Determinants (2×2 and 3×3)

The determinant is a single number distilled from a square matrix, and it answers the most important question you can ask about that matrix: is it invertible? A non-zero determinant means yes (the matrix represents a reversible transformation and a system $A\mathbf{x} = \mathbf{b}$Ax=b has a unique solution); a zero determinant means no (the matrix is singular, collapses space, and the corresponding system fails to have a unique solution). Geometrically the determinant is a signed area or volume scale factor: it tells you both how much a transformation magnifies regions and whether it flips orientation. This lesson computes determinants of $2\times2$2×2 and $3\times3$3×3 matrices, establishes the key properties, and reads off their geometric meaning. The determinant deserves its central place: of all the numbers you can extract from a matrix, it is the one that controls invertibility, solvability and orientation simultaneously, and it threads through every remaining topic in the strand. Get the computation reliable and the interpretation secure, and the inverse, the linear-systems and the eigenvalue lessons will follow almost mechanically.

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1. Where this sits in AQA 7367

Determinants are compulsory pure content for Papers 1 and 2. The computation is AO1; the interpretation (singularity, area/volume scale factor, orientation) and the use of the multiplicative property $\det(AB) = \det A \det B$det(AB)=detAdetB are AO2/AO3. The determinant is the gateway to inverses (lesson 4: $A^{-1}$A−1 exists $\iff \det A \neq 0$⟺detA=0), to solving systems (lesson 5: unique solution $\iff \det A \neq 0$⟺detA=0), and to eigenvalues (lesson 8: $\det(A - \lambda I) = 0$det(A−λI)=0). It is one of the most reused tools in the whole matrices strand.

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2. The 2×2 determinant

For $A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$A=(ac​bd​), the determinant is

$\det(A) = |A| = ad - bc.$det(A)=∣A∣=ad−bc.

"Product of the main diagonal minus product of the off-diagonal." Two quick instances:

$$\det\begin{pmatrix} 3 & 2 \\ 1 & 5 \end{pmatrix} = 3\cdot5 - 2\cdot1 = 13, \qquad \det\begin{pmatrix} 4 & 6 \\ 2 & 3 \end{pmatrix} = 4\cdot3 - 6\cdot2 = 0.$$det(31​25​)=3⋅5−2⋅1=13,det(42​63​)=4⋅3−6⋅2=0.

The second is singular ($\det = 0$det=0) — its rows $(4,6)$(4,6) and $(2,3)$(2,3) are proportional, a visible sign of dependence. Note the notation $|A|$∣A∣ means determinant, not "absolute value"; a determinant can be negative.

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3. The 3×3 determinant by cofactor expansion

For $A = \begin{pmatrix} a & b & c \\ d & e & f \\ g & h & k \end{pmatrix}$A=​adg​beh​cfk​​, expanding along the first row:

$\det(A) = a\begin{vmatrix} e & f \\ h & k \end{vmatrix} - b\begin{vmatrix} d & f \\ g & k \end{vmatrix} + c\begin{vmatrix} d & e \\ g & h \end{vmatrix} = a(ek - fh) - b(dk - fg) + c(dh - eg).$det(A)=a​eh​fk​​−b​dg​fk​​+c​dg​eh​​=a(ek−fh)−b(dk−fg)+c(dh−eg).

Each $2\times2$2×2 determinant is the minor of the entry it multiplies — the determinant of what remains after deleting that entry's row and column. The alternating $+,-,+$+,−,+ signs come from the cofactor sign pattern:

$$\begin{pmatrix} + & - & + \\ - & + & - \\ + & - & + \end{pmatrix}.$$​+−+​−+−​+−+​​.

You may expand along any row or column using this pattern — choose the one with the most zeros to minimise work.

It is worth being precise about the vocabulary, because exam questions use it. The minor $M_{ij}$Mij​ of entry $a_{ij}$aij​ is the determinant of the $(n-1)\times(n-1)$(n−1)×(n−1) matrix obtained by deleting row $i$i and column $j$j. The cofactor $C_{ij} = (-1)^{i+j}M_{ij}$Cij​=(−1)i+jMij​ is the minor with its checkerboard sign attached. The expansion along row $i$i is then $\det A = \sum_j a_{ij}C_{ij}$detA=∑j​aij​Cij​, and along column $j$j it is $\det A = \sum_i a_{ij}C_{ij}$detA=∑i​aij​Cij​. The sign $(-1)^{i+j}$(−1)i+j is $+$+ when $i + j$i+j is even and $-$− when odd — which is exactly the checkerboard, starting with $+$+ in the top-left corner. This recursive structure (a $3\times3$3×3 reduces to three $2\times2$2×2s, a $4\times4$4×4 to four $3\times3$3×3s) is precisely how determinants of any size are defined; the cofactors will reappear in the next lesson as the building blocks of the adjugate matrix used to invert $A$A.

Worked Example 1 — 3×3 determinant, first-row expansion (with mark scheme)

Find $\det\begin{pmatrix} 2 & 1 & 3 \\ 0 & -1 & 2 \\ 1 & 4 & -1 \end{pmatrix}$det​201​1−14​32−1​​.

$$\begin{aligned} \det &= 2\big((-1)(-1) - (2)(4)\big) - 1\big((0)(-1) - (2)(1)\big) + 3\big((0)(4) - (-1)(1)\big) \\ &= 2(1 - 8) - 1(0 - 2) + 3(0 + 1) \\ &= 2(-7) - 1(-2) + 3(1) = -14 + 2 + 3 = -9. \end{aligned}$$det​=2((−1)(−1)−(2)(4))−1((0)(−1)−(2)(1))+3((0)(4)−(−1)(1))=2(1−8)−1(0−2)+3(0+1)=2(−7)−1(−2)+3(1)=−14+2+3=−9.​

Mark scheme: M1 for a correct cofactor expansion structure (three $2\times2$2×2 minors with the correct $+,-,+$+,−,+ signs); A1 for the three minors $-7,\,-2,\,1$−7,−2,1 correct; A1 for $\det = -9$det=−9. The minus sign in front of the middle term is the most common error — losing it gives $-18$−18.

Worked Example 2 — expanding along a column of zeros (with mark scheme)

Find $\det\begin{pmatrix} 1 & 2 & 0 \\ 3 & -1 & 0 \\ 2 & 5 & 4 \end{pmatrix}$det​132​2−15​004​​.

The third column is $(0, 0, 4)^{T}$(0,0,4)T, so expanding down it kills two terms. The $(3,3)$(3,3) cofactor sign is $+$+:

$$\det = 4\begin{vmatrix} 1 & 2 \\ 3 & -1 \end{vmatrix} = 4\big(1\cdot(-1) - 2\cdot3\big) = 4(-7) = -28.$$det=4​13​2−1​​=4(1⋅(−1)−2⋅3)=4(−7)=−28.

Mark scheme: M1 for choosing the column (or row) with most zeros and expanding correctly; A1 for the surviving minor $\begin{vmatrix} 1 & 2 \\ 3 & -1 \end{vmatrix} = -7$​13​2−1​​=−7; A1 for $\det = -28$det=−28. Choosing the zero-rich line is rewarded implicitly through fewer chances to slip.

Using row operations to simplify first

For a dense $3\times3$3×3 with no zeros, it is often quicker to create a zero first. Adding a multiple of one row to another does not change the determinant, so we can engineer a convenient entry. Consider

$$\det\begin{pmatrix} 2 & 3 & 1 \\ 4 & 1 & 2 \\ 1 & 5 & 3 \end{pmatrix}.$$det​241​315​123​​.

Subtract $2 \times$2× row 3 from row 1 (leaving the determinant unchanged): row 1 becomes $(2 - 2, \, 3 - 10, \, 1 - 6) = (0, -7, -5)$(2−2,3−10,1−6)=(0,−7,−5). Now expand along the new zero in column 1:

$$\det = \begin{vmatrix} 0 & -7 & -5 \\ 4 & 1 & 2 \\ 1 & 5 & 3 \end{vmatrix} = -(-7)\begin{vmatrix} 4 & 2 \\ 1 & 3 \end{vmatrix} + (-5)\begin{vmatrix} 4 & 1 \\ 1 & 5 \end{vmatrix} = 7(10) - 5(19) = 70 - 95 = -25.$$det=​041​−715​−523​​=−(−7)​41​23​​+(−5)​41​15​​=7(10)−5(19)=70−95=−25.

(Here the row-1 cofactor signs are $+, -, +$+,−,+, so the $-7$−7 entry carries a $-$− and the $-5$−5 entry a $+$+.) Direct first-row expansion of the original matrix gives the same $-25$−25 — a good consistency check, and proof that row operations are a legitimate labour-saver.

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4. Properties of determinants

Property	Statement
$\det(I) = 1$det(I)=1	The identity has determinant $1$1
$\det(AB) = \det(A)\det(B)$det(AB)=det(A)det(B)	Determinants are multiplicative
$\det(A^{T}) = \det(A)$det(AT)=det(A)	Transpose leaves the determinant unchanged
$\det(kA) = k^{n}\det(A)$det(kA)=kndet(A)	For an $n \times n$n×n matrix
Row/column swap	Multiplies the determinant by $-1$−1
Two equal rows/columns	Determinant is $0$0
$\det(A^{-1}) = \dfrac{1}{\det(A)}$det(A−1)=det(A)1​	When $A$A is invertible
$\det(A) = 0$det(A)=0	$\iff A$⟺A is singular (no inverse)

Two consequences worth internalising. First, $\det(AB) = \det A \det B$det(AB)=detAdetB means a product is singular precisely when a factor is — this is why the zero-divisor example in the multiplication lesson had two singular factors. Second, $\det(kA) = k^n \det A$det(kA)=kndetA carries the dimension $n$n as an exponent: scaling a $3\times3$3×3 matrix by $2$2 multiplies its determinant by $2^3 = 8$23=8, reflecting that volume scales with the cube of length.

The multiplicative property also pins down $\det(A^{-1})$det(A−1). Since $AA^{-1} = I$AA−1=I, taking determinants gives $\det(A)\det(A^{-1}) = \det(I) = 1$det(A)det(A−1)=det(I)=1, so $\det(A^{-1}) = 1/\det(A)$det(A−1)=1/det(A) — which already shows that an inverse can only exist when $\det A \neq 0$detA=0 (otherwise we would be dividing by zero). The same trick gives $\det(A^n) = (\det A)^n$det(An)=(detA)n for any positive integer $n$n, and indeed for negative $n$n when $A$A is invertible. These are not separate facts to memorise but quick corollaries of one identity — a theme of good Further-Maths technique is to derive rather than store.

Worked Example 3 — the multiplicative property as a shortcut (with mark scheme)

Given $\det A = 5$detA=5 and $\det B = -2$detB=−2 for $3\times3$3×3 matrices, find $\det(2A B^{T})$det(2ABT).

$$\det(2AB^{T}) = 2^{3}\det(A)\det(B^{T}) = 8 \cdot \det(A)\det(B) = 8 \cdot 5 \cdot (-2) = -80.$$det(2ABT)=23det(A)det(BT)=8⋅det(A)det(B)=8⋅5⋅(−2)=−80.

Mark scheme: M1 for using $\det(kA) = k^n\det A$det(kA)=kndetA with $n = 3$n=3; M1 for $\det(AB^{T}) = \det A\,\det B^{T}$det(ABT)=detAdetBT and $\det B^{T} = \det B$detBT=detB; A1 for $-80$−80. A frequent slip is using $2$2 rather than $2^3$23 for the scalar factor.

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5. Geometric interpretation

For a $2\times2$2×2 matrix acting as a linear transformation:

• $|\det(A)|$∣det(A)∣ is the area scale factor;
• $\det(A) > 0$det(A)>0 preserves orientation; $\det(A) < 0$det(A)<0 reverses it (a reflection is involved);
• $\det(A) = 0$det(A)=0 collapses the plane onto a line or point (area becomes $0$0).

For a $3\times3$3×3 matrix, $|\det(A)|$∣det(A)∣ is the volume scale factor, with the same sign convention for orientation.

For example, $\begin{pmatrix} 2 & 0 \\ 0 & 3 \end{pmatrix}$(20​03​) has determinant $6$6 and stretches every area by a factor of $6$6 (it doubles in $x$x, triples in $y$y). The unit square $0 \le x,y \le 1$0≤x,y≤1, area $1$1, maps to a $2\times3$2×3 rectangle, area $6$6 — exactly the determinant. A reflection such as $\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}$(10​0−1​) has determinant $-1$−1: area is preserved in size but orientation flips.

The interpretation extends to any region, not just the unit square, because a linear map scales all areas by the same factor. So if a triangle of area $7$7 is transformed by a matrix with $\det = -4$det=−4, its image has area $7 \times |{-4}| = 28$7×∣−4∣=28, with the vertices traversed in the opposite (clockwise vs anticlockwise) sense because the determinant is negative. This is genuinely useful: to find how a transformation rescales an awkward shape you need only its determinant, never the shape's own area formula.

Two standard transformations to recognise from their determinants: a pure rotation has determinant $+1$+1 (it preserves both area and orientation), and a pure reflection has determinant $-1$−1 (it preserves area but reverses orientation). A determinant with $|\det| \neq 1$∣det∣=1 therefore signals that some genuine stretching or shrinking is taking place on top of any rotation or reflection.

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6. Singular matrices

A matrix is singular when $\det(A) = 0$det(A)=0. Equivalent statements (all examined):

• $A$A has no inverse;
• the rows (equivalently columns) are linearly dependent;
• the transformation collapses a dimension (2D onto a line; 3D onto a plane, line or point);
• the system $A\mathbf{x} = \mathbf{0}$Ax=0 has non-trivial solutions.