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1.4 Matrices

Syllabus
9231–2028–2029
Topic
1.4
Level
AS

Learning objectives

1.4.1Matrix operations• carry out operations of matrix addition, subtraction and multiplication, and recognise the terms zero matrix and identity (or unit) matrix Including non-square matrices. Matrices will have at most 3 rows and columns.1.4.2Determinants and inverses• recall the meaning of the terms 'singular' and 'non-singular' as applied to square matrices and, for 22# and 33# matrices, evaluate determinants and find inverses of non-singular matrices The notations det M for the determinant of a matrix M, and I for the identity matrix, will be used.1.4.3Inverse matrix products• understand and use the result, for non-singular matrices, (AB)-1 = B-1A-1 Extension to the product of more than two matrices may be required.1.4.4Matrix transformations• understand the use of 22# matrices to represent certain geometric transformations in the x-y plane, in particular - understand the relationship between the transformations represented by A and A-1 - recognise that the matrix product AB represents the transformation that results from the transformation represented by B followed by the transformation represented by A - recall how the area scale factor of a transformation is related to the determinant of the corresponding matrix - find the matrix that represents a given transformation or sequence of transformations Understanding of the terms 'rotation', 'reflection', 'enlargement', 'stretch' and 'shear' for 2D transformations will be required. Other 2D transformations may be included, but no particular knowledge of them is expected.1.4.5Invariant points and lines• understand the meaning of 'invariant' as applied to points and lines in the context of transformations represented by matrices, and solve simple problems involving invariant points and invariant lines. e.g. to locate the invariant points of the transformation represented by 6 2 5 3eo, or to find the invariant lines through the origin for 4 2 1 1 - eo, or to show that any line with gradient 1 is invariant for 2 1 0 1eo.

Matrix dimensions decide which operations exist

Operation Condition Result
A±BA\pm B A and B have the same dimensions Add/subtract corresponding entries; same dimensions
ABAB columns of A = rows of B If A is m×nm\times n and B is n×pn\times p, AB is m×pm\times p
AOAO or OAOA dimensions make the product valid Zero matrix of the resulting dimensions
AIAI or IAIA I has the matching square order A

Entry (i,j)(i,j) of ABAB is row i of A dotted with column j of B. This rule works for non-square matrices and is not entry-by-entry multiplication. AB may exist when BA does not; if both exist, they need not be equal.

A=\begin{pmatrix}1&2&0\-1&3&1\end{pmatrix},\quad B=\begin{pmatrix}2&1\0&-2\4&3\end{pmatrix}AB=\begin{pmatrix}1(2)+2(0)+0(4)&1(1)+2(-2)+0(3)\-1(2)+3(0)+1(4)&-1(1)+3(-2)+1(3)\end{pmatrix}=\begin{pmatrix}2&-3\2&-4\end{pmatrix}.

The zero matrix is shape-specific, while the identity matrix is square. Always write dimensions before multiplying; matching the visible number of entries is not enough.

The determinant is the gate to a matrix inverse

A square matrix M is non-singular exactly when det⁡M≠0\det M\ne0; then one unique inverse satisfies MM−1=M−1M=IMM^{-1}=M^{-1}M=I. If det⁡M=0\det M=0, M is singular and no inverse exists.

ForForM=\begin{pmatrix}a&b\c&d\end{pmatrix},\qquad \det M=ad-bc,and, when $ad-bc\ne0$,M^{-1}=\frac1{ad-bc}\begin{pmatrix}d&-b\-c&a\end{pmatrix}.

For a 3×33\times3 matrix, evaluate the determinant by a signed cofactor expansion. Find the inverse either by M−1=adj⁡(M)/det⁡MM^{-1}=\operatorname{adj}(M)/\det M (cofactor matrix, then transpose) or by row-reducing [M∣I][M\mid I] to [I∣M−1][I\mid M^{-1}]. Finish by multiplying back to I.

ForForM=\begin{pmatrix}1&1&0\0&1&1\0&0&2\end{pmatrix},\quad \det M=2,soMisnon−singular.Rowreductiongivesso M is non-singular. Row reduction givesM^{-1}=\begin{pmatrix}1&-1&1/2\0&1&-1/2\0&0&1/2\end{pmatrix},and direct multiplication gives $MM^{-1}=I$.

Do not take reciprocals entry by entry. A negative determinant still permits an inverse; only zero makes the matrix singular.

Undo every matrix factor in reverse order

Fornon−singularAandB,For non-singular A and B,(AB)^{-1}=B^{-1}A^{-1},becausebecause(AB)(B^{-1}A^{-1})=A(BB^{-1})A^{-1}=I.Moregenerally,More generally,(A_1A_2\cdots A_k)^{-1}=A_k^{-1}\cdots A_2^{-1}A_1^{-1}.

A product applies the rightmost action first. Its inverse must undo the last applied action first, so every factor and the whole order reverse. This is not a commutativity shortcut; associativity lets the neighbouring inverse pairs cancel.

LetLetA=\begin{pmatrix}1&2\0&1\end{pmatrix},\quad B=\begin{pmatrix}3&0\0&1\end{pmatrix},\quad AB=\begin{pmatrix}3&2\0&1\end{pmatrix}. ThenThenB^{-1}A^{-1}=\begin{pmatrix}1/3&0\0&1\end{pmatrix}\begin{pmatrix}1&-2\0&1\end{pmatrix}=\begin{pmatrix}1/3&-2/3\0&1\end{pmatrix}=(AB)^{-1}.

The rule requires square non-singular factors. Do not write A−1B−1A^{-1}B^{-1} unless that product independently happens to equal the correct reverse-order result.

Read a 2D transformation from its basis images and product order

For column vectors, the columns of a 2×22\times2 matrix M are the images of (1,0)T(1,0)^T and (0,1)T(0,1)^T. This gives a direct way to build or identify the transformation, including unfamiliar linear transformations.

Transformation about the origin Matrix
Rotation anticlockwise by θ\theta (cos⁡θ−sin⁡θsin⁡θcos⁡θ)\begin{pmatrix}\cos\theta&-\sin\theta\\\sin\theta&\cos\theta\end{pmatrix}
Reflection in x-axis / line y=xy=x (100−1)\begin{pmatrix}1&0\\0&-1\end{pmatrix} / (0110)\begin{pmatrix}0&1\\1&0\end{pmatrix}
Enlargement factor k kIkI
Stretch parallel to x-axis, factor k (k001)\begin{pmatrix}k&0\\0&1\end{pmatrix}
Shear x′=x+kyx'=x+ky (1k01)\begin{pmatrix}1&k\\0&1\end{pmatrix}

If B acts first and A second, the composite matrix is AB. For a stretch parallel to the x-axis by 3 followed by reflection in $y=x$,A=\begin{pmatrix}0&1\1&0\end{pmatrix},\quad B=\begin{pmatrix}3&0\0&1\end{pmatrix},\quad AB=\begin{pmatrix}0&1\3&0\end{pmatrix}.

If M is non-singular, M−1M^{-1} represents the inverse transformation. The area scale factor is ∣det⁡M∣|\det M|; a negative determinant records orientation reversal, not negative area. A zero determinant collapses area and has no inverse transformation.

Products act right-to-left on column vectors: AB means B then A. A stretch is directional, whereas an enlargement scales every direction equally.

Separate fixed points from invariant lines

Object Meaning under M Equation
Invariant (fixed) point vv The point itself does not move Mv=vMv=v, so (M−I)v=0(M-I)v=0
Invariant line Every point of the line maps somewhere on the same line For a line through the origin, its direction v satisfies Mv=λvMv=\lambda v for some scalar λ\lambda
Line of invariant points Every point on the line is fixed Mv=vMv=v for every direction/point on it, so the relevant eigenvalue is 1

Every linear matrix transformation fixes the origin. Non-zero fixed points exist only when M−IM-I is singular. Solve the simultaneous equations from (M−I)(x,y)T=0(M-I)(x,y)^T=0 and describe the full set, which may be just the origin or an entire line of fixed points.

For $M=\begin{pmatrix}a&b\\c&d\end{pmatrix}$ and a candidate line $y=mx$, the direction $(1,m)^T$ maps to $(a+bm,c+dm)^T$. Invariance requires parallel directions:c+dm=m(a+bm). Also check a vertical line separately using direction $(0,1)^T$.

ForForM=\begin{pmatrix}4&-1\2&1\end{pmatrix},2+m=m(4-m)\iff m^2-3m+2=0,so $m=1$ or $m=2$. Thus $y=x$ and $y=2x$ are invariant lines through the origin; points on them are generally scaled, not fixed.

Do not impose Mv=vMv=v when the question asks only for an invariant line: that would find fixed points and miss directions that remain on the same line while being stretched or reversed.