Chapter 2 — Algebra

What is an algebraic expression?
An algebraic expression is a combination of numbers, variables and arithmetic operations (addition, subtraction, multiplication, division and powers). Variables such as x and y represent numbers that may change. Expressions are not equations — they do not include an equals sign — but they can be simplified and manipulated to form equations.

Removing brackets and using distributive law
One of the first steps in simplification is to remove brackets correctly. Use the distributive law a(b + c) = ab + ac. When a negative sign precedes a bracket, remember to change signs of all terms inside: -(x - 2) = -x + 2. For nested brackets, work from inner to outer brackets.

Collecting like terms
Like terms have exactly the same variable factors with identical powers, for example 3x^2 and -5x^2. Combine like terms by adding coefficients: 3x^2 - 5x^2 = -2x^2. Do not combine unlike terms such as x^2 and x.

Factorisation as reverse of expansion
Simplification often prepares an expression for factorisation. Spot common factors first: 6x^2 - 9x = 3x(2x - 3). For rational expressions (fractions), factor numerator and denominator fully and cancel common factors — but only after factoring, never across sums unless factored.

Handling rational expressions
When simplifying fractions with algebraic expressions, find common denominators, perform addition/subtraction of fractions by combining like terms in numerators, and simplify result by factoring. Always note domain restrictions: values making denominators zero are not allowed.

Use of substitution
Substitution (putting a number for a variable) helps verify simplification steps and check special cases. It is also helpful to test if two expressions are equivalent by trying several values of the variable.

Common pitfalls and checks
Do not cancel terms across a plus or minus sign without factoring. Watch signs when distributing negatives. After simplification, expand the result mentally or by substitution to check whether you get back the original expression; this validates steps and reduces careless mistakes.

Practice advice
Start each problem by scanning for a greatest common factor, then remove brackets and combine like terms. Recognise standard patterns such as (a + b)^2 or a^2 - b^2 which allow quick simplification. Clear presentation — showing each algebraic step — gives full credit in examinations and helps avoid sign errors.

What is a polynomial?
A polynomial is an algebraic expression made of terms of the form ax^n where a is a coefficient and n is a whole number (0,1,2,...). Polynomials include constants (degree 0), linear expressions (degree 1), quadratics (degree 2) and higher degrees. The degree of a polynomial is the highest exponent that appears with a non-zero coefficient. For example, 4x^3 - x + 7 is a polynomial of degree 3.

Basic operations
Addition and subtraction of polynomials combine like terms. Multiplying polynomials uses distributive law repeatedly: multiply each term of the first polynomial by each term of the second, then collect like terms. Division of polynomials can be done by long division or synthetic division when dividing by a linear factor; long division is useful to simplify rational expressions.

Zeros (roots) and factors
A number a is a zero or root of polynomial p(x) if p(a) = 0. If a is a root, then (x - a) is a factor. This relationship allows factorisation: knowing one root reduces degree by one when dividing the polynomial by (x - a). To find rational roots quickly, check factors of the constant term divided by factors of the leading coefficient (Rational Root Theorem).

Factorisation strategies
Begin by looking for a greatest common factor (GCF). If GCF is present, factor it out to reduce coefficients and simplify the polynomial. For polynomials with four terms try grouping: group terms pairwise and factor each pair. For quadratics look for two numbers whose product equals ac and whose sum equals b (for ax^2 + bx + c). Synthetic division is a fast method to divide by (x - r) and check for remainder zero; it works well when testing integer roots.

Special forms and patterns
Recognise perfect square trinomials like a^2 + 2ab + b^2 and difference of squares a^2 - b^2 which factorises as (a - b)(a + b). Cubic polynomials may sometimes be factored as sum/difference of cubes: a^3 ± b^3 = (a ± b)(a^2 ∓ ab + b^2). Substitutions can help: for polynomials in x^2, use t = x^2 to reduce degree and factor in t, then back-substitute.

Applications and verification
Polynomials model many algebraic relationships. After factorising, multiply the factors to verify you obtain the original polynomial — this check prevents careless errors. For repeated roots check multiplicity by trying division again. When solving equations, factorisation gives roots directly and simplifies further algebraic work.

Exam tips
Write the chosen method (GCF, grouping, quadratic formula, synthetic division) and proceed stepwise. When rational roots are tested, list candidates from factors of constant/leading coefficient. Use neat arithmetic and show checks, such as substituting roots back into the polynomial, to secure full marks.

What are algebraic identities?
Algebraic identities are equalities that hold for all values of the variables involved. They are patterns that simplify calculation, help in factorisation and provide shortcuts for expansion. Common identities are simple but powerful; recognising them inside complicated expressions saves time.

Key identities and recognition
Memorise and understand standard identities: (a + b)^2, (a - b)^2, (a + b)^3, (a - b)^3, sum and difference of cubes, and the difference of squares. For example, (a + b)^2 = a^2 + 2ab + b^2. Often exam problems rely on spotting these patterns inside larger expressions or using them in reverse to factorise.

Using identities to expand and simplify
To expand (x + 3)^2 quickly, apply identity: (x + 3)^2 = x^2 + 6x + 9. For more complex tasks, substitute a composite expression for a or b: treat (2x - 1) as a in (a + b)^2 to expand (2x - 1 + y)^2 if needed. Mastering substitution avoids long term-by-term multiplication.

Factorisation using identities
Use identities in reverse for factorisation. Recognise a^2 - b^2 to immediately write (a - b)(a + b). For cubic expressions like x^3 + y^3, apply sum of cubes factorisation x^3 + y^3 = (x + y)(x^2 - xy + y^2). These techniques reduce work in solving equations and simplifying rational expressions.

Proof by expansion
Every identity can be verified by expansion. Understanding proofs helps remember identities and know when they apply. For example, (a + b)^3 proof is by expanding (a + b)(a + b)^2. Knowing proofs improves flexibility in applying identities to unfamiliar forms.

Advanced usage and substitution
Identities can be combined: use (a + b)^2 and difference of squares together. For expressions involving symmetric terms such as x^2 + y^2 + 2xy, identity recognition gives immediate simplification. Sometimes substitution like a = x + y, b = x - y converts expression into known identity form enabling factorisation.

Exam strategy
Practise spotting identities in different disguises and always consider if expanding or factoring via identities simplifies the problem. Write the identity you use in solution to make scoring clear. Keep attention to signs and coefficients; small mistakes in sign create wrong answers. With routine practice, identities become a quick tool in problem solving.

Definition and standard form
A quadratic equation is an equation of degree two: ax^2 + bx + c = 0 where a ≠ 0. Solutions (roots) may be real or complex but in Class 10 we focus on real roots. Quadratics are central in algebra because many problems reduce to quadratic form.

Methods of solution
There are three standard methods: factorisation, completing the square and using the quadratic formula. Factorisation is quick when roots are integers; write ax^2 + bx + c = (px + q)(rx + s) and solve. Completing the square rewrites the quadratic into a perfect square plus constant and directly shows the vertex. The quadratic formula x = [-b ± sqrt(b^2 - 4ac)]/(2a) gives roots for any quadratic and is derived by completing the square.

Discriminant and its meaning
The discriminant D = b^2 - 4ac decides the nature of roots. If D > 0, two distinct real roots exist; if D = 0, there is one repeated real root (a double root); if D < 0, real roots do not exist (complex roots). In many practical problems D ≥ 0 is required for real solutions.

Sum and product relations
If α and β are roots of ax^2 + bx + c = 0 then α + β = -b/a and αβ = c/a. These relations are useful when forming a quadratic from given roots or solving symmetric equations, and they can be applied to problems asking for expressions involving roots without finding the roots explicitly.

Completing the square insight
Completing the square rewrites ax^2 + bx + c = a[(x + b/(2a))^2 - (b^2 - 4ac)/(4a^2)]. This shows the vertex and the minimum or maximum value of quadratic function y = ax^2 + bx + c. The x-coordinate of vertex is -b/(2a). Use this to solve optimization questions where quadratic models appear.

Application examples
Quadratics model projectile motion, area problems, and simple optimization tasks. Converting a word problem into a quadratic equation is a key skill: define variables clearly, write the algebraic relation, simplify, and solve by the most suitable method. Always check solutions in the original context to discard extraneous roots (for example negative lengths are not acceptable).

Exam tips
Show all steps: state discriminant, compute carefully and simplify square roots. If factorisation isn’t obvious, use the quadratic formula. When asked for vertex or axis of symmetry, compute -b/(2a) and the corresponding y-value by substitution or completing the square.

Motivation for completing the square
Completing the square is a method that transforms a quadratic ax^2 + bx + c into a square of a linear expression plus a constant: a(x - h)^2 + k. This form, called vertex or canonical form, reveals the vertex (h, k) of the parabola y = ax^2 + bx + c and makes graphing and optimisation straightforward.

Step-by-step method
For ax^2 + bx + c, start by dividing by a (so coefficient of x^2 becomes 1). For x^2 + (b/a)x + c/a, take half of (b/a), i.e. b/(2a), square it and add and subtract this square inside the expression to form a perfect square trinomial. Algebraically: ax^2 + bx + c = a[(x + b/(2a))^2 - b^2/(4a^2)] + c. Simplify constants to obtain a(x - h)^2 + k.

Deriving quadratic formula
The quadratic formula is derived by applying completing the square to ax^2 + bx + c = 0. Rearranging and completing the square gives x + b/(2a) = ±sqrt(b^2 - 4ac)/(2a) which yields x = [-b ± sqrt(b^2 - 4ac)]/(2a). Understanding this derivation helps recall the formula and its dependence on the discriminant.

Vertex interpretation
In y = a(x - h)^2 + k, the point (h, k) is the vertex; if a > 0 the parabola opens upwards and vertex is minimum value k; if a < 0 it opens downwards and vertex is maximum. The axis of symmetry is x = h. Calculate h directly by -b/(2a) and k by substituting h into the quadratic or by formula k = c - b^2/(4a).

Applications
Completing the square is used to find minimum or maximum values in optimization problems without calculus. It is also useful in converting quadratic expressions for integration or to find ranges of quadratic functions. In geometry, it helps to identify circle equation forms by completing squares in x and y.

Practical steps and mistakes to avoid
Keep fractions under control by simplifying early. When dividing by a, check that a ≠ 0. Care with signs when adding and subtracting the square term prevents common errors. After completing the square, always expand back mentally to check equivalence if unsure.

Standard form and solutions
A linear equation in two variables is written as ax + by + c = 0 where a and b are not both zero. Each such equation represents a straight line on the Cartesian plane. A solution is any ordered pair (x, y) that satisfies the equation; the set of all solutions is the set of points on the line.

Graphing methods
To graph ax + by + c = 0, find two easy solutions: set x = 0 to find y-intercept, set y = 0 to find x-intercept. Plot these two points and draw the line joining them. Alternatively rearrange to slope-intercept form y = mx + c where slope m = -a/b and c is the y-intercept; slope indicates steepness and direction.

Slope and interpretation
Slope m = rise/run = (y2 - y1)/(x2 - x1). Positive slope means line rises left to right; negative slope means it falls. Parallel lines have equal slopes; perpendicular lines have slopes that multiply to -1 (if neither is vertical).

Systems of two linear equations
A pair of linear equations in two variables corresponds to two lines. Their intersection gives common solution(s): a single intersection gives a unique solution, parallel non-coincident lines give no solution, and coincident lines give infinitely many solutions. Solve systems by substitution (replace one variable using the other equation), elimination (add equations after multiplying to cancel a variable), or graphically (approximate and visual).

Application modelling
Many word problems translate to linear equations: cost problems, mixing problems, speed-time relations. Set variables clearly, derive equations from conditions, and solve algebraically. Check units and logical plausibility of solutions (e.g., negative quantities often invalid in a real context).

Exam techniques
Show stepwise algebra and, if using graph, label axes and points. For elimination, choose multipliers that minimise arithmetic complexity. For checking, substitute the solution back into original equations to verify correctness.

Understanding system behaviour
A system of two linear equations in two unknowns can have a unique solution, no solution, or infinitely many solutions. The algebraic methods correspond to geometric cases: intersecting lines (unique), parallel lines (no solution) and coincident lines (infinitely many).

Solution techniques
Substitution: Solve one equation for a variable and substitute into the other; effective when one variable has coefficient 1 or -1. Elimination (addition/subtraction): multiply equations if necessary to create opposite coefficients for one variable, then add or subtract to eliminate that variable. This is usually fastest for exam problems.

Cramer's rule and determinants
For system ax + by = e and cx + dy = f with coefficient matrix A = [[a b],[c d]], determinant D = ad - bc. If D ≠ 0, unique solution exists and Cramer's rule gives x = D_x/D, y = D_y/D where D_x replaces first column by constants (e,f) and D_y replaces second column. Cramer's rule is handy for small systems; it fails when D = 0.

Interpretation of D = 0
If D = 0 and D_x = D_y = 0 the system has infinitely many solutions (coincident lines); if D = 0 but at least one of D_x or D_y ≠ 0 there is no solution (parallel lines). State these conditions clearly in answers to get full marks.

Word problems and modelling
Translate word problems (mixture, age, distance-time) into linear equations by defining variables and forming equations from given relationships. Solve and interpret results in context (e.g., non-negative ages). For mixtures, use proportions; for cost problems equate total costs; for speed problems use distance = speed × time.

Exam tips
Write variables definitions, show formation of equations, then solve by chosen method. If using Cramer's rule, compute determinants stepwise and comment on D value. Check solutions by substitution back into original equations to confirm correctness.

Basic ideas and symbols
Inequalities compare two expressions using symbols <, >, ≤, ≥. They describe ranges of acceptable values rather than precise points. Solving inequalities produces intervals or unions of intervals as answers. Many algebraic rules mirror those for equations, but with one important caveat: multiplying or dividing both sides by a negative number reverses the inequality symbol.

Solving linear inequalities
To solve a linear inequality such as ax + b < 0, isolate x by usual algebraic steps while remembering the sign reversal rule for negative multiplication/division. Represent the result on a number line: use an open circle for strict inequalities (< or >) and a closed circle for non-strict (≤ or ≥). For compound inequalities like a < x ≤ b, solve the pair of inequalities together and show the intersection as the solution interval.

Absolute value inequalities
Absolute value measures distance from zero. |x - a| < b means x lies within distance b of a, so -b < x - a < b, which simplifies to a - b < x < a + b. For |x - a| ≥ b the solution is x ≤ a - b or x ≥ a + b (a union of two intervals). Solve absolute inequalities by splitting into the appropriate cases and solve linear inequalities in each case. Check endpoints according to strictness.

Quadratic inequalities
For ax^2 + bx + c < 0 or > 0, first find roots of ax^2 + bx + c = 0. Denote roots r1 and r2 with r1 ≤ r2. The sign of the quadratic on intervals (-∞, r1), (r1, r2), and (r2, ∞) is determined by the leading coefficient a and by testing any point in each interval. If a > 0 the parabola opens upwards: expression < 0 between roots (r1, r2) and ≥ 0 outside. If a < 0 reverse this. Represent solution intervals accordingly, and remember to include endpoints only when inequality is non-strict and the corresponding root satisfies equality.

Rational inequalities
To solve inequalities involving fractions, bring all terms to one side and form a single rational expression. Factor numerator and denominator completely. The critical points are zeros of numerator and denominator. Construct a sign table across intervals determined by these critical points, remembering that values making denominator zero are excluded even if numerator also zero. Choose test points in each interval to check sign and combine intervals where sign meets inequality requirement.

Inequalities in two variables and graphical interpretation
An inequality in x and y, such as y ≥ mx + c, represents a half-plane bounded by the line y = mx + c. To solve systems of linear inequalities, graph each boundary line (solid for ≥ or ≤, dashed for > or <) and shade the region satisfying each inequality; the solution set is the intersection (overlap) of shaded regions. This method is useful for feasibility questions and forms the starting idea for linear programming.

Practical tips and checking
Always list critical points and open/closed status of each. Use number-line sketches or sign tables to show reasoning; examiners award marks for correct method even if arithmetic slips. Check border cases: substitute endpoint values into the original inequality to decide inclusion. For word problems, interpret interval solutions in context (for example ensure physical quantities like length stay non-negative).

Definition and nth term
An arithmetic progression (AP) is a sequence in which the difference between consecutive terms is constant; this constant is the common difference d. If the first term is a, then the nth term is a_n = a + (n - 1)d. This simple linear pattern makes AP easy to handle and appears in many practical contexts like evenly spaced amounts or fixed incremental sequences.

Sum of first n terms
The sum of the first n terms S_n is given by S_n = n/2 [2a + (n - 1)d]. This formula can also be written as S_n = n/2 (a + l) where l is the nth term. Derive the first form by pairing terms: write the sequence forwards and backwards and add termwise to see each pair sums to (a + l), repeated n/2 times when n is even. For odd n one term remains central but the formula still holds.

Applications and problem types
APs appear in age problems, where ages increase uniformly, and in arithmetic arrangements and payments. Problems typically ask for nth term, number of terms for a given value, or sum of certain terms. When asked to find n, solve the linear equation given by the nth term formula; ensure result is an integer and check context restrictions like positivity.

Properties and sums of special APs
Sum of first n natural numbers is n(n + 1)/2 which is a special AP result with a = 1, d = 1. Sum of first n odd numbers equals n^2 — useful for quick checks and contest-style questions. Differences and averages in AP have useful relationships: the average of first and last term equals average of any symmetric pair of terms and equals S_n/n.

Mixed problems and decomposition
Some problems mix AP with other sequences or ask for partial sums (sum of terms from p to q). Use S_q - S_{p-1} = sum from p to q. For AP appearing alongside geometric sequences in a problem, separate parts and apply respective formulas, then combine algebraically to solve for unknowns.

Exam tips
State the values of a, d, n clearly before using formulas. When solving for n, demonstrate algebraic steps: set a + (n - 1)d = given term and solve for n. For sums, pick the form of S_n formula most convenient and simplify carefully to avoid arithmetic mistakes.

Definition and nth term
A geometric progression (GP) is a sequence where each term after the first is obtained by multiplying the previous by a fixed non-zero number r called the common ratio. The general term is a_n = ar^{n - 1} where a is the first term. GPs model multiplicative growth such as compound interest and certain population models.

Deriving the finite sum
To derive the sum S_n = a + ar + ar^2 + ... + ar^{n-1}, multiply both sides by r to get rS_n = ar + ar^2 + ... + ar^n. Subtract: S_n - rS_n = a - ar^n, so S_n(1 - r) = a(1 - r^n). For r ≠ 1 this gives S_n = a(1 - r^n)/(1 - r). This derivation is a key algebraic manipulation and must be memorised for quick use in exams.

Infinite GP and convergence
If |r| < 1 the terms tend to 0 as n increases and the infinite series converges to S_∞ = a/(1 - r). If |r| ≥ 1 the infinite series diverges and no finite sum exists. Recognising whether |r| < 1 is a necessary first step before applying the infinite sum formula; show this condition explicitly in answers.

Behaviour and sign of r
If r > 1 terms grow rapidly. If 0 < r < 1 terms decrease toward zero. If r is negative, the terms alternate sign and may oscillate in magnitude depending on |r|. These behaviours affect the pattern of partial sums and determine convergence for infinite series.

Solving problems with r^n
When a problem gives S_n or a_n and asks for n or r, algebraic rearrangement often leads to r^n = expression. Take logarithms when necessary: n log r = log(expression) so n = log(expression)/log r, provided r > 0 and r ≠ 1. Use change-of-base with calculators. For integer n required by context, check that computed n is an integer and satisfies original equations.

Applications and examples
Common applications include compound interest: amount after n periods equals principal × (1 + rate)^n, which is a GP. Another is repeating decimals: 0.333... can be written as geometric series 3/10 + 3/100 + ... and summed to a rational fraction using S_∞. In exam problems, identify whether the context calls for finite or infinite sum and state convergence conditions when using infinite formula.

Practical tips
State whether r = 1 separately; if r = 1 then S_n = na. For sums with alternating signs keep careful track of signs. Simplify expressions algebraically before plugging into calculators to reduce rounding error. Show the derivation or key steps briefly to earn method marks in exams.

Statement of the theorem
The binomial theorem expands (a + b)^n for a non-negative integer n as a sum of terms involving binomial coefficients: (a + b)^n = sum_{k=0}^{n} C(n,k) a^{n-k} b^k, where C(n,k) = n!/(k!(n-k)!). Each term corresponds to choosing k of the n places to supply b rather than a, which connects combinatorics to algebra.

Pascal's triangle
Binomial coefficients are arranged in Pascal's triangle where each entry is the sum of the two above it. Row n gives the coefficients for (a + b)^n with symmetry C(n,k) = C(n,n-k). The sum of coefficients in row n equals 2^n because substituting a = b = 1 gives (1 + 1)^n = 2^n.

Finding specific terms and coefficients
To find the coefficient of a particular term like a^{n-k}b^k, compute C(n,k). For expressions like (x + 2)^5 use coefficients 1,5,10,10,5,1 and multiply each by appropriate powers of x and 2. For (a - b)^n alternate signs since b is replaced by -b.

Computation techniques
Compute C(n,k) efficiently using cancellation to avoid large factorials: C(n,k) = n(n-1)...(n-k+1)/k!. Use symmetry to compute smaller k. For problems requesting coefficients or specific terms, show combination calculation clearly and simplify numerator and denominator before multiplying numbers.

Applications and identities
Binomial theorem helps expand polynomial powers, evaluate sums, and solve problems connecting algebra to counting. It provides coefficients needed in probability problems (distribution of successes) and is used in deriving identities such as sum of binomial coefficients and alternating sums. Newton's generalised binomial (for non-integer exponents) is beyond Class 10 but the integer case is fundamental.

Exam tips
Use Pascal's triangle for small n quickly. For questions asking for a particular coefficient, state the value of k and compute C(n,k). If the term includes constants (like 2^k), include them in arithmetic carefully. Always present the expansion in decreasing powers of the variable unless asked otherwise.

Counting basics
Permutation and combination are counting techniques. Use the multiplication principle: if one step has m choices and the next has n choices, there are m×n sequences. Distinguish between order-important (permutations) and order-not-important (combinations) problems.

Permutations
Number of ordered arrangements (permutations) of r distinct objects chosen from n distinct objects is P(n,r) = n(n - 1)...(n - r + 1) = n!/(n - r)!. For arranging all n objects, number is n!. For circular arrangements, adjust formulas to account for rotations being equivalent (often (n - 1)! for n objects around a circle).

Combinations
Combinations count selections where order does not matter: C(n,r) = n!/[r!(n - r)!]. Relations: P(n,r) = C(n,r) × r!. Use combinations for forming committees, choosing teams or selecting items. For problems with identical objects or restrictions, divide by factorials of repeats or use stars-and-bars reasoning when distribution problems appear (careful: stars-and-bars is usually beyond simple ICSE combinations but basic restricted selection appears).

Handling repeats and restrictions
When objects repeat, divide total permutations by factorials of repeats: for word with repeated letters, anagram count = total letters ! / product of factorials of each repeated letter count. When restrictions exist (e.g., certain items must be together), treat grouped items as one object then adjust for internal arrangements.

Problem translation and strategies
Read questions carefully: words like 'arrange', 'order', 'different ways' usually indicate permutations; words like 'select' or 'choose' point to combinations. Break complex problems into cases and sum results. For multi-step choices, multiply counts of independent steps and add counts for mutually exclusive cases.

Exam tips
Write down whether order matters before applying formula. Use factorial cancellation to simplify computations. For large numbers, simplify algebraically where possible rather than compute huge factorials. Show reasoning when grouping or applying restrictions to receive full marks.

Matrix basics
A matrix is a rectangular array of numbers arranged in rows and columns. Notation: an m × n matrix has m rows and n columns, and element in row i, column j is a_{ij}. Special matrices include zero matrix (all entries zero), identity matrix I_n (diagonal ones), diagonal and scalar matrices. Matrices give a compact way to represent linear systems and transformations.

Matrix operations
Addition and scalar multiplication are entry-wise and defined only for matrices of the same order. Matrix multiplication is defined when the number of columns of the first matrix equals the number of rows of the second; the (i,j) entry of the product AB is sum_k a_{ik} b_{kj}. Multiplication is associative and distributive over addition but not commutative in general (AB need not equal BA).

Transpose and properties
Transpose A^T flips rows to columns: (A^T)_{ij} = A_{ji}. For square matrices, symmetry (A = A^T) is an important property. The identity matrix I acts like 1 under multiplication: AI = IA = A for compatible sizes.

Determinants for 2×2 and 3×3
For 2×2 matrix [[a b],[c d]] determinant is det = ad - bc. Determinant is a scalar giving information about invertibility: matrix is invertible iff determinant ≠ 0. For 3×3 matrices, compute determinant by expansion of minors or use Sarrus rule (a quick mnemonic) for exams. Determinants change sign when two rows are swapped and scale by scalar factor when a row is scaled.

Inverse of a 2×2 matrix and solving systems
If det ≠ 0, inverse of 2×2 matrix A = [[a b],[c d]] is (1/det)[[d -b],[-c a]]. Use inverse to solve linear system Ax = b by x = A^{-1}b when A is small and invertible. For larger systems Cramer's rule or other methods may be used but computational cost increases rapidly.

Applications and exam technique
Matrices model transformations, systems of equations and data. For exam problems show matrix multiplication steps, determinant calculation and check invertibility by checking determinant. When using inverse or Cramer's rule, compute determinants carefully and substitute back to verify solutions.

Determinant meaning
The determinant of a square matrix is a scalar derived from its entries. It encodes geometric information such as area or volume scaling under the linear transformation represented by the matrix, and determines invertibility: a matrix is invertible exactly when its determinant is non-zero.

Computing determinants
For 2×2 matrix [[a b],[c d]], det = ad - bc. For 3×3 matrices, use expansion by minors or Sarrus rule: for [[a b c],[d e f],[g h i]], det = aei + bfg + cdh - ceg - bdi - afh. For larger matrices use row/column operations to simplify into triangular form or use cofactor expansion. Keep track of sign changes when swapping rows and factor out scalars when scaling rows.

Cramer's rule
For a system of n linear equations with n unknowns written in matrix form Ax = b, if det(A) ≠ 0 then the unique solution for each variable x_i is x_i = det(A_i)/det(A) where A_i is obtained by replacing the i-th column of A with the column vector b. For 2×2 and 3×3 systems, Cramer's rule yields direct formulas and is often used in exam answers because it gives explicit determinants to compute.

Special cases and interpretation
If det(A) = 0 the system may have infinitely many solutions (if determinants of augmented matrices also vanish) or no solution (if augmented determinants not all zero). State these cases clearly when answering. Cramer's rule cannot be used if det(A) = 0.

Computation tips
When calculating determinants simplify rows or columns first by common factors to reduce arithmetic. Use sign pattern carefully in cofactor expansions: signs alternate starting with plus at (1,1). For 3×3 Sarrus is quick but works only for 3×3; for larger sizes do row operations. Always show determinant steps or at least state intermediate simplified matrices to secure marks.

Exam examples and checking
After obtaining solution by Cramer's rule substitute values back into original equations to verify. In case of fractions simplify to lowest terms. If a determinant equals zero during work, pause and consider whether system is dependent or inconsistent rather than continuing with Cramer's rule.

What are relations and functions?
A relation between two sets associates elements of the first set to elements of the second. A function is a special relation that assigns exactly one output to each input. Denote a function as y = f(x). Domain is the set of all allowable input values of x; range is the set of outputs y produced by f from the domain.

Function evaluation and notation
Evaluate functions by substituting values: if f(x) = x^2 + 1, then f(2) = 5. Use clear notation when composing functions: (f ∘ g)(x) = f(g(x)). For inverse functions f^{-1}(x) solve y = f(x) for x in terms of y and then interchange variables (only if f is one-to-one on the chosen domain).

Domain and range determination
To find domain, identify values that make expressions undefined such as division by zero or negative arguments of even-indexed roots or logarithms. For polynomials domain is all real numbers. Range can be found by algebraic methods: e.g., complete the square for quadratics to determine minimal or maximal y-values, or reason about monotonicity for linear functions.

Types of functions
Linear functions y = mx + c have domain and range all real numbers; quadratic functions y = ax^2 + bx + c have range y ≥ vertex value when a > 0. Rational functions have restrictions in domain where denominator is zero and may have horizontal or vertical asymptotes determining long-term behaviour of range.

Inverse and composition
Inverse functions exist only for one-to-one functions on their domain; for quadratics restrict domain to make the function one-to-one before finding inverse. Composition combines functions and requires careful attention to domains: domain of (f ∘ g) is values x in domain of g for which g(x) lies in domain of f.

Applications and exam approach
Functions model relations between quantities in physics, economics and other subjects. In exams, state domain and range explicitly, show steps when finding inverses, and illustrate results with simple graphs where required. Use algebraic manipulation such as completing square to find range precisely when needed.

Exponents — rules and understanding
Exponents represent repeated multiplication: a^n denotes a multiplied by itself n times for integer n ≥ 0. Key laws simplify computations: a^m · a^n = a^{m+n}, (a^m)^n = a^{mn}, (ab)^n = a^n b^n, a^0 = 1 for a ≠ 0 and a^{-n} = 1/a^n. These rules allow manipulation of powers in algebraic expressions and help solve equations where variable is in exponent.

Logarithms — inverse of exponentiation
Logarithm log_a b is the exponent to which base a must be raised to get b: if a^x = b then log_a b = x. Common log (base 10) and natural log (base e) are special, but for Class 10 focus is often on base 10 and change-of-base use for solving equations. Laws of logs: log(ab) = log a + log b, log(a/b) = log a - log b, log(a^k) = k log a. These convert multiplicative problems into additive ones and turn powers into multipliers.

Solving exponential equations
If variable appears in exponent, take logarithms to bring it down: solve a^{f(x)} = b by f(x) = log_a b or use natural or common logs with change-of-base: f(x) = ln b / ln a. For equations like 3^x = 20 use x = log 20 / log 3. If bases are same, equate exponents: if a^m = a^n then m = n provided a ≠ 0,1.

Domain issues and applications
Logarithm arguments must be positive: log_a x defined only for x > 0 (and a > 0, a ≠ 1). Exponential and logarithmic models appear in growth/decay problems, compound interest, and decay processes. Understand when to use discrete compounding (powers) versus continuous models (involving e and natural logs in higher classes).

Practical tips
When solving logarithmic equations combine terms using log laws before exponentiating to avoid mistakes. Use change-of-base formula log_a b = log_c b / log_c a to compute logs with calculator keys. Check solutions in original equation to ensure domain conditions (positive arguments) are satisfied.

Exam strategy
Show application of log laws in steps, state domain restrictions explicitly, and give numerical answers to required decimal places. For contest-style problems be comfortable manipulating exponents and logs algebraically without relying solely on calculator approximations.