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NDA Mathematics · Formula sheet

Matrices & Determinants formulas

28 formulas, 1 reference table and 15 common traps for NDA Mathematics Matrices & Determinants, grouped by subtopic.

Full notes with worked examples

Matrices: Order, Algebra, Powers & Equations

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Matrix multiplication and conformability

Order of a product

Am×n Bn×p=(AB)m×pA_{m\times n}\, B_{n\times p} = (AB)_{m\times p}

Transpose and its rules

Transpose rules

(AB)T=BTAT(AB)^T = B^T A^T

Powers of a matrix

A2=A⋅AAn=An−1AA^2 = A\cdot A \qquad A^n = A^{n-1}A

Matrix polynomials and equations

Cayley–Hamilton (2×2)

A2−(a+d) A+(ad−bc) I=OA^2 - (a+d)\,A + (ad - bc)\,I = O

Matrix algebra — where numbers' rules break

Non-commutative expansions

(A+B)2=A2+AB+BA+B2(A+B)(A−B)=A2−AB+BA−B2(A+B)^2 = A^2 + AB + BA + B^2 \qquad (A+B)(A-B) = A^2 - AB + BA - B^2

Common traps

Matrix multiplication is NOT commutative — AB≠BAAB \neq BA

Order matters: ABAB and BABA are generally different matrices (and one may not even exist). Never reorder factors inside a product. If a question gives ABAB, compute ABAB — answering with BABA is the classic trap.

(AB)T=BTAT(AB)^T = B^T A^T — the order REVERSES

The transpose of a product flips the factors: (AB)T=BTAT(AB)^T = B^T A^T, not ATBTA^T B^T. Sum and scalar transposes keep their order ((A+B)T=AT+BT(A+B)^T = A^T+B^T); only the PRODUCT reverses. Writing ATBTA^T B^T is the trap.

Don't import a2−b2=(a+b)(a−b)a^2 - b^2 = (a+b)(a-b) into matrices

Every number identity that secretly uses commutativity can break for matrices. (A+B)(A−B)(A+B)(A-B), (A+B)2(A+B)^2, (AB)2=A2B2(AB)^2 = A^2B^2 all FAIL unless AB=BAAB = BA. When an option assumes one of these, it's almost always the trap answer.

AB=OAB = O does NOT force A=OA = O or B=OB = O

Unlike numbers, matrices have zero divisors: two non-zero matrices can multiply to the zero matrix (e.g. (1000)(0001)=O\begin{pmatrix}1&0\\0&0\end{pmatrix}\begin{pmatrix}0&0\\0&1\end{pmatrix} = O). So you may NOT 'cancel' a matrix; concluding A=OA=O or B=OB=O from AB=OAB=O is the trap.

Special Matrices and Their Tell-Tale Properties

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Symmetric and skew-symmetric matrices

Symmetric + skew decomposition

A=12(A+AT)⏟symmetric+12(A−AT)⏟skew-symmetricA = \underbrace{\tfrac12(A + A^T)}_{\text{symmetric}} + \underbrace{\tfrac12(A - A^T)}_{\text{skew-symmetric}}

Diagonal, scalar, and identity matrices

Diagonal determinant and inverse

det⁡D=∏idi,D−1=diag⁡ ⁣(1d1,…,1dn)Dk=diag⁡(d1k,…,dnk)det⁡(kIn)=kn\det D = \prod_i d_i, \qquad D^{-1} = \operatorname{diag}\!\left(\tfrac{1}{d_1}, \dots, \tfrac{1}{d_n}\right) \qquad D^k = \operatorname{diag}(d_1^k, \dots, d_n^k) \qquad \det(kI_n) = k^n

Orthogonal matrices

Orthogonality

AAT=I  ⟹  A−1=AT,det⁡A=±1AAT=ATA=IAA^T = I \;\Longrightarrow\; A^{-1} = A^T,\quad \det A = \pm 1 \qquad AA^T = A^T A = I

Rotation matrices

Rotation composition

R(θ) R(ϕ)=R(θ+ϕ),R(θ)n=R(nθ)R(\theta)\,R(\phi) = R(\theta + \phi), \qquad R(\theta)^n = R(n\theta)

Idempotent and involutory matrices

Idempotent, involutory, all-ones, nilpotent

A2=A⇒An=AA2=I⇒A−1=AJn2=nJnAk=O (nilpotent)A^2 = A \Rightarrow A^n = A \qquad A^2 = I \Rightarrow A^{-1} = A \qquad J_n^2 = nJ_n \qquad A^k = O \text{ (nilpotent)}

The special-matrix catalog

TypeDefining propertyKey consequence
SymmetricAT=AA^T = Aaij=ajia_{ij} = a_{ji}; inverse (if any) is symmetric
Skew-symmetricAT=−AA^T = -Adiagonal all 0; odd order ⇒det⁡=0\Rightarrow \det = 0
Odd-order skew-symmetric is ALWAYS singular (det 0). Even-order need not be.
Diagonaloff-diagonal all 0det⁡=\det = product of diagonal entries
OrthogonalAAT=IAA^T = IA−1=ATA^{-1} = A^T; det⁡=±1\det = \pm 1
IdempotentA2=AA^2 = Adet⁡∈{0,1}\det \in \{0, 1\}
InvolutoryA2=IA^2 = IA−1=AA^{-1} = A; det⁡=±1\det = \pm 1
Hermitian(Aˉ)T=A(\bar{A})^T = Acomplex analogue of symmetric; diagonal entries are real; A+(Aˉ)TA+(\bar{A})^T is always Hermitian
For a REAL matrix, Hermitian == symmetric. The conjugate-transpose (Aˉ)T(\bar{A})^T is also written A∗A^{*} or A†A^{\dagger}.
Skew-Hermitian(Aˉ)T=−A(\bar{A})^T = -Acomplex analogue of skew-symmetric; diagonal entries are 0 or purely imaginary
Recognise the defining equation first; the determinant and inverse follow immediately.

Common traps

Odd-order skew-symmetric ⇒det⁡=0\Rightarrow \det = 0; diagonal entries are 0

From AT=−AA^T = -A: the diagonal satisfies aii=−aiia_{ii} = -a_{ii}, so every diagonal entry is 0. And for ODD order, det⁡A=(−1)ndet⁡A=−det⁡A\det A = (-1)^n\det A = -\det A, forcing det⁡A=0\det A = 0. Don't assume the determinant is unknown — for odd order it is always 0. (Even order need not be.)

The cross-product matrix of vv kills the outer product vvTvv^T

The 3×33\times3 skew-symmetric matrix PP with Px=v×xPx = v\times x (entries 0,−c,b; c,0,−a; −b,a,00, -c, b;\ c, 0, -a;\ -b, a, 0 for v=(a,b,c)v=(a,b,c)) and the symmetric matrix Q=vvTQ = vv^T satisfy PQ=QP=OPQ = QP = O: each column of QQ is a multiple of vv, and v×v=0v\times v = 0. A statement-check asking whether PQPQ is zero, symmetric, or equal to QPQP has all three true.

Determinants: Evaluation & Properties

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Evaluating 2×2 and 3×3 determinants

2×2 determinant

∣abcd∣=ad−bc\begin{vmatrix} a & b \\ c & d \end{vmatrix} = ad - bc

Determinant of products, scalars, and powers

Multiplicativity and scaling

det⁡(AB)=det⁡A det⁡B,det⁡(kA)=k ndet⁡Adet⁡(AT)=det⁡Adet⁡(Am)=(det⁡A)mdet⁡(A−1)=1det⁡Adet⁡(B−1AB)=det⁡A\det(AB) = \det A\,\det B, \qquad \det(kA) = k^{\,n}\det A \qquad \det(A^T) = \det A \qquad \det(A^m) = (\det A)^m \qquad \det(A^{-1}) = \frac{1}{\det A} \qquad \det(B^{-1}AB) = \det A

Core row and column properties

Core row/column properties

Ri↔Rj⇒det⁡→−det⁡two identical/proportional rows⇒det⁡=0det⁡(kRi-scaled)=kdet⁡AR_i \leftrightarrow R_j \Rightarrow \det \to -\det \qquad \text{two identical/proportional rows} \Rightarrow \det = 0 \qquad \det(kR_i\text{-scaled}) = k\det A

Singular matrices and determinant equations

Singular matrix

A singular  ⟺  ∣A∣=0∣A∣=0⇒A−1 does not existA \text{ singular} \iff |A| = 0 \qquad |A| = 0 \Rightarrow A^{-1} \text{ does not exist}

Factor-theorem and Vandermonde determinants

Vandermonde (3×3)

∣111abca2b2c2∣=(a−b)(b−c)(c−a)\begin{vmatrix}1&1&1\\a&b&c\\a^2&b^2&c^2\end{vmatrix} = (a-b)(b-c)(c-a)

Cyclic determinants

Cyclic determinant

∣abcbcacab∣=−(a3+b3+c3−3abc)\begin{vmatrix}a&b&c\\b&c&a\\c&a&b\end{vmatrix} = -(a^3+b^3+c^3-3abc)

Sums and sequences of determinants

Determinant is not additive

det⁡(A+B)≠det⁡A+det⁡B (in general)\det(A+B) \neq \det A + \det B \text{ (in general)}

Differentiating a Determinant

Derivative of a 3-row determinant

ddx∣R1R2R3∣=∣R1′R2R3∣+∣R1R2′R3∣+∣R1R2R3′∣\frac{d}{dx}\begin{vmatrix}R_1\\R_2\\R_3\end{vmatrix} = \begin{vmatrix}R_1'\\R_2\\R_3\end{vmatrix} + \begin{vmatrix}R_1\\R_2'\\R_3\end{vmatrix} + \begin{vmatrix}R_1\\R_2\\R_3'\end{vmatrix}

Binomial-Coefficient Determinants (Pascal's Identity)

Pascal's identity (the only tool needed)

(nr)+(nr+1)=(n+1r+1)\binom{n}{r} + \binom{n}{r+1} = \binom{n+1}{r+1}

Common traps

det⁡(kA)=kndet⁡A\det(kA) = k^n \det A, NOT kdet⁡Ak\det A

Pulling a scalar out of a determinant works one ROW at a time. Factoring kk from the whole n×nn\times n matrix factors it from each of the nn rows → knk^n. The single most common determinant mistake in this bank.

det⁡(I+AAT)=1+ATA\det(I+AA^T) = 1 + A^TA — don't forget the +1+1

For a column vector AA, det⁡(I+AAT)=1+ATA=1+∑ai2\det(I+AA^T)=1+A^TA=1+\sum a_i^2. The common slip is to report just ∑ai2\sum a_i^2 (dropping the +1+1), or to expand the full 3×33\times3 by hand.

det⁡(A+B)≠det⁡A+det⁡B\det(A+B) \neq \det A + \det B

Determinant is multiplicative (det⁡(AB)=det⁡A det⁡B\det(AB)=\det A\,\det B) but NOT additive. det⁡(A+B)\det(A+B) has no shortcut — you must add the matrices first, then take ONE determinant. Splitting it as det⁡A+det⁡B\det A + \det B is the trap.

It is a SUM of determinants, not one determinant with every row differentiated

ddxdet⁡≠\dfrac{d}{dx}\det\neq the determinant of all-rows-differentiated. Differentiate exactly one row per term and add. For higher derivatives, apply the rule again to each surviving term.

Don't evaluate the coefficients — use Pascal to collapse a column

Computing (94)\binom{9}{4}, (116)\binom{11}{6}, … and expanding is slow and error-prone. The whole design is that one column is the Pascal sum of two others; the right move is a single column operation.

Special Determinants: Trig, Complex, ω, Polynomial

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Determinants with complex entries

Powers of i (period 4)

i=i,i2=−1,i3=−i,i4=1i = i,\quad i^2 = -1,\quad i^3 = -i,\quad i^4 = 1

Cube-root-of-unity determinants

Cube roots of unity

ω3=1,1+ω+ω2=0\omega^3 = 1, \qquad 1 + \omega + \omega^2 = 0

Cofactors, Adjoint & Inverse

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Minors, cofactors, and expansion

Cofactor expansion vs alien cofactors

∑jaijCij=det⁡A,∑jaijCkj=0 (k≠i)\sum_j a_{ij}C_{ij} = \det A, \qquad \sum_j a_{ij}C_{kj} = 0\ (k \neq i)

The adjoint (adjugate)

Adjoint identity

A(adj⁡A)=(adj⁡A)A=∣A∣ InA(\operatorname{adj}A) = (\operatorname{adj}A)A = |A|\,I_n

Adjoint power formulas

Adjoint of an n×n matrix

∣adj⁡A∣=∣A∣ n−1,adj⁡(adj⁡A)=∣A∣ n−2A|\operatorname{adj}A| = |A|^{\,n-1}, \qquad \operatorname{adj}(\operatorname{adj}A) = |A|^{\,n-2}A

Inverse via the adjoint

Inverse formula

A−1=1∣A∣adj⁡A(∣A∣≠0)A^{-1} = \frac{1}{|A|}\operatorname{adj}A \quad (|A| \neq 0)

Inverse properties and the reversal law

Reversal law for inverses

(AB)−1=B−1A−1(AB)^{-1} = B^{-1}A^{-1}

Inverses of special matrices

diag⁡(di)−1=diag⁡(1/di)orthogonal: A−1=ATR(θ)−1=R(−θ)involutory: A−1=A\operatorname{diag}(d_i)^{-1} = \operatorname{diag}(1/d_i) \qquad \text{orthogonal: } A^{-1} = A^T \qquad R(\theta)^{-1} = R(-\theta) \qquad \text{involutory: } A^{-1} = A

Common traps

∣adj⁡A∣=∣A∣ n−1|\operatorname{adj}A| = |A|^{\,n-1}, NOT ∣A∣|A| or ∣A∣n|A|^n

The determinant of the adjoint carries the exponent n−1n-1, where nn is the ORDER. For a 3×33\times3 matrix it is ∣A∣2|A|^2 — students who answer ∣A∣|A| (forgetting the power) or ∣A∣3|A|^3 (over-counting) walk into the two standard distractors.

adj⁡(AB)=adj⁡B adj⁡A\operatorname{adj}(AB) = \operatorname{adj}B\,\operatorname{adj}A — the order REVERSES

Like transpose and inverse, the adjoint of a product flips the factors: adj⁡(AB)=adj⁡B adj⁡A\operatorname{adj}(AB) = \operatorname{adj}B\,\operatorname{adj}A, not adj⁡A adj⁡B\operatorname{adj}A\,\operatorname{adj}B. Keeping the original order is the trap.

(AB)−1=B−1A−1(AB)^{-1} = B^{-1}A^{-1}, NOT A−1B−1A^{-1}B^{-1}

The inverse of a product reverses the order of factors — the single most-missed inverse fact. It must, so that (AB)(B−1A−1)=A(BB−1)A−1=I(AB)(B^{-1}A^{-1}) = A(BB^{-1})A^{-1} = I. Writing A−1B−1A^{-1}B^{-1} is the trap.

(AT)−1=(A−1)T(A^T)^{-1} = (A^{-1})^T — don't drop the transpose

Transpose and inverse commute: (AT)−1=(A−1)T(A^T)^{-1} = (A^{-1})^T. The slip is to compute A−1A^{-1} and forget to transpose it (or vice-versa), reporting plain A−1A^{-1} as the answer.

Linear Systems: Consistency & Cramer's Rule

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Cramer's rule

x=ΔxΔ,y=ΔyΔ,z=ΔzΔx = \frac{\Delta_x}{\Delta}, \quad y = \frac{\Delta_y}{\Delta}, \quad z = \frac{\Delta_z}{\Delta}

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