GATE GUIDE

GATE Engineering Mathematics Formula Sheet for ECE, EE, ME and CE

By MD ANISH AHAMADUpdated 9 Oct 20269 min read
GATE Engineering Mathematics Formula Sheet for ECE, EE, ME and CE

Engineering Mathematics is the one section the ECE, EE, ME and Civil papers share, and it mostly tests whether you know a few dozen results exactly. This sheet gives them group by group, each with a formula table, a short worked example and the trap that costs marks. If you write the CSE paper, the CSE engineering mathematics formula sheet covers the discrete mathematics that paper adds.

In this guide
  1. Key takeaways
  2. The words this sheet uses
  3. How many marks Engineering Mathematics carries in each paper
  4. Linear algebra: determinants, rank and eigenvalues
  5. Calculus: limits, series, extrema and integrals
  6. Vector calculus: gradient, divergence, curl and the three theorems
  7. Differential equations: the solution forms to know
  8. Complex variables: Cauchy–Riemann, Cauchy's formula and residues
  9. Probability and statistics: distributions and Bayes
  10. Numerical methods: Newton–Raphson, trapezoidal and Simpson's rules
  11. Quick revision

Key takeaways

The words this sheet uses

How many marks Engineering Mathematics carries in each paper

Each branch book counted every question of its papers against the official answer keys, placing a borderline question where it leans. The totals are:

Paper Papers counted Engineering Mathematics marks
ME 2022 (both sets) to 2026, six papers 13 in every paper
CE 2019 to 2026, two sets a year, sixteen papers 13 in twelve; range 10 to 15
EE 2022 to 2026, five papers 15, 13, 13, 15, 18
ECE 2010, 2019, 2020, 2025, 2026 10, 13, 13, 15, 9

So 13 marks is the settled figure for ME and CE, EE has run higher, and ECE has swung either side. These are past counts, not a forecast; check the current notification. For the branch subjects themselves, use the ECE, EE, ME and Civil formula sheets.

Remember: The four syllabi differ in detail. EE files vector calculus under Calculus, ECE lists Vector Analysis on its own, and only ME and CE name numerical methods. Read Section 1 of your own paper's syllabus before you revise.

Linear algebra: determinants, rank and eigenvalues

Most matrix questions need two eigenvalue checks and one rank comparison. In the table, AA is n×nn \times n.

Result Formula
Scaled matrix det⁡(kA)=kndet⁡A\det(kA) = k^{n}\det A
Product and adjoint det⁡(AB)=det⁡A⋅det⁡B\det(AB) = \det A \cdot \det B; det⁡(adj⁡A)=(det⁡A)n−1\det(\operatorname{adj} A) = (\det A)^{n-1}
Eigenvalue checks ∑λi=tr⁡A\sum \lambda_i = \operatorname{tr} A; ∏λi=det⁡A\prod \lambda_i = \det A
Spectral mapping AkA^{k} has λk\lambda^{k}; A−1A^{-1} has 1λ\displaystyle \frac{1}{\lambda}; A+cIA + cI has λ+c\lambda + c
Special matrices triangular: the diagonal entries; real symmetric: real eigenvalues; orthogonal: ∣λ∣=1\vert \lambda \vert = 1
Cayley–Hamilton AA satisfies its own characteristic equation
Unique solution rank⁡A=rank⁡[A∣b]=n\operatorname{rank} A = \operatorname{rank}[A \mid \mathbf{b}] = n
Infinitely many solutions rank⁡A=rank⁡[A∣b]<n\operatorname{rank} A = \operatorname{rank}[A \mid \mathbf{b}] < n
No solution rank⁡A<rank⁡[A∣b]\operatorname{rank} A < \operatorname{rank}[A \mid \mathbf{b}]
Homogeneous Ax=0A\mathbf{x} = \mathbf{0} always consistent, with n−rank⁡An - \operatorname{rank} A free parameters

Worked example. Take A=[3113]A = \begin{bmatrix} 3 & 1 \\ 1 & 3 \end{bmatrix}. Its trace is 6 and its determinant is 8, so λ2−6λ+8=0\lambda^{2} - 6\lambda + 8 = 0 and the eigenvalues are 2 and 4. By Cayley–Hamilton, A2−6A+8I=0A^{2} - 6A + 8I = 0. Multiply through by A−1A^{-1}:

A−6I+8A−1=0A−1=18(6I−A)=18[3−1−13]\begin{aligned} A - 6I + 8A^{-1} &= 0 \\ A^{-1} &= \frac{1}{8}(6I - A) = \frac{1}{8}\begin{bmatrix} 3 & -1 \\ -1 & 3 \end{bmatrix} \end{aligned}

Trap: A zero determinant only tells you the solution is not unique. The system may then have no solution or infinitely many, and a question with a parameter usually wants one of the two. Compare rank⁡A\operatorname{rank} A with the rank of the augmented matrix to decide.

Calculus: limits, series, extrema and integrals

Know these results by heart and the working stays short.

Result Formula
Standard limits sin⁡xx→1\displaystyle \frac{\sin x}{x} \to 1, 1−cos⁡xx2→12\displaystyle \frac{1 - \cos x}{x^{2}} \to \frac{1}{2}, (1+x)1/x→e(1 + x)^{1/x} \to e as x→0x \to 0
L'Hôpital for 00\displaystyle \frac{0}{0} or ∞∞\displaystyle \frac{\infty}{\infty} only: lim⁡fg=lim⁡f′g′\displaystyle \lim \frac{f}{g} = \lim \frac{f'}{g'}
1∞1^{\infty} form lim⁡fg=elim⁡g (f−1)\lim f^{g} = e^{\lim g\,(f - 1)}
Taylor series f(a+h)=f(a)+hf′(a)+h22!f′′(a)+⋯\displaystyle f(a + h) = f(a) + h f'(a) + \frac{h^{2}}{2!} f''(a) + \cdots
Common series sin⁡x=x−x33!+⋯\displaystyle \sin x = x - \frac{x^{3}}{3!} + \cdots; ln⁡(1+x)=x−x22+x33−⋯\displaystyle \ln(1 + x) = x - \frac{x^{2}}{2} + \frac{x^{3}}{3} - \cdots
Extrema, one variable f′=0f' = 0; f′′>0f'' > 0 minimum, f′′<0f'' < 0 maximum; on [a,b][a, b] also compare the end values
Extrema, two variables r=fxxr = f_{xx}, s=fxys = f_{xy}, t=fyyt = f_{yy}; rt−s2>0rt - s^{2} > 0 gives a minimum if r>0r > 0, a maximum if r<0r < 0; rt−s2<0rt - s^{2} < 0 is a saddle
Gamma integral ∫0∞xne−ax dx=n!an+1\displaystyle \int_0^{\infty} x^{n} e^{-ax}\,dx = \frac{n!}{a^{n+1}}; Γ(12)=π\displaystyle \Gamma(\tfrac{1}{2}) = \sqrt{\pi}
Symmetry ∫−aaf dx=0\int_{-a}^{a} f\,dx = 0 for odd ff, and 2∫0af dx2\int_0^{a} f\,dx for even ff
Improper integral ∫1∞dxxp\displaystyle \int_1^{\infty} \frac{dx}{x^{p}} converges only for p>1p > 1
Leibniz rule ddx∫a(x)b(x)f(t) dt=f(b(x)) b′(x)−f(a(x)) a′(x)\displaystyle \frac{d}{dx}\int_{a(x)}^{b(x)} f(t)\,dt = f(b(x))\,b'(x) - f(a(x))\,a'(x)

A Taylor series often settles a limit faster than repeated L'Hôpital. Since x−sin⁡x=x36−⋯\displaystyle x - \sin x = \frac{x^{3}}{6} - \cdots, the ratio x−sin⁡xx3\displaystyle \frac{x - \sin x}{x^{3}} tends to 16\displaystyle \frac{1}{6}. The Leibniz rule is just as quick: ddx∫0x2sin⁡t dt=2xsin⁡(x2)\displaystyle \frac{d}{dx}\int_0^{x^{2}} \sin t\,dt = 2x\sin(x^{2}).

Worked example. Find the extrema of f=x3+y3−3xyf = x^{3} + y^{3} - 3xy. Setting fx=3x2−3y=0f_x = 3x^{2} - 3y = 0 and fy=3y2−3x=0f_y = 3y^{2} - 3x = 0 gives the real points (0,0)(0, 0) and (1,1)(1, 1). Then:

(1,1):r=6, s=−3, t=6, rt−s2=27>0, r>0  ⇒  minimum f=−1(0,0):r=0, s=−3, t=0, rt−s2=−9<0  ⇒  saddle point\begin{aligned} (1, 1):\quad & r = 6,\ s = -3,\ t = 6,\ rt - s^{2} = 27 > 0,\ r > 0 \;\Rightarrow\; \text{minimum } f = -1 \\ (0, 0):\quad & r = 0,\ s = -3,\ t = 0,\ rt - s^{2} = -9 < 0 \;\Rightarrow\; \text{saddle point} \end{aligned}

Trap. Check the form before using L'Hôpital. As x→0x \to 0, 1+cos⁡x1+x\displaystyle \frac{1 + \cos x}{1 + x} is simply 21=2\displaystyle \frac{2}{1} = 2. Differentiating top and bottom gives −sin⁡x1→0\displaystyle \frac{-\sin x}{1} \to 0, which is wrong, because the limit was never 00\displaystyle \frac{0}{0}.

Vector calculus: gradient, divergence, curl and the three theorems

The three integral theorems trade a hard integral for an easier one, given the right orientation and a closed boundary.

Result Formula
Gradient ∇ϕ=(ϕx,ϕy,ϕz)\nabla\phi = (\phi_x, \phi_y, \phi_z), normal to the surface ϕ=c\phi = c
Directional derivative ∇ϕ⋅u^\nabla\phi \cdot \hat{u} with u^\hat{u} a unit vector; greatest value ∣∇ϕ∣\vert \nabla\phi \vert
Divergence ∇⋅F=∂F1∂x+∂F2∂y+∂F3∂z\displaystyle \nabla \cdot \mathbf{F} = \frac{\partial F_1}{\partial x} + \frac{\partial F_2}{\partial y} + \frac{\partial F_3}{\partial z}; solenoidal if zero
Curl ∇×F=0\nabla \times \mathbf{F} = \mathbf{0} means irrotational, and then F=∇ϕ\mathbf{F} = \nabla\phi
Identities ∇×(∇ϕ)=0\nabla \times (\nabla\phi) = \mathbf{0}; ∇⋅(∇×F)=0\nabla \cdot (\nabla \times \mathbf{F}) = 0; ∇⋅r=3\nabla \cdot \mathbf{r} = 3
Green ∮C(M dx+N dy)=∬R(∂N∂x−∂M∂y)dx dy\displaystyle \oint_C (M\,dx + N\,dy) = \iint_R \left(\frac{\partial N}{\partial x} - \frac{\partial M}{\partial y}\right) dx\,dy, anticlockwise
Stokes ∮CF⋅dr=∬S(∇×F)⋅n^ dS\oint_C \mathbf{F} \cdot d\mathbf{r} = \iint_S (\nabla \times \mathbf{F}) \cdot \hat{n}\,dS
Gauss (divergence) ∬SF⋅n^ dS=∭V∇⋅F dV\iint_S \mathbf{F} \cdot \hat{n}\,dS = \iiint_V \nabla \cdot \mathbf{F}\,dV, for a closed surface SS

Worked example. Find the outward flux of F=(x2,y2,z2)\mathbf{F} = (x^{2}, y^{2}, z^{2}) through the surface of the unit cube 0≤x,y,z≤10 \le x, y, z \le 1. Instead of six face integrals, use Gauss. Here ∇⋅F=2x+2y+2z\nabla \cdot \mathbf{F} = 2x + 2y + 2z, and each of xx, yy, zz averages 12\displaystyle \frac{1}{2} over the cube:

∭V(2x+2y+2z) dV=2(12+12+12)=3\iiint_V (2x + 2y + 2z)\,dV = 2\left(\tfrac{1}{2} + \tfrac{1}{2} + \tfrac{1}{2}\right) = 3

Trap. Make the direction a unit vector first. For ϕ=x2y+z\phi = x^{2}y + z at (1,2,0)(1, 2, 0), ∇ϕ=(4,1,1)\nabla\phi = (4, 1, 1). Along (1,2,2)(1, 2, 2) the dot product is 8, but the unit vector is 13(1,2,2)\displaystyle \frac{1}{3}(1, 2, 2), so the directional derivative is 83\displaystyle \frac{8}{3}.

Differential equations: the solution forms to know

Recognise the standard form, and the solution follows a fixed recipe.

Equation or case Solution rule
Linear first order y′+P(x) y=Q(x)y' + P(x)\,y = Q(x) IF=e∫P dx\text{IF} = e^{\int P\,dx}; y⋅IF=∫Q⋅IF dx+Cy \cdot \text{IF} = \int Q \cdot \text{IF}\,dx + C
Exact M dx+N dy=0M\,dx + N\,dy = 0 exact only if ∂M∂y=∂N∂x\displaystyle \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}
Constant coefficients, distinct real roots C1em1x+C2em2xC_1 e^{m_1 x} + C_2 e^{m_2 x}
Repeated root mm (C1+C2x) emx(C_1 + C_2 x)\,e^{mx}
Complex roots α±iβ\alpha \pm i\beta eαx(C1cos⁡βx+C2sin⁡βx)e^{\alpha x}(C_1 \cos\beta x + C_2 \sin\beta x)
Cauchy–Euler x2y′′+axy′+by=0x^{2}y'' + axy' + by = 0 try y=xmy = x^{m}: m(m−1)+am+b=0m(m - 1) + am + b = 0
Particular integral for eaxe^{ax} eaxf(a)\displaystyle \frac{e^{ax}}{f(a)}, or x eaxf′(a)\displaystyle \frac{x\,e^{ax}}{f'(a)} when f(a)=0f(a) = 0
Particular integral for sin⁡ax\sin ax, cos⁡ax\cos ax replace D2D^{2} by −a2-a^{2}
Particular integral for a polynomial expand 1f(D)\displaystyle \frac{1}{f(D)} in powers of DD
General solution complementary function plus particular integral

Worked example. Solve y′′−3y′+2y=exy'' - 3y' + 2y = e^{x}. Here f(D)=D2−3D+2f(D) = D^{2} - 3D + 2, with roots 1 and 2. Since f(1)=0f(1) = 0, the plain rule fails and you use the derivative form:

CF=C1ex+C2e2xf′(D)=2D−3,f′(1)=−1PI=x exf′(1)=−x exy=C1ex+C2e2x−x ex\begin{aligned} \text{CF} &= C_1 e^{x} + C_2 e^{2x} \\ f'(D) = 2D - 3, \quad f'(1) &= -1 \\ \text{PI} &= \frac{x\,e^{x}}{f'(1)} = -x\,e^{x} \\ y &= C_1 e^{x} + C_2 e^{2x} - x\,e^{x} \end{aligned}

Trap. A repeated root needs its extra factor. For the Cauchy–Euler equation x2y′′−3xy′+4y=0x^{2}y'' - 3xy' + 4y = 0, the auxiliary equation m2−4m+4=0m^{2} - 4m + 4 = 0 gives m=2m = 2 twice. The solution is (C1+C2ln⁡x) x2(C_1 + C_2 \ln x)\,x^{2}: the factor is ln⁡x\ln x, not xx.

Each branch book's last-minute sheet carries the full Engineering Mathematics table for its own syllabus, including the Laplace transforms, PDE classification and Fourier series this sheet leaves out. It comes with the GATE 2027 books.

Complex variables: Cauchy–Riemann, Cauchy's formula and residues

ECE and EE write jj for −1\sqrt{-1}; the results are the same with ii. Contours run anticlockwise unless the question says otherwise.

Result Formula
Cauchy–Riemann f=u+ivf = u + iv analytic gives ux=vyu_x = v_y and uy=−vxu_y = -v_x
Derivative f′(z)=ux+i vxf'(z) = u_x + i\,v_x
Cauchy's theorem ∮Cf(z) dz=0\oint_C f(z)\,dz = 0 if ff is analytic on and inside CC
Cauchy's integral formula ∮Cf(z)z−a dz=2πi f(a)\displaystyle \oint_C \frac{f(z)}{z - a}\,dz = 2\pi i\,f(a)
Derivative form ∮Cf(z)(z−a)n+1 dz=2πin! f(n)(a)\displaystyle \oint_C \frac{f(z)}{(z - a)^{n+1}}\,dz = \frac{2\pi i}{n!}\,f^{(n)}(a)
Residue theorem ∮Cf(z) dz=2πi×\oint_C f(z)\,dz = 2\pi i \times (sum of residues inside CC)
Simple pole lim⁡z→a(z−a)f(z)\lim_{z \to a} (z - a) f(z), or p(a)q′(a)\displaystyle \frac{p(a)}{q'(a)} for f=pq\displaystyle f = \frac{p}{q}
Pole of order mm 1(m−1)!lim⁡z→adm−1dzm−1[(z−a)mf(z)]\displaystyle \frac{1}{(m - 1)!}\lim_{z \to a} \frac{d^{m-1}}{dz^{m-1}}\left[(z - a)^{m} f(z)\right]

Worked example. Evaluate the integral of f(z)=z2+1z(z−3)\displaystyle f(z) = \frac{z^{2} + 1}{z(z - 3)} round ∣z∣=2\vert z \vert = 2. The poles are z=0z = 0, inside the circle, and z=3z = 3, outside it:

Resz=0=lim⁡z→0z2+1z−3=−13∮∣z∣=2f(z) dz=2πi(−13)=−2πi3\begin{aligned} \text{Res}_{z=0} &= \lim_{z \to 0} \frac{z^{2} + 1}{z - 3} = -\frac{1}{3} \\ \oint_{\vert z \vert = 2} f(z)\,dz &= 2\pi i\left(-\frac{1}{3}\right) = -\frac{2\pi i}{3} \end{aligned}

Trap: The pole at z=3z = 3 has residue 103\displaystyle \frac{10}{3}, and adding it gives 6πi6\pi i, the wrong answer. A pole counts only when it lies strictly inside the contour, and a clockwise contour flips the sign.

Probability and statistics: distributions and Bayes

Most probability questions name a distribution and ask for a mean, a variance or one probability.

Distribution PMF or PDF Mean Variance
Binomial(n,p)(n, p) (nk)pk(1−p)n−k\binom{n}{k}p^{k}(1 - p)^{n-k} npnp np(1−p)np(1 - p)
Poisson(λ)(\lambda) e−λλkk!\displaystyle \frac{e^{-\lambda}\lambda^{k}}{k!} λ\lambda λ\lambda
Uniform(a,b)(a, b) 1b−a\displaystyle \frac{1}{b - a} on (a,b)(a, b) a+b2\displaystyle \frac{a + b}{2} (b−a)212\displaystyle \frac{(b - a)^{2}}{12}
Exponential(λ)(\lambda) λe−λx\lambda e^{-\lambda x}, x≥0x \ge 0 1λ\displaystyle \frac{1}{\lambda} 1λ2\displaystyle \frac{1}{\lambda^{2}}
Normal(μ,σ2)(\mu, \sigma^{2}) 1σ2πe−(x−μ)2/(2σ2)\displaystyle \frac{1}{\sigma\sqrt{2\pi}}e^{-(x - \mu)^{2}/(2\sigma^{2})} μ\mu σ2\sigma^{2}

Two rules sit beside the table: Var(X)=E[X2]−(E[X])2\text{Var}(X) = E[X^{2}] - (E[X])^{2} and Var(aX+b)=a2 Var(X)\text{Var}(aX + b) = a^{2}\,\text{Var}(X). Bayes' theorem reverses a conditional probability, with the total probability of the evidence below the line:

P(Ai∣B)=P(B∣Ai) P(Ai)∑jP(B∣Aj) P(Aj)P(A_i \mid B) = \frac{P(B \mid A_i)\,P(A_i)}{\sum_j P(B \mid A_j)\,P(A_j)}

Worked example. Supplier S1S_1 makes 70% of the parts, 1% of them defective. Supplier S2S_2 makes 30%, 4% defective. For a defective part:

P(D)=0.7×0.01+0.3×0.04=0.019P(S2∣D)=0.0120.019=1219≈0.632\begin{aligned} P(D) &= 0.7 \times 0.01 + 0.3 \times 0.04 = 0.019 \\ P(S_2 \mid D) &= \frac{0.012}{0.019} = \frac{12}{19} \approx 0.632 \end{aligned}

Trap. Rescale a Poisson rate to the interval asked. Four arrivals an hour means λ=2\lambda = 2 over 30 minutes, so P(X≥1)=1−e−2≈0.865P(X \ge 1) = 1 - e^{-2} \approx 0.865. Using λ=4\lambda = 4 gives the wrong answer.

Numerical methods: Newton–Raphson, trapezoidal and Simpson's rules

The ME and CE syllabi name numerical methods; the 2027 EC and EE syllabi do not. Questions test each rule's order as often as its value.

Method Formula Order and condition
Newton–Raphson xn+1=xn−f(xn)f′(xn)\displaystyle x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)} quadratic at a simple root; fails when f′(xn)≈0f'(x_n) \approx 0
Trapezoidal h2[y0+2(y1+⋯+yn−1)+yn]\displaystyle \frac{h}{2}\left[y_0 + 2(y_1 + \cdots + y_{n-1}) + y_n\right] error of order h2h^{2}; exact for straight lines
Simpson's 1/3 h3[y0+4(y1+y3+⋯ )+2(y2+y4+⋯ )+yn]\displaystyle \frac{h}{3}\left[y_0 + 4(y_1 + y_3 + \cdots) + 2(y_2 + y_4 + \cdots) + y_n\right] nn even; error of order h4h^{4}; exact for cubics
Euler's method yn+1=yn+h f(xn,yn)y_{n+1} = y_n + h\,f(x_n, y_n) global error of order hh

Worked example. Find 3\sqrt{3} with Newton–Raphson on f(x)=x2−3f(x) = x^{2} - 3 from x0=2x_0 = 2, then integrate x3x^{3} from 0 to 2 with h=1h = 1, whose ordinates are 0, 1 and 8:

x1=2−14=1.75,x2=1.75−0.06253.5≈1.732143Trapezoidal=12[0+2(1)+8]=5Simpson’s 13=13[0+4(1)+8]=4\begin{aligned} x_1 &= 2 - \frac{1}{4} = 1.75, \qquad x_2 = 1.75 - \frac{0.0625}{3.5} \approx 1.732143 \\ \text{Trapezoidal} &= \frac{1}{2}\left[0 + 2(1) + 8\right] = 5 \\ \text{Simpson's } \tfrac{1}{3} &= \frac{1}{3}\left[0 + 4(1) + 8\right] = 4 \end{aligned}

Two Newton steps match 3≈1.732051\sqrt{3} \approx 1.732051 to four decimal places, and Simpson's 4 is exact because the rule is exact for cubics.

In one line: Halving hh cuts the trapezoidal error to about a quarter and Simpson's error to about a sixteenth, and Simpson's 1/3 rule works only with an even number of sub-intervals.

Quick revision

  1. Eigenvalues: sum equals the trace, product equals the determinant; det⁡(kA)=kndet⁡A\det(kA) = k^{n}\det A.
  2. Unique solution only when rank⁡A=rank⁡[A∣b]=n\operatorname{rank} A = \operatorname{rank}[A \mid \mathbf{b}] = n; a zero determinant means none or infinitely many.
  3. L'Hôpital only on 00\displaystyle \frac{0}{0} or ∞∞\displaystyle \frac{\infty}{\infty}; for two variables, the sign of rt−s2rt - s^{2} decides.
  4. Leibniz: ddx∫a(x)b(x)f(t) dt=f(b) b′−f(a) a′\displaystyle \frac{d}{dx}\int_{a(x)}^{b(x)} f(t)\,dt = f(b)\,b' - f(a)\,a'.
  5. Gauss turns a closed-surface flux into ∭∇⋅F dV\iiint \nabla \cdot \mathbf{F}\,dV; a directional derivative needs a unit vector.
  6. The particular integral for eaxe^{ax} picks up a factor xx when f(a)=0f(a) = 0; a repeated Cauchy–Euler root brings ln⁡x\ln x.
  7. Only poles strictly inside the contour count; at a simple pole the residue is p(a)q′(a)\displaystyle \frac{p(a)}{q'(a)}.
  8. Binomial variance np(1−p)np(1 - p), Poisson mean and variance λ\lambda, exponential 1λ\displaystyle \frac{1}{\lambda} and 1λ2\displaystyle \frac{1}{\lambda^{2}}; trapezoidal error h2h^{2}, Simpson's h4h^{4}.

Frequently asked questions

How many marks is Engineering Mathematics worth in GATE ME, CE, EE and ECE?

In the books' own counts of past papers, ME carried 13 marks in all six papers from 2022 to 2026, and CE carried 13 in twelve of its sixteen papers from 2019 to 2026. EE ranged from 13 to 18 marks between 2022 and 2026, and ECE from 9 to 15 in its five counted papers. These are past counts, not a forecast, so check the current pattern at gate2027.iitm.ac.in.

Is the Engineering Mathematics syllabus the same for every GATE branch?

No. The core is shared: linear algebra, calculus, differential equations, complex variables, and probability and statistics. The details differ. The 2027 ME and CE syllabi name numerical methods, while the EC and EE syllabi do not, and EC lists vector analysis as a topic of its own. Read Section 1 of your own paper's official syllabus before you plan your revision.

What is the Cayley-Hamilton theorem used for in GATE?

It says every square matrix satisfies its own characteristic equation. For a 2 by 2 matrix with trace TT and determinant DD, that gives A2−TA+DI=0A^{2} - TA + DI = 0, so A−1=(TI−A)/DA^{-1} = (TI - A)/D whenever DD is not zero. The same equation lets you rewrite a high power of AA as a combination of AA and II without multiplying matrices.

What are the error orders of the trapezoidal and Simpson's rules?

The composite trapezoidal rule has a global error of order h2h^{2}, so halving the step cuts the error to about a quarter. Simpson's 1/3 rule has a global error of order h4h^{4}, so halving the step cuts it to about a sixteenth. Simpson's rule is exact for polynomials up to degree three, and it needs an even number of sub-intervals.

How do you find the residue of a function at a pole?

At a simple pole aa, the residue is the limit of (z−a)f(z)(z - a)f(z) as zz tends to aa, or p(a)/q′(a)p(a)/q'(a) when f=p/qf = p/q. At a pole of order mm, differentiate (z−a)mf(z)(z - a)^{m} f(z) m−1m - 1 times, divide by (m−1)!(m - 1)! and take the limit. Add only the residues of poles inside the contour, then multiply by 2πi2\pi i.

Sources

Dates, fees and the syllabus are set by the GATE 2027 organising institute and can change. Always confirm at gate2027.iitm.ac.in.

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