(RS) = 0 Confirming this is a Reflection.

Example 8.7: Finding the inverse of a 3×3 matrix

Section titled “Example 8.7: Finding the inverse of a 3×3 matrix”

Problem. Find A−1\mathbf{A}^{-1} where A=(10201−1110)\mathbf{A} = \begin{pmatrix}1&0&2\\0&1&-1\\1&1&0\end{pmatrix}.

Solution. det⁡(A)=1(0+1)−0+2(0−1)=1−2=−1≠0\det(\mathbf{A}) = 1(0+1) - 0 + 2(0-1) = 1 - 2 = -1 \neq 0.

A−1=1−1(12−2−1−21−1−11)=(−1−2212−111−1)\mathbf{A}^{-1} = \dfrac{1}{-1}\begin{pmatrix}1&2&-2\\-1&-2&1\\-1&-1&1\end{pmatrix} = \boxed{\begin{pmatrix}-1&-2&2\\1&2&-1\\1&1&-1\end{pmatrix}}


PitfallCorrect Approach
Assuming all matrices are diagonalisableCheck geometric multiplicity equals algebraic multiplicity for each eigenvalue
Forgetting that (AB)−1=B−1A−1(\mathbf{AB})^{-1} = \mathbf{B}^{-1}\mathbf{A}^{-1}The order reverses
Computing det⁡(A+B)=det⁡(A)+det⁡(B)\det(\mathbf{A}+\mathbf{B}) = \det(\mathbf{A})+\det(\mathbf{B}), det⁡(A+B)≠det⁡(A)+det⁡(B)\det(\mathbf{A}+\mathbf{B}) \neq \det(\mathbf{A})+\det(\mathbf{B})
Mixing up row and column operationsColumn operations change the determinant differently from row operations

Find the eigenvalues and eigenvectors of A=(4−121)\mathbf{A} = \begin{pmatrix}4&-1\\2&1\end{pmatrix}. Hence Write down P\mathbf{P} and D\mathbf{D} such that P−1AP=D\mathbf{P}^{-1}\mathbf{A}\mathbf{P} = \mathbf{D}.

Solution

det⁡(A−λI)=(4−λ)(1−λ)+2=λ2−5λ+6=0\det(\mathbf{A}-\lambda\mathbf{I}) = (4-\lambda)(1-\lambda)+2 = \lambda^2-5\lambda+6 = 0. λ=2,3\lambda = 2, 3.

λ=2\lambda = 2: (2−12−1)v=0  ⟹  2v1−v2=0  ⟹  v=(1,2)\begin{pmatrix}2&-1\\2&-1\end{pmatrix}\mathbf{v}=\mathbf{0} \implies 2v_1-v_2=0 \implies \mathbf{v}=(1,2).

λ=3\lambda = 3: (1−12−2)v=0  ⟹  v1=v2  ⟹  v=(1,1)\begin{pmatrix}1&-1\\2&-2\end{pmatrix}\mathbf{v}=\mathbf{0} \implies v_1=v_2 \implies \mathbf{v}=(1,1).

P=(1121)\mathbf{P} = \begin{pmatrix}1&1\\2&1\end{pmatrix} D=(2003)\mathbf{D} = \begin{pmatrix}2&0\\0&3\end{pmatrix}.

Prove that if λ\lambda is an eigenvalue of A\mathbf{A} with eigenvector v\mathbf{v} Then λ2\lambda^2 is an eigenvalue of A2\mathbf{A}^2 with the same eigenvector.

Solution

Av=λv\mathbf{Av} = \lambda\mathbf{v}.

A2v=A(Av)=A(λv)=λ(Av)=λ(λv)=λ2v\mathbf{A}^2\mathbf{v} = \mathbf{A}(\mathbf{Av}) = \mathbf{A}(\lambda\mathbf{v}) = \lambda(\mathbf{Av}) = \lambda(\lambda\mathbf{v}) = \lambda^2\mathbf{v}.

Therefore λ2\lambda^2 is an eigenvalue of A2\mathbf{A}^2 with eigenvector v\mathbf{v}. ■\blacksquare

The transformation represented by M=(abcd)\mathbf{M} = \begin{pmatrix}a&b\\c&d\end{pmatrix} maps the unit Square to a parallelogram of area 6. Given a+d=5a+d = 5 and ad−bc=6ad-bc = 6 Find the eigenvalues of M\mathbf{M}.

Solution

The characteristic equation: λ2−(a+d)λ+det⁡(M)=λ2−5λ+6=0\lambda^2 - (a+d)\lambda + \det(\mathbf{M}) = \lambda^2 - 5\lambda + 6 = 0.

(λ−2)(λ−3)=0(\lambda-2)(\lambda-3) = 0.

λ=2 and λ=3\boxed{\lambda = 2 \text{ and } \lambda = 3}


Complex eigenvalues lead to rotation-scaling transformations. See Complex Numbers.

The cross product can be written as a matrix multiplication. See Vectors in 3D.

Cayley—Hamilton connects matrices to polynomial algebra. See Further Algebra.


ResultFormula/Condition
InvertibilityA\mathbf{A} is invertible   ⟺  \iff det⁡(A)≠0\det(\mathbf{A}) \neq 0
(AB)−1(\mathbf{AB})^{-1}B−1A−1\mathbf{B}^{-1}\mathbf{A}^{-1}
(AB)T(\mathbf{AB})^TBTAT\mathbf{B}^T\mathbf{A}^T
det⁡(AB)\det(\mathbf{AB})det⁡(A)det⁡(B)\det(\mathbf{A})\det(\mathbf{B})
Trace of producttr(AB)=tr(BA)\text{tr}(\mathbf{AB}) = \text{tr}(\mathbf{BA})
Cayley—HamiltonA\mathbf{A} satisfies its own characteristic equation
DiagonalisabilityAll eigenvalues must have geometric multiplicity == algebraic multiplicity
Area scaling∥det⁡(T)∥×\| \det(\mathbf{T}) \| \times original area == image area

Find the matrix representing a stretch of scale factor 3 parallel to the yy-axis followed by a Reflection in the xx-axis.

Solution

Stretch: S=(1003)S = \begin{pmatrix}1&0\\0&3\end{pmatrix}. Reflection: R=(100−1)R = \begin{pmatrix}1&0\\0&-1\end{pmatrix}.

Combined: RS=(100−1)(1003)=(100−3)RS = \begin{pmatrix}1&0\\0&-1\end{pmatrix}\begin{pmatrix}1&0\\0&3\end{pmatrix} = \begin{pmatrix}1&0\\0&-3\end{pmatrix}.

(100−3)\boxed{\begin{pmatrix}1&0\\0&-3\end{pmatrix}}

Prove that det⁡(AT)=det⁡(A)\det(\mathbf{A}^T) = \det(\mathbf{A}) for any square matrix A\mathbf{A}.

Solution

The determinant can be computed by cofactor expansion along any row or column. Expanding det⁡(A)\det(\mathbf{A}) along row ii and det⁡(AT)\det(\mathbf{A}^T) along column ii (which is row ii of A\mathbf{A}) gives the same expression, since the cofactors are the same.

Formally: this follows from the permutation definition of the determinant. ■\blacksquare


14.1 Eigenvalues and the characteristic polynomial

Section titled “14.1 Eigenvalues and the characteristic polynomial”

The characteristic polynomial of A\mathbf{A} is p(λ)=det⁡(A−λI)p(\lambda) = \det(\mathbf{A}-\lambda\mathbf{I}).

Properties:

  • The sum of eigenvalues equals the trace: ∑λi=tr(A)\sum \lambda_i = \mathrm{tr}(\mathbf{A})
  • The product of eigenvalues equals the determinant: ∏λi=det⁡(A)\prod \lambda_i = \det(\mathbf{A})
  • A\mathbf{A} is invertible iff no eigenvalue is zero

Not all matrices are diagonalisable. The Jordan normal form is a generalisation where the diagonal Matrix D\mathbf{D} may have 1s on the superdiagonal (Jordan blocks).

For example, if A\mathbf{A} has a repeated eigenvalue λ\lambda with only one eigenvector:

P−1AP=(λ10λ)\mathbf{P}^{-1}\mathbf{AP} = \begin{pmatrix}\lambda&1\\0&\lambda\end{pmatrix}

A real symmetric matrix A\mathbf{A} can always be diagonalised by an orthogonal matrix: A=QDQT\mathbf{A} = \mathbf{Q}\mathbf{D}\mathbf{Q}^T where QT=Q−1\mathbf{Q}^T = \mathbf{Q}^{-1}.

The Spectral Theorem states that A=∑i=1nλiqiqiT\mathbf{A} = \sum_{i=1}^{n} \lambda_i \mathbf{q}_i\mathbf{q}_i^T where λi\lambda_i are eigenvalues and qi\mathbf{q}_i are orthonormal eigenvectors.

The Frobenius norm: ∥A∥F=∑i,jaij2=tr(ATA)\|\mathbf{A}\|_F = \sqrt{\sum_{i,j} a_{ij}^2} = \sqrt{\mathrm{tr}(\mathbf{A}^T\mathbf{A})}.

The spectral norm: ∥A∥2=σmax⁡\|\mathbf{A}\|_2 = \sigma_{\max} (largest singular value).


Find the eigenvalues and eigenvectors of A=(111111111)\mathbf{A} = \begin{pmatrix}1&1&1\\1&1&1\\1&1&1\end{pmatrix}.

Solution

det⁡(A−λI)=(1−λ)3−3(1−λ)−2+3(1−λ)=(1−λ)3−3(1−λ)+3(1−λ)−2\det(\mathbf{A}-\lambda\mathbf{I}) = (1-\lambda)^3 - 3(1-\lambda) - 2 + 3(1-\lambda) = (1-\lambda)^3 - 3(1-\lambda) + 3(1-\lambda) - 2.

Wait, let me compute directly. det⁡(1−λ1111−λ1111−λ)\det\begin{pmatrix}1-\lambda&1&1\\1&1-\lambda&1\\1&1&1-\lambda\end{pmatrix}.

=(1−λ)[(1−λ)2−1]−1[(1−λ)−1]+1[1−(1−λ)]= (1-\lambda)[(1-\lambda)^2-1] - 1[(1-\lambda)-1] + 1[1-(1-\lambda)]

=(1−λ)[1−2λ+λ2−1]+λ+λ=(1−λ)(λ2−2λ)+2λ= (1-\lambda)[1-2\lambda+\lambda^2-1] + \lambda + \lambda = (1-\lambda)(\lambda^2-2\lambda) + 2\lambda

=(1−λ)λ(λ−2)+2λ=λ[(1−λ)(λ−2)+2]=λ[λ−2−λ2+2λ+2]=λ[−λ2+3λ]=λ2(3−λ)= (1-\lambda)\lambda(\lambda-2) + 2\lambda = \lambda[(1-\lambda)(\lambda-2)+2] = \lambda[\lambda-2-\lambda^2+2\lambda+2] = \lambda[-\lambda^2+3\lambda] = \lambda^2(3-\lambda).

Eigenvalues: λ=0\lambda = 0 (double) and λ=3\lambda = 3.

λ=0\lambda = 0: (111111111)v=0  ⟹  v1+v2+v3=0\begin{pmatrix}1&1&1\\1&1&1\\1&1&1\end{pmatrix}\mathbf{v}=\mathbf{0} \implies v_1+v_2+v_3=0. Two Independent eigenvectors: (1,−1,0)(1,-1,0) and (1,0,−1)(1,0,-1).

λ=3\lambda = 3: (−2111−2111−2)v=0  ⟹  v1=v2=v3\begin{pmatrix}-2&1&1\\1&-2&1\\1&1&-2\end{pmatrix}\mathbf{v}=\mathbf{0} \implies v_1=v_2=v_3. Eigenvector: (1,1,1)(1,1,1).

Prove that similar matrices have the same eigenvalues.

Solution

If B=P−1AP\mathbf{B} = \mathbf{P}^{-1}\mathbf{AP} Then:

det⁡(B−λI)=det⁡(P−1AP−λI)=det⁡(P−1(A−λI)P)\det(\mathbf{B}-\lambda\mathbf{I}) = \det(\mathbf{P}^{-1}\mathbf{AP}-\lambda\mathbf{I}) = \det(\mathbf{P}^{-1}(\mathbf{A}-\lambda\mathbf{I})\mathbf{P})

=det⁡(P−1)det⁡(A−λI)det⁡(P)=det⁡(A−λI)= \det(\mathbf{P}^{-1})\det(\mathbf{A}-\lambda\mathbf{I})\det(\mathbf{P}) = \det(\mathbf{A}-\lambda\mathbf{I}).

Since the characteristic polynomials are identical, the eigenvalues are the same. ■\blacksquare


Any square matrix A\mathbf{A} can be decomposed as A=LU\mathbf{A} = \mathbf{L}\mathbf{U} where L\mathbf{L} is lower triangular and U\mathbf{U} is upper triangular. This is used for efficient Numerical solution of systems Ax=b\mathbf{Ax} = \mathbf{b}.

16.2 The Cayley—Hamilton theorem, applications

Section titled “16.2 The Cayley—Hamilton theorem, applications”

Since A\mathbf{A} satisfies p(A)=Op(\mathbf{A}) = \mathbf{O} where pp is the characteristic Polynomial:

  • A−1\mathbf{A}^{-1} can be computed from An\mathbf{A}^n terms
  • An\mathbf{A}^n for large nn can be reduced using the recurrence

Any m×nm \times n matrix A\mathbf{A} can be written as A=UΣVT\mathbf{A} = \mathbf{U}\boldsymbol{\Sigma}\mathbf{V}^T where U\mathbf{U} and V\mathbf{V} are Orthogonal and Σ\boldsymbol{\Sigma} is diagonal with non-negative singular values.

A symmetric matrix A\mathbf{A} is positive definite if xTAx>0\mathbf{x}^T\mathbf{A}\mathbf{x} > 0 for all x≠0\mathbf{x} \neq \mathbf{0}.

Equivalent conditions: all eigenvalues positive, all leading principal minors positive.


Find det⁡(A)\det(\mathbf{A}) where A=(123456780)\mathbf{A} = \begin{pmatrix}1&2&3\\4&5&6\\7&8&0\end{pmatrix}.

Solution

det⁡(A)=1(0−48)−2(0−42)+3(32−35)=−48+84−9=27\det(\mathbf{A}) = 1(0-48) - 2(0-42) + 3(32-35) = -48 + 84 - 9 = \boxed{27}.

Prove that the trace of a matrix equals the sum of its eigenvalues (with multiplicity).

Solution

The characteristic polynomial is p(λ)=(−1)n[λn−(tr A)λn−1+⋯+(−1)ndet⁡(A)]p(\lambda) = (-1)^n[\lambda^n - (\text{tr}\,\mathbf{A})\lambda^{n-1} + \cdots + (-1)^n\det(\mathbf{A})].

By Vieta’s formulae, the coefficient of λn−1\lambda^{n-1} equals −(λ1+λ2+⋯+λn)-(\lambda_1 + \lambda_2 + \cdots + \lambda_n).

Therefore tr A=λ1+λ2+⋯+λn\text{tr}\,\mathbf{A} = \lambda_1 + \lambda_2 + \cdots + \lambda_n. ■\blacksquare

  • Complex Numbers — Complex eigenvalues lead to rotation-scaling transformations, connecting matrix theory to the geometry of complex arithmetic.
  • Further Algebra — The Cayley-Hamilton theorem links matrices to polynomial algebra through the characteristic equation.
  • Further Calculus — Matrix methods underpin systems of differential equations solved using eigenvalues and eigenvectors.
  • Vectors in 3D — The cross product can be written as a matrix multiplication, connecting vector algebra to linear transformations.