GATE DA Formula Sheet 2027: Key Formulas by Subject

Q: Can I take a formula sheet into the GATE DA exam?

Plan as if you cannot. Every formula you need should be in memory before the exam, together with the condition under which it holds. The exam gives you an on-screen virtual calculator, but a calculator cannot tell you which expression to use. Confirm the current exam-day rules, including what you may carry, at gate2027.iitm.ac.in.

Q: Should I use log base 2 or the natural log in GATE DA?

It depends on the quantity. Entropy and information gain in bits use log₂. Likelihoods and the logistic regression loss use the natural logarithm ln. The two differ by a factor of ln 2 ≈ 0.693, so the wrong base gives a wrong numerical answer even when the method is right. Read which unit the question asks for.

Q: Does sample variance divide by n or n minus 1?

The unbiased sample variance divides by n - 1: s² = 1/(n-1) Σ (xᵢ - x)². Dividing by n gives the population form. GATE DA questions can ask for either, so check the wording. Also remember that Var(aX + b) = a² Var(X), which is the other variance slip that often costs marks.

Q: What is the closed-form solution of ridge regression?

Ridge regression minimises lVert y - Xβ rVert² + λ lVert β rVert², which gives β = (X^(top) X + λ I)⁻¹ X^(top) y. The matrix is invertible for every λ > 0. As λ grows, the coefficients shrink towards zero but not exactly to zero, bias rises and variance falls. Setting λ = 0 gives ordinary least squares.

By MD ANISH AHAMADUpdated 4 Oct 20268 min read

GATE DA Formula Sheet 2027: Key Formulas by Subject

This GATE DA formula sheet collects the key formulas of every technical section of the 2027 syllabus, with the trap to watch for beside each one. It is a free selection from the last-minute sheet in the GATE DA 2027 book. Use it to test your recall, then practise each formula on previous-year questions.

In this guide

Key takeaways
How to read this sheet
Probability and Statistics
Linear Algebra
Calculus and Optimisation
Programming, Data Structures and Algorithms
Database Management and Warehousing
Machine Learning
Search, Logic and Reasoning Under Uncertainty
General Aptitude essentials
How to use this sheet in the last weeks
Subject-wise formula sheets
Quick revision

Key takeaways

A formula is only safe with its condition: independence, finite variance, non-negative edge weights or an admissible heuristic.
Many lost marks come from four slips: the wrong variance, the wrong conditional, the wrong logarithm and an off-by-one range.
Probability, linear algebra and calculus feed straight into machine learning, so learn them as one block.
Python and SQL rules sit on the sheet beside formulas, because there a rule decides the answer.
You must carry every formula in memory. Confirm the current exam-day rules at gate2027.iitm.ac.in.

How to read this sheet

A formula here is a result you apply directly. Its condition is what must be true for it to hold. Write both from memory, or you have learned only half the entry.

Notation follows the book. $\ln$ is the natural logarithm; $\log_2$ is written out wherever bits are meant. $P(A \mid B)$ is the probability of $A$ given $B$ . Vectors are columns and $x^\top$ is the transpose. $n$ is the number of observations or elements unless stated.

Probability and Statistics

Formula	Watch out for
$\displaystyle P(A \mid B) = \frac{P(A \cap B)}{P(B)}$ , $P(B) > 0$	$P(A \mid B)$ is not $P(B \mid A)$
$\displaystyle P(A_i \mid B) = \frac{P(B \mid A_i)\,P(A_i)}{\sum_j P(B \mid A_j)\,P(A_j)}$	The $A_j$ must form a partition
Independent: $P(A \cap B) = P(A)\,P(B)$ ; mutually exclusive: $P(A \cap B) = 0$	Events with non-zero probability cannot be both
$\operatorname{Var}(X) = E[X^2] - (E[X])^2$ ; $\operatorname{Var}(aX + b) = a^2 \operatorname{Var}(X)$	$a^2$ , not $a$ ; the shift $b$ drops out
$\operatorname{Var}(X + Y) = \operatorname{Var}(X) + \operatorname{Var}(Y) + 2\operatorname{Cov}(X, Y)$	The covariance term vanishes only if uncorrelated
$\displaystyle \rho = \frac{\operatorname{Cov}(X, Y)}{\sigma_X \sigma_Y}$ , $-1 \le \rho \le 1$	Uncorrelated does not imply independent
$\displaystyle s^2 = \frac{1}{n - 1}\sum (x_i - \bar{x})^2$	Divide by $n$ only for the population form
$E[X] = E\big[E[X \mid Y]\big]$ ; $\operatorname{Var}(X) = E[\operatorname{Var}(X \mid Y)] + \operatorname{Var}(E[X \mid Y])$	Two terms in the total variance, not one

Distribution	Mean	Variance
$\text{Binomial}(n, p)$	$np$	$np(1 - p)$
$\text{Poisson}(\lambda)$	$\lambda$	$\lambda$
Geometric, trial of first success	$\displaystyle \frac{1}{p}$	$\displaystyle \frac{1 - p}{p^2}$
$\text{Uniform}(a, b)$	$\displaystyle \frac{a + b}{2}$	$\displaystyle \frac{(b - a)^2}{12}$
$\text{Exponential}(\lambda)$	$\displaystyle \frac{1}{\lambda}$	$\displaystyle \frac{1}{\lambda^2}$
$\chi^2$ with $k$ degrees of freedom	$k$	$2k$

For inference, the central limit theorem gives $\displaystyle \bar{X} \approx N\!\left(\mu, \frac{\sigma^2}{n}\right)$ for independent draws with finite variance. A 95 per cent interval with known $\sigma$ is $\displaystyle \bar{x} \pm 1.96\,\frac{\sigma}{\sqrt{n}}$ . With $\sigma$ unknown, use $t$ with $n - 1$ degrees of freedom. The chi-squared statistic is $\displaystyle \sum \frac{(O - E)^2}{E}$ , and a test of independence has $(r - 1)(c - 1)$ degrees of freedom.

A one-line check of your own: if $\operatorname{Var}(X) = 4$ , then $\operatorname{Var}(3X + 2) = 9 \times 4 = 36$ .

Trap: The exponential variance is $1/\lambda^2$ , not $1/\lambda$ . Only the exponential (continuous) and the geometric (discrete) are memoryless.

Linear Algebra

Formula	Watch out for
$\operatorname{rank} A + \operatorname{nullity} A = \text{number of columns}$	Columns, not rows
$Ax = b$ has a unique solution iff $\operatorname{rank} A = \operatorname{rank}[A \,\vert\, b] = n$	Equal ranks below $n$ give infinitely many
$\sum \lambda_i = \operatorname{tr} A$ , $\prod \lambda_i = \det A$ ; $2 \times 2$ : $\lambda^2 - (\operatorname{tr} A)\lambda + \det A = 0$	Count repeated eigenvalues
$A^k$ has $\lambda^k$ ; $A^{-1}$ has $\displaystyle \frac{1}{\lambda}$ ; $A + cI$ has $\lambda + c$	Eigenvectors stay the same
$\det(kA) = k^n \det A$ for $n \times n$	Not $k \det A$
$P = A(A^\top A)^{-1}A^\top$ ; $P^\top = P$ , $P^2 = P$	Needs independent columns
Idempotent: eigenvalues $0$ or $1$ , $\operatorname{rank} = \operatorname{tr}$	$I - P$ is idempotent too
Positive definite iff all $\lambda > 0$ iff all leading principal minors $> 0$	Leading minors $\ge 0$ do not prove semidefinite
$A = U\Sigma V^\top$ ; $\sigma_i^2$ are eigenvalues of $A^\top A$ ; $\lVert A \rVert_2 = \sigma_1$	Rank is the number of non-zero $\sigma_i$

Calculus and Optimisation

Standard limits as $x \to 0$ : $\displaystyle \frac{\sin x}{x} \to 1$ , $\displaystyle \frac{1 - \cos x}{x^2} \to \frac{1}{2}$ , $\displaystyle \frac{e^x - 1}{x} \to 1$ . Use L'Hôpital only for $\displaystyle \frac{0}{0}$ or $\displaystyle \frac{\infty}{\infty}$ .
Differentiable implies continuous, never the reverse: $\lvert x \rvert$ and $\max(x, 0)$ are continuous at 0 but not differentiable there.
Sigmoid: $\displaystyle \sigma(x) = \frac{1}{1 + e^{-x}}$ , $\sigma'(x) = \sigma(x)\big(1 - \sigma(x)\big)$ , largest value $\displaystyle \frac{1}{4}$ at $x = 0$ .
Taylor: $\displaystyle f(x) = \sum_{n \ge 0} \frac{f^{(n)}(a)}{n!}(x - a)^n$ ; $\displaystyle \ln(1 + x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \cdots$ for $-1 < x \le 1$ .
Second-derivative test: $f''(c) > 0$ minimum, $f''(c) < 0$ maximum, $f''(c) = 0$ no conclusion. On $[a, b]$ , compare $f(a)$ , $f(b)$ and the interior critical points.
Gradient descent $x \leftarrow x - \eta f'(x)$ on $f = ax^2$ ( $a > 0$ ) converges iff $\displaystyle 0 < \eta < \frac{1}{a}$ . Newton's step is $\displaystyle x \leftarrow x - \frac{f'(x)}{f''(x)}$ .

Programming, Data Structures and Algorithms

Formula	Watch out for
Binary search worst case: $\lfloor \log_2 n \rfloor + 1$ probes	The array must be sorted
Selection sort: always $\displaystyle \frac{n(n - 1)}{2}$ comparisons, at most $n - 1$ swaps	Same count on sorted input
Insertion sort: $n - 1$ comparisons sorted, $\displaystyle \frac{n(n - 1)}{2}$ reverse-sorted	Shifts equal the number of inversions
Merging lists of lengths $m$ and $n$ : at most $m + n - 1$ comparisons	Not $m + n$
Quicksort worst: $T(n) = T(n - 1) + \Theta(n) = \Theta(n^2)$	Sorted input with an end pivot
Chaining: expected search $1 + \alpha$ , $\displaystyle \alpha = \frac{n}{m}$ ; open addressing: at most $\displaystyle \frac{1}{1 - \alpha}$ probes unsuccessful	$\alpha$ is keys over slots
Height $h$ in edges: $h + 1 \le \text{nodes} \le 2^{h+1} - 1$	Check whether height counts edges or levels
Binary tree shapes on $n$ nodes: $\displaystyle \frac{1}{n + 1}\binom{2n}{n}$	1, 2, 5, 14, 42 for $n = 1$ to 5
Dijkstra $O((V + E)\log V)$ ; Bellman–Ford $O(VE)$ ; Floyd–Warshall $O(V^3)$	Dijkstra needs non-negative weights

Python 3 rules decide answers too: -7 // 2 is -4, -7 % 2 is 1, round(2.5) is 2, and range(a, b) stops before b.

Database Management and Warehousing

Lossless binary decomposition iff $R_1 \cap R_2 \to R_1$ or $R_1 \cap R_2 \to R_2$ . BCNF decomposition is always lossless, but not always dependency-preserving.
$\lvert R \times S \rvert = \lvert R \rvert \cdot \lvert S \rvert$ ; a natural join with no common attribute is a Cartesian product.
B+ tree internal node with $p$ pointers of $P$ bytes and $p - 1$ keys of $K$ bytes in a block of $B$ bytes: $\displaystyle p = \left\lfloor \frac{B + K}{P + K} \right\rfloor$ .
Min-max normalisation to $[a, b]$ : $\displaystyle v' = \frac{v - \min}{\max - \min}(b - a) + a$ ; z-score: $\displaystyle v' = \frac{v - \bar{v}}{\sigma}$ ; decimal scaling: $\displaystyle v' = \frac{v}{10^j}$ with the smallest $j$ that makes every $\lvert v' \rvert < 1$ .
Cuboids: $2^n$ for $n$ dimensions without hierarchies; $\prod_i (L_i + 1)$ with $L_i$ levels. CUBE forms $2^n$ groupings, ROLLUP forms $n + 1$ .
COUNT(*) counts rows, while COUNT(col) and AVG ignore NULLs.

This page is a selection. The book's last-minute sheet covers every topic in all seven technical sections, with each result's condition printed beside it. It is part of the GATE DA 2027 book, which also has 907 questions with worked solutions and 10 full mock tests.

Machine Learning

Formula	Watch out for
$\displaystyle \beta_1 = \frac{\sum (x_i - \bar{x})(y_i - \bar{y})}{\sum (x_i - \bar{x})^2}$ , $\beta_0 = \bar{y} - \beta_1 \bar{x}$	The line passes through $(\bar{x}, \bar{y})$
$\hat{\beta} = (X^\top X)^{-1}X^\top y$ ; ridge: $(X^\top X + \lambda I)^{-1}X^\top y$	Ridge shrinks, never to exactly zero
$P(y = 1 \mid x) = \sigma(w^\top x + b)$	The boundary $w^\top x + b = 0$ is linear
$P(C \mid x) \propto P(C)\prod_j P(x_j \mid C)$	Features independent given the class
SVM margin $\displaystyle = \frac{2}{\lVert w \rVert}$	Only support vectors fix $w$
$H = -\sum_k p_k \log_2 p_k$ ; $\text{Gini} = 1 - \sum_k p_k^2$	Entropy in bits uses $\log_2$
Precision $\displaystyle \frac{TP}{TP + FP}$ ; recall $\displaystyle \frac{TP}{TP + FN}$ ; $\displaystyle F_1 = \frac{2PR}{P + R}$	Do not swap FP and FN
Expected error $= \text{bias}^2 + \text{variance} + \text{noise}$	Larger $k$ in k-NN raises bias
PCA share of component $k$ : $\displaystyle \frac{\lambda_k}{\sum_j \lambda_j}$	Centre the data first

A worked line of my own: a node with 3 positive and 1 negative example has $H = -(0.75\log_2 0.75 + 0.25\log_2 0.25) \approx 0.811$ bits and Gini $= 1 - (0.5625 + 0.0625) = 0.375$ .

Remember: Layers of sizes $n_0, n_1, \ldots, n_L$ have $\sum_l n_{l-1} n_l$ weights plus $\sum_{l \ge 1} n_l$ biases. A 10–5–2 network has 60 weights and 67 parameters.

Search, Logic and Reasoning Under Uncertainty

Formula	Watch out for
BFS time and space $O(b^d)$ ; DFS time $O(b^m)$ , space $O(bm)$ ; iterative deepening time $O(b^d)$ , space $O(bd)$	$d$ is goal depth, $m$ maximum depth
A: $f(n) = g(n) + h(n)$ ; admissible $h(n) \le h^(n)$ ; consistent $h(n) \le c(n, n') + h(n')$	Tree search needs admissible, graph search consistent
Alpha-beta: prune when $\alpha \ge \beta$ ; best ordering $O(b^{m/2})$	The root value equals minimax
$p \to q \equiv \neg p \lor q \equiv \neg q \to \neg p$	The converse is not equivalent
$KB \models \alpha$ iff $KB \land \neg\alpha$ is unsatisfiable	An unsatisfiable KB entails everything
$\neg\forall x\, P(x) \equiv \exists x\, \neg P(x)$	"Some A is B" is $\exists x\,(A(x) \land B(x))$
$P(x_1, \ldots, x_n) = \prod_i P(x_i \mid \text{parents}(X_i))$	A Boolean node with $k$ Boolean parents needs $2^k$ numbers
Likelihood weighting: weight $= \prod_i P(e_i \mid \text{parents}(E_i))$	No finite sample is exact

Trap: In a collider $A \to B \leftarrow C$ , $A$ and $C$ are independent until $B$ , or any descendant of $B$ , is observed. Observing it makes them dependent.

General Aptitude essentials

General Aptitude is worth 15 marks in GATE DA. The DA papers have leaned on number sense rather than commercial arithmetic. The General Aptitude preparation guide covers the whole section.

Geometric series: $\displaystyle a + ar + ar^2 + \cdots = \frac{a}{1 - r}$ for $\lvert r \rvert < 1$ .
Exponents and logs: $a^m a^n = a^{m+n}$ , $(a^m)^n = a^{mn}$ , $\log(xy) = \log x + \log y$ , $\log x^k = k \log x$ .
A number is divisible by 3 iff its digit sum is.
An inscribed angle is half the central angle on the same arc.
Scaling one side of a rectangle by $k$ scales the area by $k$ ; scaling both scales it by $k^2$ .

How to use this sheet in the last weeks

Cover the right-hand column and recall each condition before you check. Mark every entry you miss and work one small example of your own for it. If you also know the CS syllabus, GATE CS vs GATE DA shows where the two papers overlap, so you can see which formulas you already own.

Practise your arithmetic on the GATE virtual calculator, and carry full precision until the final answer. The MCQ, MSQ and NAT marking scheme, common to every GATE paper, has negative marks only for MCQs. Check the exam-day rules before you travel.

In one line: Write the condition next to every formula, and the four classic slips have nowhere to hide.

Subject-wise formula sheets

Each of these goes deeper on one part of the GATE DA paper, with a worked example and the trap for every formula:

Quick revision

Sample variance divides by $n - 1$ , and $\operatorname{Var}(aX + b) = a^2 \operatorname{Var}(X)$ .
Write the conditional the question asks for, $P(A \mid B)$ or $P(B \mid A)$ , before substituting.
Entropy uses $\log_2$ ; the logistic loss uses $\ln$ .
$\operatorname{rank} A + \operatorname{nullity} A$ equals the number of columns.
Gradient descent on $ax^2$ converges only for $\displaystyle 0 < \eta < \frac{1}{a}$ .
Dijkstra needs non-negative weights; Bellman–Ford handles negative ones.
A* tree search needs an admissible heuristic; graph search needs a consistent one.
range(a, b) and s[i:j] stop one short of the end.

Frequently asked questions

Can I take a formula sheet into the GATE DA exam?

Plan as if you cannot. Every formula you need should be in memory before the exam, together with the condition under which it holds. The exam gives you an on-screen virtual calculator, but a calculator cannot tell you which expression to use. Confirm the current exam-day rules, including what you may carry, at gate2027.iitm.ac.in.

Should I use log base 2 or the natural log in GATE DA?

It depends on the quantity. Entropy and information gain in bits use $\log_2$ . Likelihoods and the logistic regression loss use the natural logarithm $\ln$ . The two differ by a factor of $\ln 2 \approx 0.693$ , so the wrong base gives a wrong numerical answer even when the method is right. Read which unit the question asks for.

Does sample variance divide by n or n minus 1?

The unbiased sample variance divides by $n - 1$ : $\displaystyle s^2 = \frac{1}{n-1}\sum (x_i - \bar{x})^2$ . Dividing by $n$ gives the population form. GATE DA questions can ask for either, so check the wording. Also remember that $\operatorname{Var}(aX + b) = a^2 \operatorname{Var}(X)$ , which is the other variance slip that often costs marks.

What is the closed-form solution of ridge regression?

Ridge regression minimises $\lVert y - X\beta \rVert^2 + \lambda \lVert \beta \rVert^2$ , which gives $\hat{\beta} = (X^\top X + \lambda I)^{-1} X^\top y$ . The matrix is invertible for every $\lambda > 0$ . As $\lambda$ grows, the coefficients shrink towards zero but not exactly to zero, bias rises and variance falls. Setting $\lambda = 0$ gives ordinary least squares.

When is A* search optimal?

A* tree search is optimal when the heuristic is admissible, meaning $h(n)$ never exceeds the true cost to the goal. A* graph search needs a consistent heuristic, $h(n) \le c(n, n') + h(n')$ with $h(\text{goal}) = 0$ . Every consistent heuristic is admissible. With $h = 0$ , A* becomes uniform-cost search.

Sources

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