Determinant, transpose and inverse – Matrices and Determinants of Order 2 (Senior 4 Mathematics) | Ganzaa

Matrices and Determinants of Order 2

Lessons

Senior 4 · Mathematics

Matrices and Determinants of Order 2

5 lessons

Lesson 4 of 5

Determinant, transpose and inverse

25 min

How much does a matrix stretch area?

The matrix $M = \begin{pmatrix} 3 & 1 \\ 1 & 2 \end{pmatrix}$ sends the unit square to the parallelogram built on its columns $(3, 1)$ and $(1, 2)$ . The dashed rectangle around the parallelogram measures $4 \times 3 = 12$ . Cut away the two triangles of area $\frac{1}{2}(3)(1)$ , the two of area $\frac{1}{2}(1)(2)$ and the two corner unit squares, and what is left is

12 - 3 - 2 - 2 = 5 = 3 \times 2 - 1 \times 1 .

Every region of the plane has its area multiplied by the same factor 5, because $M$ maps the whole grid of unit squares to a grid of identical parallelograms. The number $3 \times 2 - 1 \times 1$ is the determinant of $M$ . It also answers a second question: can the transformation be undone? If the determinant were 0, the plane would be flattened onto a line, and no transformation could rebuild it.

Definition

Determinant of order 2: for $A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$ ,

\det A = \begin{vmatrix} a & b \\ c & d \end{vmatrix} = ad - bc .

(Straight bars mean the determinant, a number; round brackets mean the matrix.) $|\det A|$ is the factor by which the transformation multiplies areas. A negative determinant means the transformation also reverses orientation, as a reflection does.

Example: $\begin{vmatrix} 3 & 1 \\ 2 & 2 \end{vmatrix} = 6 - 2 = 4$ ; $\begin{vmatrix} 1 & 0 \\ 0 & -1 \end{vmatrix} = -1$ (a reflection keeps areas but flips the plane); $\begin{vmatrix} 2 & -3 \\ 4 & -6 \end{vmatrix} = -12 + 12 = 0$ .

Properties of the determinant

Property	Why
$\det I = 1$	$1 \times 1 - 0 \times 0$ ; the identity changes no area.
$\det(AB) = \det A \cdot \det B$	Doing $B$ then $A$ multiplies areas by $\det B$ , then by $\det A$ . (Expanding both sides also proves it.)
$\det(kA) = k^2 \det A$	Each of the two columns is multiplied by $k$ : $(k a) (k d) - (k b) (k c) = k^{2} (a d - b c)$ .
$\det A^{\mathsf{T}} = \det A$	$a d - c b = a d - b c$ (the transpose is defined below).
Swapping the two columns (or rows) changes the sign	$\begin{vmatrix} b & a \\ d & c \end{vmatrix} = bc - ad$ .
$\det A = 0 \iff$ the columns are linearly dependent	This is the test $a d - b c = 0$ from the unit on vector spaces.

Worked example 1: using the properties

$A = \begin{pmatrix} 2 & 1 \\ 1 & 3 \end{pmatrix}$ , $B = \begin{pmatrix} 1 & -1 \\ 0 & 2 \end{pmatrix}$ . (a) Find $\det(AB)$ without multiplying the matrices. (b) Find $\det(3A)$ . (c) A triangle of area $6\,\mathrm{cm^2}$ is mapped by $A$ . Find the area of its image.

Solution. $\det A = 6 - 1 = 5$ and $\det B = 2 - 0 = 2$ .

(a) $\det(AB) = 5 \times 2 = 10$ . (Check: $AB = \begin{pmatrix} 2 & 0 \\ 1 & 5 \end{pmatrix}$ , whose determinant is $10$ .) ✓

(b) $\det(3A) = 3^2 \times 5 = 45$ , not $15$ .

Definition

Transpose: the transpose $A^{\mathsf{T}}$ of a matrix $A$ is obtained by writing its rows as columns: $(A^{\mathsf{T}})_{ij} = a_{ji}$ . For order 2,

\begin{pmatrix} a & b \\ c & d \end{pmatrix}^{\mathsf{T}} = \begin{pmatrix} a & c \\ b & d \end{pmatrix}.

Properties: $(A^{\mathsf{T}})^{\mathsf{T}} = A$ , $(A + B)^{\mathsf{T}} = A^{\mathsf{T}} + B^{\mathsf{T}}$ , $(kA)^{\mathsf{T}} = kA^{\mathsf{T}}$ and, with the order reversed, $(AB)^{\mathsf{T}} = B^{\mathsf{T}}A^{\mathsf{T}}$ . A matrix with $A^{\mathsf{T}} = A$ is symmetric.

Example: $\begin{pmatrix} 1 & 4 \\ -2 & 0 \end{pmatrix}^{\mathsf{T}} = \begin{pmatrix} 1 & -2 \\ 4 & 0 \end{pmatrix}$ ; $\begin{pmatrix} 5 & 7 \\ 7 & 1 \end{pmatrix}$ is symmetric.

Definition

Inverse matrix: a square matrix $A$ is invertible (non-singular) if there is a matrix $A^{- 1}$ with

AA^{-1} = A^{-1}A = I .

A matrix with no inverse is singular. If $A$ is the matrix of $f$ , then $A^{- 1}$ is the matrix of the inverse transformation $f^{- 1}$ , which undoes $f$ .

Example: the rotation $R = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$ has inverse $R^{-1} = \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}$ , the quarter-turn the other way: $RR^{-1} = I$ .

Important

Theorem (inverse of a $2 \times 2$ matrix). $A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$ is invertible if and only if $\det A \ne 0$ , and then

A^{-1} = \frac{1}{ad - bc}\begin{pmatrix} d & -b \\ -c & a \end{pmatrix}.

Proof. Multiply out:

\begin{pmatrix} a & b \\ c & d \end{pmatrix}\begin{pmatrix} d & -b \\ -c & a \end{pmatrix} = \begin{pmatrix} ad - bc & 0 \\ 0 & ad - bc \end{pmatrix} = (\det A)\,I,

and the product in the other order gives the same. If $\det A \ne 0$ , divide by it. If $\det A = 0$ and some $B$ had $A B = I$ , then $\det A \cdot \det B = \det I = 1$ , which is impossible. $\blacksquare$

In words: swap the diagonal entries, change the sign of the other two, divide by the determinant.

Worked example 2: computing inverses

Find the inverse, if it exists, of (a) $A = \begin{pmatrix} 5 & 3 \\ 3 & 2 \end{pmatrix}$ (b) $B = \begin{pmatrix} 4 & 2 \\ 3 & 1 \end{pmatrix}$ (c) $C = \begin{pmatrix} 2 & -3 \\ 4 & -6 \end{pmatrix}$ .

Solution.

(a) $\det A = 10 - 9 = 1$ , so $A^{-1} = \begin{pmatrix} 2 & -3 \\ -3 & 5 \end{pmatrix}$ . Check: $\begin{pmatrix} 5 & 3 \\ 3 & 2 \end{pmatrix}\begin{pmatrix} 2 & -3 \\ -3 & 5 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$ . ✓

(b) $\det B = 4 - 6 = -2$ , so $B^{-1} = -\frac{1}{2}\begin{pmatrix} 1 & -2 \\ -3 & 4 \end{pmatrix} = \begin{pmatrix} -0.5 & 1 \\ 1.5 & -2 \end{pmatrix}$ .

(c) $\det C = -12 + 12 = 0$ : $C$ is singular and has no inverse. Its columns $(2, 4)$ and $(- 3, - 6)$ are parallel, so $C$ squashes the plane onto the line $y = 2 x$ .

Worked example 3: when is a matrix singular? Solving $A X = B$

(a) For which $k$ is $\begin{pmatrix} k & 3 \\ 4 & k - 4 \end{pmatrix}$ singular? (b) Solve $A X = B$ for $A = \begin{pmatrix} 2 & 1 \\ 5 & 3 \end{pmatrix}$ , $B = \begin{pmatrix} 1 & 0 \\ 2 & 4 \end{pmatrix}$ .

Solution. (a) $k (k - 4) - 12 = k^{2} - 4 k - 12 = (k - 6) (k + 2) = 0$ : singular for $k = 6$ and $k = - 2$ .

(b) $\det A = 6 - 5 = 1$ , $A^{-1} = \begin{pmatrix} 3 & -1 \\ -5 & 2 \end{pmatrix}$ . Multiply both sides on the left by $A^{- 1}$ : $X = A^{-1}B = \begin{pmatrix} 3 - 2 & 0 - 4 \\ -5 + 4 & 0 + 8 \end{pmatrix} = \begin{pmatrix} 1 & -4 \\ -1 & 8 \end{pmatrix}$ .

Seki Takakazu (died 1708), in an ink drawing kept by a family in Ichinoseki, Japan. In a work of 1683 Seki studied determinants, about ten years before Leibniz used them independently in Europe. Drawing: unknown artist, public domain, via Wikimedia Commons.

Warning

Order matters for inverses. $(AB)^{-1} = B^{-1}A^{-1}$ , because to undo "first $B$ , then $A$ " you undo $A$ first. Likewise $A X = B$ gives $X = A^{-1}B$ , while $X A = B$ gives $X = BA^{-1}$ ; these are usually different. And never "divide" by a matrix: $\frac{B}{A}$ has no meaning.

Tip

For the inverse, say "swap, negate, divide": swap $a$ and $d$ , negate $b$ and $c$ , divide by $a d - b c$ . Then check one entry of $AA^{-1}$ : it takes ten seconds and catches most slips.

The Organic Chemistry Tutor: Inverse of a 2x2 matrix (about 10 min). Watch for the swap-and-negate pattern and for the check that the product is the identity.

✦

Practice

Work each one, then open the answers.

Evaluate (a) $\begin{vmatrix} 7 & -2 \\ 3 & 4 \end{vmatrix}$ (b) $\begin{vmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{vmatrix}$ (c) $\begin{vmatrix} x & x^2 \\ 1 & x \end{vmatrix}$ .
$\det A = -3$ and $\det B = 4$ for $2 \times 2$ matrices. Find $\det(AB)$ , $\det(2A)$ , $\det(A^{\mathsf{T}})$ and $\det(A^{-1})$ .
Find the inverse of (a) $\begin{pmatrix} 2 & 1 \\ 5 & 3 \end{pmatrix}$ (b) $\begin{pmatrix} 6 & 4 \\ 2 & 1 \end{pmatrix}$ .
Show that $M = \begin{pmatrix} 3 & -2 \\ 4 & -3 \end{pmatrix}$ is its own inverse.
Find every $k$ for which $\begin{pmatrix} k + 1 & 2 \\ 6 & k \end{pmatrix}$ is singular.
A square of area $9\,\mathrm{cm^2}$ is transformed by $\begin{pmatrix} 4 & -1 \\ 2 & 3 \end{pmatrix}$ . Find the area of the image.
For $A = \begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix}$ and $B = \begin{pmatrix} 3 & 0 \\ 1 & 1 \end{pmatrix}$ , verify that $(AB)^{\mathsf{T}} = B^{\mathsf{T}}A^{\mathsf{T}}$ .
Solve $X A = B$ for $A = \begin{pmatrix} 2 & 1 \\ 5 & 3 \end{pmatrix}$ and $B = \begin{pmatrix} 1 & 0 \\ 0 & 2 \end{pmatrix}$ .

Show answersHide answers

(a) $28 + 6 = 34$ (b) $\cos^2\theta + \sin^2\theta = 1$ (c) $x^{2} - x^{2} = 0$ .
$\det(AB) = -12$ ; $\det(2A) = 4(-3) = -12$ ; $\det A^{\mathsf{T}} = -3$ ; $\det A^{-1} = -\frac{1}{3}$ .
(a) $\det = 1$ : $\begin{pmatrix} 3 & -1 \\ -5 & 2 \end{pmatrix}$ (b) $\det = 6 - 8 = -2$ : $\begin{pmatrix} -0.5 & 2 \\ 1 & -3 \end{pmatrix}$ .
$M^2 = \begin{pmatrix} 9 - 8 & -6 + 6 \\ 12 - 12 & -8 + 9 \end{pmatrix} = I$ , so $M^{-1} = M$ .
$(k + 1) k - 12 = k^{2} + k - 12 = (k + 4) (k - 3) = 0$ : $k = 3$ or $k = - 4$ .
$\det = 12 + 2 = 14$ ; area $9 \times 14 = 126\,\mathrm{cm^2}$ .
$AB = \begin{pmatrix} 5 & 2 \\ 1 & 1 \end{pmatrix}$ , so $(AB)^{\mathsf{T}} = \begin{pmatrix} 5 & 1 \\ 2 & 1 \end{pmatrix}$ ; $B^{\mathsf{T}}A^{\mathsf{T}} = \begin{pmatrix} 3 & 1 \\ 0 & 1 \end{pmatrix}\begin{pmatrix} 1 & 0 \\ 2 & 1 \end{pmatrix} = \begin{pmatrix} 5 & 1 \\ 2 & 1 \end{pmatrix}$ . ✓
$X = BA^{-1} = \begin{pmatrix} 1 & 0 \\ 0 & 2 \end{pmatrix}\begin{pmatrix} 3 & -1 \\ -5 & 2 \end{pmatrix} = \begin{pmatrix} 3 & -1 \\ -10 & 4 \end{pmatrix}$ .

Quick check

Key points

$\det\begin{pmatrix} a & b \\ c & d \end{pmatrix} = ad - bc$ : the area scale factor, with a sign for orientation.
$\det(AB) = \det A \det B$ , $\det(kA) = k^2\det A$ , $\det A^{\mathsf{T}} = \det A$ .
$A^{\mathsf{T}}$ turns rows into columns; $(AB)^{\mathsf{T}} = B^{\mathsf{T}}A^{\mathsf{T}}$ .
$A^{-1} = \frac{1}{ad - bc}\begin{pmatrix} d & -b \\ -c & a \end{pmatrix}$ exists exactly when $\det A \ne 0$ ; $(AB)^{-1} = B^{-1}A^{-1}$ .

Sources and credits

Written by Ganzaa: the explanations, the diagram, the worked examples, the practice items and the questions. The unit follows the topic list of REB's S4 Core Mathematics (2026–2027); no REB text is reproduced.
History: Takakazu Seki Kowa and Matrices and determinants, J. J. O'Connor and E. F. Robertson, MacTutor History of Mathematics, University of St Andrews.
Drawing: Seki Kowa, unknown artist, Wikimedia Commons, public domain.
Video: Inverse of a 2x2 Matrix, The Organic Chemistry Tutor, embedded from YouTube.

1How much does a matrix stretch area?

2Properties of the determinant

How much does a matrix stretch area?

Properties of the determinant