GATE EC Engineering Mathematics Formula Sheet

Organised by the official GATE 2027 EC syllabus, Section 1: Engineering Mathematics. Each worked example was checked in code. Practise these: GATE EC Engineering Mathematics previous year questions.

Linear algebra

Syllabus: Vector space, basis, linear dependence and independence, matrix algebra, eigen values and eigen vectors, rank, solution of linear equations: existence and uniqueness.

Eigenvalues, trace and determinant

∑iλi=tr⁡(A),∏iλi=det⁡(A)\sum_i \lambda_i = \operatorname{tr}(A), \qquad \prod_i \lambda_i = \det(A)
∑iλi=tr⁡(A),∏iλi=det⁡(A)\begin{aligned} \sum_i \lambda_i &= \operatorname{tr}(A), \\ \prod_i \lambda_i &= \det(A) \end{aligned}
Symbols
  • λi\lambda_i: eigenvalues of the square matrix A
  • tr⁡(A)\operatorname{tr}(A): sum of the diagonal entries
Applies when
Any square matrix, counting repeated eigenvalues as often as they repeat. For a 2 × 2 matrix the eigenvalues are the roots of λ² − tr(A) λ + det(A) = 0.
Common mistake
Adding the eigenvalues to get the determinant. The sum gives the trace; the product gives the determinant.
Worked example
A = [[4, 1], [2, 3]]: the larger eigenvalue: 5 (the other is 2; 5 + 2 = 7 = tr A, 5 × 2 = 10 = det A)

Existence and uniqueness of solutions

ρ(A)=ρ([A∣b])=n⇒unique,ρ(A)=ρ([A∣b])<n⇒infinitely many,ρ(A)<ρ([A∣b])⇒none\rho(A) = \rho([A \mid b]) = n \Rightarrow \text{unique}, \quad \rho(A) = \rho([A \mid b]) < n \Rightarrow \text{infinitely many}, \quad \rho(A) < \rho([A \mid b]) \Rightarrow \text{none}
ρ(A)=ρ([A∣b])=n:unique solutionρ(A)=ρ([A∣b])<n:infinitely manyρ(A)<ρ([A∣b]):none\begin{gathered} \rho(A) = \rho([A \mid b]) = n: \\ \text{unique solution} \\ \rho(A) = \rho([A \mid b]) < n: \\ \text{infinitely many} \\ \rho(A) < \rho([A \mid b]): \text{none} \end{gathered}
Symbols
  • ρ(A)\rho(A): rank of the coefficient matrix
  • [A∣b][A \mid b]: the augmented matrix
  • nn: number of unknowns
Applies when
Any system Ax = b. With infinitely many solutions, n − ρ(A) unknowns are free.
Common mistake
Concluding 'no solution' from det(A) = 0. A singular A means no unique solution: compare the two ranks to tell 'none' from 'infinitely many'.
Worked example
x + 2y + z = 4, 2x + 4y + 2z = 8, x − y = 1: free unknowns: 1 (ρ(A) = ρ([A | b]) = 2 < 3)

Cayley-Hamilton theorem

A2−tr⁡(A) A+det⁡(A) I=0⇒A−1=tr⁡(A) I−Adet⁡(A)  (2×2)A^2 - \operatorname{tr}(A)\,A + \det(A)\,I = 0 \quad\Rightarrow\quad A^{-1} = \frac{\operatorname{tr}(A)\,I - A}{\det(A)}\ \ (2 \times 2)
A2−tr⁡(A) A+det⁡(A) I=0⇒  A−1=tr⁡(A) I−Adet⁡(A)  (2×2)\begin{aligned} &A^2 \\ &\qquad - \operatorname{tr}(A)\,A \\ &\qquad + \det(A)\,I \\ &\quad = 0 \\ \Rightarrow\; \\ A^{-1} &= \frac{\operatorname{tr}(A)\,I - A}{\det(A)}\ \ (2 \times 2) \end{aligned}
Symbols
  • AA: a square matrix
  • II: the identity matrix
Applies when
Every square matrix satisfies its own characteristic equation; the inverse form needs det(A) ≠ 0.
Common mistake
Substituting a number for λ and forgetting the I: the constant term becomes det(A) times the identity.
Worked example
A = [[2, 1], [1, 3]]: the top-left entry of A⁻¹: 0.6 (tr = 5, det = 5)

Calculus

Syllabus: Mean value theorems, theorems of integral calculus, evaluation of definite and improper integrals, partial derivatives, maxima and minima, multiple integrals, line, surface and volume integrals, Taylor series.

Lagrange's mean value theorem

f′(c)=f(b)−f(a)b−a  for some c∈(a,b)f'(c) = \frac{f(b) - f(a)}{b - a}\ \ \text{for some } c \in (a, b)
f′(c)=f(b)−f(a)b−afor some c∈(a,b)\begin{gathered} f'(c) = \frac{f(b) - f(a)}{b - a} \\ \text{for some } c \in (a, b) \end{gathered}
Symbols
  • ff: continuous on [a, b], differentiable on (a, b)
  • cc: a point strictly between a and b
Applies when
f continuous on the closed interval and differentiable on the open one. Rolle's theorem is the case f(a) = f(b).
Common mistake
Picking a c at an end point. The theorem promises a c strictly inside the interval.
Worked example
f(x) = x³ on [0, 3]: 1.732 (c = √3: f′(c) = 3c² = 9 = (27 − 0)/3)

Taylor series

f(x)=∑n=0∞f(n)(a)n! (x−a)nf(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!}\,(x - a)^n
Symbols
  • aa: the point the series is expanded about
  • f(n)(a)f^{(n)}(a): the n-th derivative at a
Applies when
f infinitely differentiable near a, and x inside the radius of convergence. With a = 0 it is the Maclaurin series.
Common mistake
Dropping the n! in the denominator, which only matters from the x² term on, so first-order checks don't catch it.
Worked example
e^0.2 from the first three terms about 0: 1.22 (the exact value is 1.2214)

Gamma integral

∫0∞xne−ax dx=n!an+1,∫−∞∞e−x2 dx=π\int_0^{\infty} x^{n} e^{-ax}\, dx = \frac{n!}{a^{n+1}}, \qquad \int_{-\infty}^{\infty} e^{-x^2}\, dx = \sqrt{\pi}
∫0∞xne−ax dx=n!an+1,∫−∞∞e−x2 dx=π\begin{aligned} &\int_0^{\infty} x^{n} e^{-ax}\, dx \\ &\quad = \frac{n!}{a^{n+1}}, \\ &\int_{-\infty}^{\infty} e^{-x^2}\, dx \\ &\quad = \sqrt{\pi} \end{aligned}
Symbols
  • nn: a non-negative integer
  • aa: a positive constant
Applies when
a > 0, so the integral converges. For non-integer n, n! becomes Γ(n + 1).
Common mistake
Writing aⁿ instead of aⁿ⁺¹ in the denominator.
Worked example
∫₀^∞ x² e^(−3x) dx: 0.0741 (2/27)

Maxima and minima of f(x, y)

fx=fy=0;rt−s2>0 with r<0⇒maximum, r>0⇒minimum;rt−s2<0⇒saddlef_x = f_y = 0;\quad rt - s^2 > 0 \text{ with } r < 0 \Rightarrow \text{maximum},\ r > 0 \Rightarrow \text{minimum};\quad rt - s^2 < 0 \Rightarrow \text{saddle}
fx=fy=0;rt−s2>0 with r<0⇒  maximum, r>0⇒  minimum;rt−s2<0⇒  saddle\begin{aligned} f_x &= f_y \\ &= 0; \\ rt - s^2 &> 0 \text{ with } r \\ &< 0 \\ &\Rightarrow\; \text{maximum},\ r \\ &\quad > 0 \\ \Rightarrow\; \text{minimum}; \\ rt - s^2 &< 0 \\ \Rightarrow\; \text{saddle} \end{aligned}
Symbols
  • r,s,tr, s, t: fxxf_{xx}, fxyf_{xy} and fyyf_{yy} at the stationary point
Applies when
f has continuous second partial derivatives. rt − s² = 0 is inconclusive.
Common mistake
Deciding from r alone. Check rt − s² first: a saddle can have r > 0.
Worked example
f = x² + xy + y² − 3x: the minimum value: -3 (at x = 2, y = −1; r = 2, s = 1, t = 2)

Differential equations

Syllabus: Linear differential equations, Euler-Cauchy equation, nonhomogeneous equations, methods of solution using variation of parameters, complementary function and particular integral, partial differential equations, variable separable method, initial and boundary value problems.

First-order linear equation

y′+P(x) y=Q(x)  ⇒  y e∫P dx=∫Q e∫P dx dx+cy' + P(x)\,y = Q(x) \;\Rightarrow\; y\,e^{\int P\,dx} = \int Q\,e^{\int P\,dx}\, dx + c
y′+P(x) y=Q(x)⇒  y e∫P dx=∫Q e∫P dx dx+c\begin{aligned} y' + P(x)\,y &= Q(x) \\ &\Rightarrow\; y\,e^{\int P\,dx} \\ &\quad = \int Q\,e^{\int P\,dx}\, dx \\ &\qquad + c \end{aligned}
Symbols
  • e∫P dxe^{\int P\,dx}: the integrating factor
  • cc: constant fixed by the initial condition
Applies when
The equation is linear in y and y′, with y′ having coefficient 1.
Common mistake
Using the integrating factor before dividing through by the coefficient of y′.
Worked example
y′ + 2y = 4, y(0) = 0: y(1): 1.729 (y = 2(1 − e^(−2x)))

Second-order constant-coefficient equation

ay′′+by′+cy=0:am2+bm+c=0;y=c1em1x+c2em2x  (m1≠m2)ay'' + by' + cy = 0:\quad am^2 + bm + c = 0;\quad y = c_1 e^{m_1 x} + c_2 e^{m_2 x}\ \ (m_1 \ne m_2)
ay′′+by′+cy=0am2+bm+c=0y=c1em1x+c2em2x (m1≠m2)\begin{gathered} ay'' + by' + cy = 0 \\ am^2 + bm + c = 0 \\ y = c_1 e^{m_1 x} + c_2 e^{m_2 x}\ (m_1 \ne m_2) \end{gathered}
Symbols
  • m1,m2m_1, m_2: roots of the auxiliary equation
Applies when
Real distinct roots as shown; a repeated root m gives (c₁ + c₂x)e^(mx), and roots α ± jβ give e^(αx)(c₁ cos βx + c₂ sin βx).
Common mistake
Writing c₁e^(mx) + c₂e^(mx) for a repeated root, which is only one independent solution.
Worked example
y″ + 3y′ + 2y = 0, y(0) = 1, y′(0) = 0: y(1): 0.6004 (y = 2e^(−x) − e^(−2x))

Euler-Cauchy equation

x2y′′+axy′+by=0:y=xm,m2+(a−1)m+b=0x^2 y'' + a x y' + b y = 0:\quad y = x^m,\quad m^2 + (a - 1)m + b = 0
x2y′′+axy′+by=0:y=xm,m2+(a−1)m+b=0\begin{aligned} &x^2 y'' + a x y' + b y \\ &\quad = 0: \\ y &= x^m, \\ &m^2 + (a - 1)m + b \\ &\quad = 0 \end{aligned}
Symbols
  • a,ba, b: constants
  • mm: roots of the indicial equation
Applies when
x > 0. The substitution x = e^t turns it into a constant-coefficient equation in t.
Common mistake
Using m² + am + b = 0, which leaves out the −m from the y″ term.
Worked example
x²y″ − 2xy′ + 2y = 0: the larger root: 2 (m² − 3m + 2 = 0, so y = c₁x + c₂x²)

Vector analysis

Syllabus: Vectors in plane and space, vector operations, gradient, divergence and curl, Gauss's, Green's and Stokes' theorems.

Directional derivative

Da^f=∇f⋅a^,max⁡Da^f=∣∇f∣D_{\hat a} f = \nabla f \cdot \hat a, \qquad \max D_{\hat a} f = |\nabla f|
Da^f=∇f⋅a^,max⁡Da^f=∣∇f∣\begin{aligned} D_{\hat a} f &= \nabla f \cdot \hat a, \\ \max D_{\hat a} f &= |\nabla f| \end{aligned}
Symbols
  • ∇f\nabla f: gradient of f
  • a^\hat a: unit vector in the chosen direction
Applies when
f differentiable at the point. The rate is largest along ∇f itself.
Common mistake
Dotting with the direction vector before making it a unit vector.
Worked example
f = x²y at (1, 2), towards (3, 4): 3.2 (∇f = (4, 1); (12 + 4)/5)

Gauss's divergence theorem

∮SF⋅dS=∫V(∇⋅F) dV\oint_S \mathbf{F} \cdot d\mathbf{S} = \int_V (\nabla \cdot \mathbf{F})\, dV
Symbols
  • SS: closed surface, outward normal
  • VV: the volume it encloses
Applies when
F has continuous first derivatives throughout V.
Common mistake
Using an inward normal, which flips the sign.
Worked example
F = (x², y, z) over the unit cube [0, 1]³: 3 (∇·F = 2x + 2, integrated over the cube)

Green's theorem in the plane

∮C(P dx+Q dy)=∬R(∂Q∂x−∂P∂y)dx dy\oint_C (P\,dx + Q\,dy) = \iint_R \left(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\right) dx\,dy
∮C(P dx+Q dy)=∬R(∂Q∂x−∂P∂y)dx dy\begin{aligned} &\oint_C (P\,dx + Q\,dy) \\ &\quad = \iint_R \left(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\right) dx\,dy \end{aligned}
Symbols
  • CC: closed curve, traversed counter-clockwise
  • RR: the region inside C
Applies when
P and Q with continuous partial derivatives on R. Stokes' theorem is the same idea on a surface in space.
Common mistake
Swapping the order to ∂P/∂y − ∂Q/∂x, which flips the sign.
Worked example
P = −y, Q = x round the circle of radius 2: 25.13 (2 × the area, 8π)

Complex analysis

Syllabus: Analytic functions, Cauchy's integral theorem, Cauchy's integral formula, sequences, series, convergence tests, Taylor and Laurent series, residue theorem.

Cauchy-Riemann equations

f=u+jv analytic  ⟺  ∂u∂x=∂v∂y,∂u∂y=−∂v∂xf = u + jv \text{ analytic} \iff \frac{\partial u}{\partial x} = \frac{\partial v}{\partial y},\quad \frac{\partial u}{\partial y} = -\frac{\partial v}{\partial x}
f=u+jv analytic  ⟺  ∂u∂x=∂v∂y,∂u∂y=−∂v∂x\begin{aligned} f &= u \\ &\quad + jv \text{ analytic} \iff \frac{\partial u}{\partial x} \\ &= \frac{\partial v}{\partial y}, \\ \frac{\partial u}{\partial y} &= -\frac{\partial v}{\partial x} \end{aligned}
Symbols
  • u,vu, v: real and imaginary parts of f(x + jy)
Applies when
Partial derivatives continuous in the region. Then u and v are both harmonic.
Common mistake
Getting the sign on the second equation wrong: it is uyu_{y} = −vxv_{x}.
Worked example
u = x² − y²: its harmonic conjugate v at (1, 2): 4 (v = 2xy, from f = z²)

Cauchy's integral formula

∮Cf(z)z−a dz=2πj f(a),∮Cf(z)(z−a)n+1 dz=2πjn! f(n)(a)\oint_C \frac{f(z)}{z - a}\, dz = 2\pi j\, f(a), \qquad \oint_C \frac{f(z)}{(z - a)^{n+1}}\, dz = \frac{2\pi j}{n!}\, f^{(n)}(a)
∮Cf(z)z−a dz=2πj f(a),∮Cf(z)(z−a)n+1 dz=2πjn! f(n)(a)\begin{aligned} &\oint_C \frac{f(z)}{z - a}\, dz \\ &\quad = 2\pi j\, f(a), \\ &\oint_C \frac{f(z)}{(z - a)^{n+1}}\, dz \\ &\quad = \frac{2\pi j}{n!}\, f^{(n)}(a) \end{aligned}
Symbols
  • CC: closed contour, counter-clockwise, enclosing a
  • ff: analytic on and inside C
Applies when
f analytic on and inside C, and a inside C. If a is outside C the integral is 0 (Cauchy's theorem).
Common mistake
Applying it when a lies outside the contour.
Worked example
∮ e^z / z³ dz round |z| = 1: the imaginary part: 3.1416 (2πj/2! × e⁰ = πj)

Residue theorem

∮Cf(z) dz=2πj∑Res⁡,Res⁡z=aN(z)D(z)=N(a)D′(a) (simple pole)\oint_C f(z)\, dz = 2\pi j \sum \operatorname{Res}, \qquad \operatorname{Res}_{z=a} \frac{N(z)}{D(z)} = \frac{N(a)}{D'(a)}\ \text{(simple pole)}
∮Cf(z) dz=2πj∑Res⁡Res⁡z=aN(z)D(z)=N(a)D′(a)(simple pole)\begin{gathered} \oint_C f(z)\, dz = 2\pi j \sum \operatorname{Res} \\ \operatorname{Res}_{z=a} \frac{N(z)}{D(z)} = \frac{N(a)}{D'(a)} \\ \text{(simple pole)} \end{gathered}
Symbols
  • Res⁡\operatorname{Res}: residues at the poles inside C
  • N,DN, D: numerator and denominator, D(a) = 0, N(a) ≠ 0
Applies when
f analytic inside C except at isolated poles, none on C. Count only the poles inside.
Common mistake
Adding the residues of poles outside the contour.
Worked example
∮ dz/(z² + 1) round |z − j| = 1: 3.1416 (only z = j is inside; residue 1/(2j), so π)

Probability and statistics

Syllabus: Mean, median, mode, standard deviation, combinatorial probability, probability distributions, binomial distribution, Poisson distribution, exponential distribution, normal distribution, joint and conditional probability, correlation and regression analysis.

Binomial distribution

P(X=k)=(nk)pk(1−p)n−k,E[X]=np,Var⁡(X)=np(1−p)P(X = k) = \binom{n}{k} p^k (1 - p)^{n-k}, \qquad E[X] = np, \qquad \operatorname{Var}(X) = np(1 - p)
P(X=k)=(nk)pk(1−p)n−k,E[X]=np,Var⁡(X)=np(1−p)\begin{aligned} P(X = k) &= \binom{n}{k} p^k (1 - p)^{n-k}, \\ E[X] &= np, \\ \operatorname{Var}(X) &= np(1 - p) \end{aligned}
Symbols
  • nn: number of independent trials
  • pp: probability of success in each
  • kk: number of successes
Applies when
Fixed n, independent trials, the same p every time.
Common mistake
Leaving out the C(n, k) factor, which counts the orders the successes can come in.
Worked example
n = 5, p = 0.5: P(X = 2): 0.3125 (10/32)

Poisson distribution

P(X=k)=λke−λk!,E[X]=Var⁡(X)=λP(X = k) = \frac{\lambda^k e^{-\lambda}}{k!}, \qquad E[X] = \operatorname{Var}(X) = \lambda
P(X=k)=λke−λk!,E[X]=Var⁡(X)=λ\begin{aligned} P(X = k) &= \frac{\lambda^k e^{-\lambda}}{k!}, \\ E[X] &= \operatorname{Var}(X) \\ &= \lambda \end{aligned}
Symbols
  • λ\lambda: mean number of events in the interval
Applies when
Events independent and at a constant average rate; also the limit of the binomial with n large, p small and np = λ.
Common mistake
Using the rate per second as λ for a 10 s interval. Scale λ to the interval first.
Worked example
λ = 2: P(X = 0): 0.1353 (e^(−2))

Exponential distribution

f(x)=λe−λx (x≥0),P(X>t)=e−λt,E[X]=1λ, Var⁡(X)=1λ2f(x) = \lambda e^{-\lambda x}\ (x \ge 0), \qquad P(X > t) = e^{-\lambda t}, \qquad E[X] = \frac{1}{\lambda},\ \operatorname{Var}(X) = \frac{1}{\lambda^2}
f(x)=λe−λx (x≥0),P(X>t)=e−λt,E[X]=1λ, Var⁡(X)=1λ2\begin{aligned} f(x) &= \lambda e^{-\lambda x}\ (x \ge 0), \\ P(X > t) &= e^{-\lambda t}, \\ E[X] &= \frac{1}{\lambda},\ \operatorname{Var}(X) \\ &= \frac{1}{\lambda^2} \end{aligned}
Symbols
  • λ\lambda: rate (1/unit time)
Applies when
Waiting times between Poisson events. Memoryless: P(X > s + t | X > s) = P(X > t).
Common mistake
Taking the mean as λ instead of 1/λ.
Worked example
λ = 0.5 per hour: P(no event in 2 hours): 0.3679 (e^(−1))

Normal distribution

f(x)=1σ2π e−(x−μ)2/2σ2,z=x−μσf(x) = \frac{1}{\sigma\sqrt{2\pi}}\, e^{-(x - \mu)^2 / 2\sigma^2}, \qquad z = \frac{x - \mu}{\sigma}
f(x)=1σ2π e−(x−μ)2/2σ2,z=x−μσ\begin{aligned} f(x) &= \frac{1}{\sigma\sqrt{2\pi}}\, e^{-(x - \mu)^2 / 2\sigma^2}, \\ z &= \frac{x - \mu}{\sigma} \end{aligned}
Symbols
  • μ\mu: mean
  • σ\sigma: standard deviation
  • zz: standardised value
Applies when
About 68%, 95% and 99.7% of the probability lies within 1, 2 and 3 σ of the mean.
Common mistake
Standardising with the variance instead of the standard deviation.
Worked example
P(X ≤ μ + σ): 0.8413

Bayes' theorem

P(A∣B)=P(B∣A) P(A)P(B),P(B)=∑iP(B∣Ai) P(Ai)P(A \mid B) = \frac{P(B \mid A)\,P(A)}{P(B)}, \qquad P(B) = \sum_i P(B \mid A_i)\,P(A_i)
P(A∣B)=P(B∣A) P(A)P(B)P(B)=∑iP(B∣Ai) P(Ai)\begin{gathered} P(A \mid B) = \frac{P(B \mid A)\,P(A)}{P(B)} \\ P(B) = \sum_i P(B \mid A_i)\,P(A_i) \end{gathered}
Symbols
  • AiA_i: mutually exclusive, exhaustive events
  • P(A)P(A): the prior
  • P(A∣B)P(A \mid B): the posterior
Applies when
P(B) > 0. The total-probability sum needs the AiA_{i} to cover every case without overlap.
Common mistake
Confusing P(A | B) with P(B | A): a test that is 99% sensitive does not make a positive result 99% reliable.
Worked example
1% have a fault; a test catches 99% of faults and flags 5% of good units: P(fault | flagged): 0.1667

Correlation and regression

r=cov⁡(x,y)σxσy,y−yˉ=r σyσx (x−xˉ)r = \frac{\operatorname{cov}(x, y)}{\sigma_x \sigma_y}, \qquad y - \bar y = r\,\frac{\sigma_y}{\sigma_x}\,(x - \bar x)
r=cov⁡(x,y)σxσy,y−yˉ=r σyσx (x−xˉ)\begin{aligned} r &= \frac{\operatorname{cov}(x, y)}{\sigma_x \sigma_y}, \\ y - \bar y &= r\,\frac{\sigma_y}{\sigma_x}\,(x - \bar x) \end{aligned}
Symbols
  • rr: correlation coefficient, between −1 and 1
  • xˉ,yˉ\bar x, \bar y: means
  • σx,σy\sigma_x, \sigma_y: standard deviations
Applies when
The regression line of y on x, fitted by least squares. The line of x on y uses r σx/σyr\,\sigma_x/\sigma_y instead.
Common mistake
Using the y-on-x line to predict x from y. The two regression lines are different unless |r| = 1.
Worked example
x = 1, 2, 3, 4 and y = 2, 4, 5, 8: the slope of y on x: 1.9

Practise these: GATE EC Engineering Mathematics previous year questions

Source for the section headings: the official GATE 2027 EC syllabus, IIT Madras. The formulas, examples and notes are Studyloaf's own.

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