Algebra Study Notes – Sets, Quadratics, Complex Numbers, Matrices | IIT JEE

📐 Mathematics · Algebra 🔥 Day 1

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Sets, Relations, Functions · Quadratic Equations · Complex Numbers

Foundation of JEE algebra — functions, roots, and the Argand plane

Sets & Set Operations

Types of Sets

Empty set ∅: no elements. Singleton: one element. Finite/Infinite by cardinality. Universal set U: all elements under discussion. Power set P(A): set of all subsets; |P(A)| = 2ⁿ for |A|=n.

Set Operations

Union A∪B: elements in A or B. Intersection A∩B: elements in both. Difference A−B: in A but not B. Complement Aᶜ: elements in U not in A. Symmetric difference A△B = (A−B)∪(B−A).

De Morgan's Laws

(A∪B)ᶜ = Aᶜ ∩ Bᶜ

(A∩B)ᶜ = Aᶜ ∪ Bᶜ

Used to convert between union and intersection when complementing. Critical for JEE probability problems.

Cardinality Formulas

n(A∪B) = n(A) + n(B) − n(A∩B)

n(A∪B∪C) = n(A)+n(B)+n(C) − n(A∩B) − n(B∩C) − n(A∩C) + n(A∩B∩C)

Relations & Functions

Concept	Definition	Key point
Relation R on A	Subset of A×A	Total relations = $2^{n^2}$ for \|A\|=n
Reflexive	(a,a) ∈ R ∀a	Diagonal must be included
Symmetric	(a,b)∈R ⟹ (b,a)∈R	R = Rᵀ (symmetric matrix analog)
Transitive	(a,b),(b,c)∈R ⟹ (a,c)∈R	Most commonly tested property
Equivalence	Reflexive + Symmetric + Transitive	Partitions set into equivalence classes
One-one (injective)	$f(x_{1})=f(x_{2}) ⟹ x_{1}=x_{2}$	Horizontal line cuts graph at most once
Onto (surjective)	Range = Codomain	Every y in codomain has a pre-image
Bijective	One-one + Onto	Invertible; \|Domain\|=\|Codomain\|

💡Composition: $(g\circ f)(x) = g(f(x))$ — apply f first, then g. Inverse function $f^{-1}$ exists iff f is bijective. Domain of $f^{-1}$ = Range of f.

Quadratic Equations

Standard form: $ax^{2} + bx + c = 0$ (a ≠ 0). The discriminant $D = b^{2} - 4ac$ determines the nature of roots.

D	Nature of roots	Comment
$D > 0$	Two distinct real roots	Rational if D is a perfect square and $a, b, c \in \mathbb{Q}$
$D = 0$	Two equal real roots (repeated)	$x = -b/(2a)$
$D < 0$	Two complex conjugate roots	$\alpha = p+qi,\ \beta = p-qi$; a, b, c real

Vieta's Formulas

For $ax^2+bx+c=0$ with roots α, β:

$\alpha + \beta = -b/a$ (sum of roots)

$\alpha\beta = c/a$ (product of roots)

Form equation from roots: $x^{2} - (\alpha+\beta)x + \alpha\beta = 0$

Useful Symmetric Functions

$$ \alpha^{2}+\beta^{2} = (\alpha+\beta)^{2} - 2\alpha\beta $$

$$ \alpha^{3}+\beta^{3} = (\alpha+\beta)^{3} - 3\alpha\beta(\alpha+\beta) $$

$\alpha^{2}-\beta^{2} = (\alpha+\beta)(\alpha-\beta)$; $\alpha-\beta = \sqrt{(\alpha+\beta)^{2}-4\alpha\beta}$

$$ (\alpha+1)(\beta+1) = \alpha\beta + \alpha + \beta + 1 $$

Sign of Roots

Both roots positive: $D\ge 0,\ \alpha+\beta>0,\ \alpha\beta>0$

Both roots negative: $D\ge 0,\ \alpha+\beta<0,\ \alpha\beta>0$

Roots of opposite signs: $\alpha\beta<0$ ($D>0$ automatic)

At least one root zero: $c=0$

Common Roots

If $ax^2+bx+c=0$ and $dx^2+ex+f=0$ share a common root α:

Subtract to get a linear equation in α, then substitute back.

Condition: $(bf-ce)^{2} = (ae-bd)(cd-af)$ (cross-multiply method)

💡Range of quadratic: For $f(x)=ax^2+bx+c$ with $a>0$, minimum value $= -D/(4a)$ at $x=-b/(2a)$. Max/min of rational expressions like $(ax^2+bx+c)/(dx^2+ex+f)$: set $y$ = expression, rearrange as quadratic in x, require $D\ge 0$ for real x.

Complex Numbers

Basic Algebra

$z = a + ib$; $i = \sqrt{-1}$, $i^2 = -1$, $i^3 = -i$, $i^4 = 1$

Modulus: $|z| = \sqrt{a^{2}+b^{2}}$

Conjugate: $\bar z = a - ib$; $z\cdot\bar z = |z|^2$

Division: $z_{1}/z_{2} = (z_{1}\cdot\bar{z}_{2})/|z_{2}|^{2}$

$z + \bar z = 2\operatorname{Re}(z)$; $z - \bar z = 2i\operatorname{Im}(z)$

Polar & Euler Form

Polar: $z = r(\cos\theta + i \sin\theta),\ r = |z|,\ \theta = \arg(z)$

Euler: $z = re^{i\theta}$

Principal argument: $\theta \in (-\pi, \pi]$

Multiplication: $|z_1 z_2| = |z_1||z_2|$; $\arg(z_1 z_2) = \arg(z_1) + \arg(z_2)$

Division: $|z_1/z_2| = |z_1|/|z_2|$; $\arg(z_1/z_2) = \arg(z_1) - \arg(z_2)$

De Moivre's Theorem

$$ (\cos\theta + i \sin\theta)^{n} = \cos(n\theta) + i \sin(n\theta) $$

Used to find: powers of complex numbers, nth roots, and trigonometric identities (e.g. $\cos 3\theta$ in terms of $\cos\theta$).

nth roots of unity: $z = e^{2\pi ik/n}, k = 0,1,\ldots,n-1$

Cube Roots of Unity

$z^{3} = 1 \implies z = 1, \omega, \omega^{2}$ where $\omega = e^{2\pi i/3} = (-1+i\sqrt3)/2$

Key identities:

$$ 1 + \omega + \omega^{2} = 0 $$

$\omega^3 = 1$; $\bar\omega = \omega^2$

$a^3+b^3+c^3-3abc = (a+b+c)(a+b\omega+c\omega^2)(a+b\omega^2+c\omega)$

Argand Plane — Key Loci

Condition	Locus
$\|z - z_{0}\| = r$	Circle centre $z_0$, radius $r$
$\|z - z_{1}\| = \|z - z_{2}\|$	Perpendicular bisector of segment $z_1 z_2$
$\|z - z_{1}\| + \|z - z_{2}\| = 2a\ (2a > \|z_{1}-z_{2}\|)$	Ellipse with foci $z_1, z_2$
$\arg(z - z_{0}) = \theta$	Ray from $z_0$ at angle θ
$\operatorname{Im}(z) = 0$	Real axis (x-axis)
$\operatorname{Re}(z) = 0$	Imaginary axis (y-axis)
$\|z\| = 1$	Unit circle

Triangle Inequalities for Complex Numbers

$||z_1| - |z_2|| \le |z_1 + z_2| \le |z_1| + |z_2|$

Equality $|z_1+z_2| = |z_1|+|z_2|$ iff $\arg(z_1) = \arg(z_2)$ (same direction). Equality $|z_1+z_2| = ||z_1|-|z_2||$ iff $\arg(z_1) = \arg(z_2)\pm\pi$ (opposite directions).

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Sets, Quadratics & Complex — Key Formulas

Inclusion-Exclusion (3 sets)

n(A∪B∪C) = n(A)+n(B)+n(C) − n(A∩B) − n(B∩C) − n(A∩C) + n(A∩B∩C)

Generalises to n sets: alternating sign, adding individual terms, subtracting pairwise, adding triple, etc.

Quadratic Formula & Discriminant

$x = (-b \pm \sqrt{b^2-4ac}) / 2a$ D = b²−4ac

D>0: 2 distinct real; D=0: repeated root −b/2a; D<0: complex conjugate pair.

Vieta's Formulas

α + β = −b/a αβ = c/a (for ax²+bx+c=0)

Symmetric Functions of Roots

$\alpha^{2}+\beta^{2} = (\alpha+\beta)^{2}-2\alpha\beta$
$\alpha^{3}+\beta^{3} = (\alpha+\beta)[(\alpha+\beta)^{2}-3\alpha\beta]$
$\alpha^{2}\beta+\alpha\beta^{2} = \alpha\beta(\alpha+\beta)$
$|\alpha-\beta| = \sqrt{(\alpha+\beta)^{2}-4\alpha\beta}$

Minimum of Quadratic $ax^2+bx+c$ ($a>0$)

min value $= -D/(4a) = (4ac-b^{2})/(4a)$ at $x = -b/(2a)$

For $a<0$: this is the maximum. Axis of symmetry $x = -b/(2a)$ always holds.

Complex Number Modulus & Argument

$|z| = \sqrt{a^2+b^2}$; $\arg(z) = \tan^{-1}(b/a)$ [adjust for quadrant]
$|z_1 z_2| = |z_1||z_2|$; $\arg(z_1 z_2) = \arg z_1 + \arg z_2$

De Moivre's Theorem

$(\cos\theta + i\sin\theta)^{n} = \cos(n\theta) + i\sin(n\theta)$
nth roots of z: $z_{k} = r^{1/n}\cdot e^{i(\theta+2\pi k)/n},\ k=0,\ldots,n-1$

Cube Roots of Unity

$\omega = (-1+i\sqrt3)/2$; $\omega^2 = (-1-i\sqrt3)/2$
$1 + \omega + \omega^2 = 0$; $\omega^3 = 1$; $\bar\omega = \omega^2$

Triangle Inequality

$||z_1|-|z_2|| \le |z_1+z_2| \le |z_1|+|z_2|$

Also: $|z_1-z_2| \ge ||z_1|-|z_2||$. Useful for finding max/min of |expression| in complex loci problems.

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Sets, Quadratics & Complex — Solved Examples

In a class of 40 students: 25 play cricket, 20 play football, 10 play both. How many play neither?

Solution: $n(C\cup F) = n(C) + n(F) - n(C\cap F) = 25 + 20 - 10 =$ 35
Play neither $= 40 - 35 =$ 5 students

Find $\alpha^2+\beta^2$ and $\alpha^3+\beta^3$ if α, β are roots of $x^2-5x+6=0$

By Vieta's: $\alpha+\beta = 5, \alpha\beta = 6$
$\alpha^{2}+\beta^{2} = (\alpha+\beta)^{2} - 2\alpha\beta = 25 - 12 =$ 13
$\alpha^{3}+\beta^{3} = (\alpha+\beta)^{3} - 3\alpha\beta(\alpha+\beta) = 125 - 3(6)(5) = 125 - 90 =$ 35

Express $(3+4i)/(1-2i)$ in the form $a+ib$

Multiply by conjugate of denominator:
$$ (3+4i)(1+2i) / [(1-2i)(1+2i)] = (3 + 6i + 4i + 8i^{2}) / (1+4) = (3 + 10i - 8) / 5 $$
$= (-5 + 10i) / 5 =$ $-1 + 2i$
Check: $|\text{result}| = \sqrt{1+4} = \sqrt5$; $|z_1|/|z_2| = \sqrt{25}/\sqrt5 = 5/\sqrt5 = \sqrt5$ ✓

Find the range of $f(x) = (x^2+x+1)/(x^2-x+1)$ for $x \in \mathbb{R}$

Set y = f(x): $y(x^{2}-x+1) = x^{2}+x+1$
$$ x^{2}(y-1) - x(y+1) + (y-1) = 0 $$
For real x, $D \ge 0$: $(y+1)^2 - 4(y-1)^2 \ge 0$
$y^2 + 2y + 1 - 4(y^2-2y+1) \ge 0$
$-3y^2 + 10y - 3 \ge 0 \to 3y^2 - 10y + 3 \le 0 \to (3y-1)(y-3) \le 0$
∴ $1/3 \le y \le 3$; Range $= [1/3, 3]$
Note: $y=1/3$ when $x=-1$; $y=3$ when $x=1$. The denominator $x^2-x+1 = (x-\tfrac{1}{2})^2+\tfrac{3}{4} > 0$ always, so no restriction from denominator. ✓

If $z = 1+i$, find $z^{10}$ using De Moivre's theorem

Convert to polar: $|z| = \sqrt2,\ \arg(z) = \pi/4$
$$ z = \sqrt2 \cdot e^{i\pi/4} = \sqrt2 (\cos \pi/4 + i \sin \pi/4) $$
$$ z^{10} = (\sqrt2)^{10} \cdot (\cos \pi/4 + i \sin \pi/4)^{10} = 2^{5} \cdot (\cos(10\pi/4) + i \sin(10\pi/4)) $$
$$ = 32 \cdot (\cos(5\pi/2) + i \sin(5\pi/2)) = 32 \cdot (\cos(\pi/2) + i \sin(\pi/2)) = 32 \cdot (0 + i) $$
∴ $z^{10} =$ $32i$

If α, β are roots of $x^2-px+q=0$ and $\alpha^4, \beta^4$ are roots of $x^2-rx+s=0$, show that the discriminant of $x^2-4qx+2q^2-r = 0$ is non-negative

From first equation: $\alpha+\beta = p, \alpha\beta = q$
From second equation: $\alpha^{4}+\beta^{4} = r, \alpha^{4}\beta^{4} = s$
$$ \alpha^{4}+\beta^{4} = (\alpha^{2}+\beta^{2})^{2} - 2\alpha^{2}\beta^{2} = [(\alpha+\beta)^{2}-2\alpha\beta]^{2} - 2(\alpha\beta)^{2} = (p^{2}-2q)^{2} - 2q^{2} $$
$$ So r = p^{4} - 4p^{2}q + 4q^{2} - 2q^{2} = p^{4} - 4p^{2}q + 2q^{2} $$
Discriminant of $x^2-4qx+(2q^2-r)=0$:
$$ D = (4q)^{2} - 4(2q^{2}-r) = 16q^{2} - 8q^{2} + 4r = 8q^{2} + 4r $$
Substituting $r = p^4-4p^2 q+2q^2$: $D = 8q^2+4p^4-16p^2 q+8q^2 = 4(p^4-4p^2 q+4q^2) = 4(p^2-2q)^2$
Since $4(p^2-2q)^2 \ge 0$ for all real p, q, the discriminant is always non-negative. ✓

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Sets, Quadratics & Complex — Key Diagrams

Click any diagram to zoom

Three-Set Venn Diagram — Inclusion-Exclusion

The 7 non-overlapping regions of a three-set Venn diagram. The inclusion-exclusion formula adds all individual regions, corrects for double-counting of pairwise intersections, then adds back the triple intersection (over-subtracted).

Argand Plane — Complex Number Representation

The Argand plane represents $z=a+ib$ as point (a, b). The modulus $r=|z|$ is the distance from origin; the argument θ is the angle from positive real axis. The conjugate $\bar z$ is the reflection across the real axis.

Quadratic Parabola — Three Cases by Discriminant

For $ax^2+bx+c$ with $a>0$: $D>0$ gives two x-intercepts; $D=0$ gives one tangent point at $x=-b/(2a)$; $D<0$ gives no real roots (parabola entirely above x-axis). Flip for $a<0$.

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Sets, Quadratics & Complex — Quick Quiz

Question 1

In a group of 70 people: 37 like coffee, 52 like tea, and each person likes at least one. How many like both?

$n(C\cup T) = n(C)+n(T)-n(C\cap T) \to 70 = 37+52-n(C\cap T) \to n(C\cap T) = 89-70 =$ 19.

Question 2

If α, β are roots of $2x^2-5x+3=0$, find the value of $\alpha/\beta + \beta/\alpha$.

$\alpha+\beta = 5/2, \alpha\beta = 3/2$. $\alpha/\beta+\beta/\alpha = (\alpha^{2}+\beta^{2})/(\alpha\beta) = [(\alpha+\beta)^{2}-2\alpha\beta]/(\alpha\beta) = [25/4-3]/\tfrac{3}{2} = [13/4]/\tfrac{3}{2} = 13/4 \times 2/3 =$ 13/6.

Question 3

What is the modulus of $(1+i)^8$?

$|1+i| = \sqrt2$, so $|(1+i)^{8}| = (\sqrt2)^{8} = 2^{4} =$ 16. Alternatively: $(1+i)^2 = 2i$, so $(1+i)^4 = (2i)^2 = -4$, so $(1+i)^8 = (-4)^2 = 16$ (a positive real number).

Question 4

If $1, \omega, \omega^2$ are cube roots of unity, the value of $(1+\omega)(1+\omega^2)(1+\omega^4)(1+\omega^8)$ is:

$\omega^3=1$, so $\omega^4=\omega, \omega^8=\omega^2$. Product $= (1+\omega)(1+\omega^2)(1+\omega)(1+\omega^2) = [(1+\omega)(1+\omega^2)]^2$. Now $(1+\omega)(1+\omega^2) = 1+\omega+\omega^2+\omega^3 = 0+1 = 1$ (since $1+\omega+\omega^2=0$ and $\omega^3=1$). So product $= 1^2 =$ 1.

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Sequences, Series & Binomial Theorem

AP · GP · HP · AGP · Special Series · Binomial Expansion

Arithmetic Progression (AP)

General term: $a_n = a + (n-1)d$, where $a$ = first term, $d$ = common difference.
Sum of n terms: $S_n = \tfrac{n}{2}\,[2a + (n-1)d] = \tfrac{n}{2}\,(\text{first} + \text{last})$.
Choosing terms in AP: 3 terms: $(a-d), a, (a+d)$ [sum = 3a]. 4 terms: $(a-3d), (a-d), (a+d), (a+3d)$ [sum = 4a]. 5 terms: $(a-2d), (a-d), a, (a+d), (a+2d)$ [sum = 5a].
Arithmetic Mean: AM of a and b $= (a+b)/2$. If A is AM between a and b, then $a, A, b$ is in AP.
Properties: If a, b, c in AP → $2b = a+c$. Sum of equidistant terms from ends = first + last.

Geometric Progression (GP)

General term: $a_n = ar^{n-1}$, where $r$ = common ratio.
Sum of n terms: $S_n = a(r^n-1)/(r-1)$ for $r\ne 1$; $S_n = na$ for $r=1$.
Sum to infinity: $S_\infty = a/(1-r)$ only when $|r| < 1$.
Geometric Mean: GM of a and b $= \sqrt{ab}$. If G is GM between a and b: $a, G, b$ is in GP → $G^2 = ab$.
Properties: If a, b, c in GP → $b^2 = ac$. Product of equidistant terms from ends = first × last.
3 terms in GP: $a/r, a, ar$ [product $= a^3$].

Harmonic Progression (HP) and AM-GM-HM

$a, b, c$ in HP iff $1/a, 1/b, 1/c$ in AP → $1/b - 1/a = 1/c - 1/b \to b = 2ac/(a+c)$.
HM of a and b: $H = 2ab/(a+b)$.
AM ≥ GM ≥ HM for positive reals. Equality iff all terms are equal.
$AH = G^2$ (i.e., $\text{AM} \times \text{HM} = (\text{GM})^2$) for two positive numbers.
Application: To maximise/minimise: use AM-GM. If product is constant, minimise sum via AM-GM; if sum is constant, maximise product via AM-GM.

Arithmetico-Geometric Progression (AGP)

AGP: terms of the form $a_n\cdot b_n$ where $\{a_n\}$ is AP and $\{b_n\}$ is GP. E.g., 1·1, 2·2, 3·4, 4·8, … = 1, 4, 12, 32, …
Sum of AGP: $S_n = ab/(1-r) + dbr/(1-r)^2 - [a+(n-1)d]\cdot br^n/(1-r)$ (for $|r|<1$, $S_\infty = ab/(1-r) + dbr/(1-r)^2$).
Method: Write S, multiply by r to get rS, subtract → telescoping sum. Easier to derive than memorise.

Special Series

Series	Sum
$\Sigma k = 1+2+\ldots+n$	$n(n+1)/2$
$\Sigma k^{2} = 1^{2}+2^{2}+\ldots+n^{2}$	$n(n+1)(2n+1)/6$
$\Sigma k^{3} = 1^{3}+2^{3}+\ldots+n^{3}$	$[n(n+1)/2]^{2} = (\Sigma k)^{2}$
$\Sigma (2k-1) = 1+3+5+\ldots+(2n-1)$	$n^2$ (sum of first n odd numbers)

Binomial Theorem

Expansion: $(a+b)^n = \sum_{r=0}^{n} C(n,r)\cdot a^{n-r}\cdot b^r$, where $C(n,r) = n!/[r!(n-r)!]$.
General term: $T_{r+1} = C(n,r)\cdot a^{n-r}\cdot b^r$. (r starts from 0.)
Middle term(s): If n is even: $T_{n/2+1}$ (single middle term). If n is odd: $T_{(n+1)/2}$ and $T_{(n+3)/2}$ (two middle terms).
Term independent of x: In $(x^a + x^{-b})^n$, set power of x = 0 → solve for r.
Coefficient sum properties:
Put $x=1$: sum of all coefficients $= 2^n$.
Put $x=-1$: alternating sum $C_0 - C_1 + C_2 - \ldots = 0$.
$$ C_{0}+C_{2}+C_{4}+\ldots = C_{1}+C_{3}+C_{5}+\ldots = 2^{n-1}. $$
Greatest term: Compare $|T_{r+1}/T_r| \ge 1$, solve for r → greatest integer r gives the greatest term.
Pascal's Identity: $C(n,r) = C(n-1,r-1) + C(n-1,r)$.
Multinomial: $(a+b+c)^n = \sum n!/(p!q!r!)\cdot a^p b^q c^r$ where $p+q+r=n$.

💡 JEE Tip: For finding the numerically greatest term in $(1+x)^n$, form the ratio $|T_{r+1}/T_r| = |(n-r+1)x/r|$. Set ≥ 1 and solve. The answer is the floor of the result. Also memorise: $(1+x)^n \approx 1+nx$ for $|x|\ll 1$ (binomial approximation used in Physics).

⚠ Common Error: When using AM ≥ GM, all terms must be positive. Also: $\Sigma k^3 = [n(n+1)/2]^2$ is often written as $(\Sigma k)^2$ — this is the "Nicomachus theorem" and a very popular JEE identity. Don't confuse $\Sigma k^2 = n(n+1)(2n+1)/6$ (the coefficient 2n+1 makes it different).

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Sequences & Binomial — Formula Sheet

AP, GP, HP Formulas

AP: $a_n = a+(n-1)d$, $S_n = \tfrac{n}{2}\,[2a+(n-1)d]$

$d$ = common difference. If last term = l: $S_n = n(a+l)/2$

GP: $a_n = ar^{n-1}$, $S_n = a(r^n-1)/(r-1)$, $S_\infty = a/(1-r)$ for $|r|<1$

$r$ = common ratio. For $r=1$: $S_n=na$. Geometric mean of a, b: $G=\sqrt{ab}$

$\text{AM} \ge \text{GM} \ge \text{HM}$ | $\text{AM}\cdot\text{HM} = \text{GM}^2$ | $\text{HM} = 2ab/(a+b)$

For two positive numbers a and b. Equality iff $a = b$.

Special Series Sums

$\Sigma k = n(n+1)/2$ | $\Sigma k^2 = n(n+1)(2n+1)/6$ | $\Sigma k^3 = [n(n+1)/2]^2$

All three from $k=1$ to $n$. Note: $\Sigma k^3 = (\Sigma k)^2$ — the Nicomachus theorem.

Binomial Theorem

$(a+b)^n = \sum_{r=0}^{n} C(n,r)\cdot a^{n-r}\cdot b^r$ | $T_{r+1} = C(n,r)\cdot a^{n-r}\cdot b^r$

General (r+1)th term; $r=0,1,\ldots,n$. Total $(n+1)$ terms in expansion.

Sum of coefficients: put $a=b=1 \to 2^n$. Alternating sum: put $a=1, b=-1 \to 0$

Even-indexed + odd-indexed $= 2^n$; even = odd $= 2^{n-1}$. $C_0+C_2+\ldots=C_1+C_3+\ldots=2^{n-1}$

$C(n,r) = C(n,n-r)$ | $C(n,r)+C(n,r-1) = C(n+1,r)$ (Pascal)

Symmetry property: C(10,3)=C(10,7). Pascal's identity builds Pascal's triangle.

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Sequences & Binomial — Solved Examples

Express 0.3̄2̄ = 0.323232… as a fraction using the sum of infinite GP.

0.323232… = 0.32 + 0.0032 + 0.000032 + … This is a GP with $a = 0.32 = 32/100, r = 1/100$.
$|r| = 0.01 < 1$, so $S_\infty = a/(1-r) = (32/100) / (1 - 1/100) = (32/100) / (99/100) = 32/99$.

✓ $0.\overline{32} = $ 32/99. Verify: 32 ÷ 99 = 0.323232… ✓. General rule: n digits repeating from decimal → put those n digits over n nines.

Find the term independent of x in the expansion of $(x + 1/x^2)^9$.

General term: $T_{r+1} = C(9,r) \cdot x^{9-r} \cdot (1/x^{2})^r = C(9,r) \cdot x^{9-r} \cdot x^{-2r} = C(9,r) \cdot x^{9-3r}.$
For term independent of x: power of x = 0 → $9-3r = 0 \to r = 3$.
$T_{4} = C(9,3) \cdot x^{0} = 84 \cdot 1 =$ 84.

✓ The term independent of x is the 4th term (r=3) with value 84. Always write the general term first, then set the x-power to 0.

Find the minimum value of $f(x) = x + 4/x$ for $x > 0$. Also find the minimum of $f(x) = x^2 + 1/x^2 + 1/x^4$ for $x > 0$.

For $x + 4/x$: Apply AM-GM: $x + 4/x \ge 2\sqrt{x \cdot 4/x} = 2\sqrt4 = $ 4. Equality when $x = 4/x \to x = 2$.
Verify: $f(2) = 2 + 4/2 = 2+2 = 4$ ✓. Minimum = 4 at $x = 2$.
For $x^2 + 1/x^2 + 1/x^4$: write as three positive terms and apply AM-GM to all three:
$(x^2 + 1/x^2 + 1/x^4)/3 \ge \sqrt[3]{x^2 \cdot 1/x^2 \cdot 1/x^4} = \sqrt[3]{1/x^4}$. This doesn't simplify nicely — use substitution $t = 1/x^2 \to$ minimise $t^{-1} + t + t^2$. Let $g(t) = 1/t + t + t^2$, $g'(t) = -1/t^2 + 1 + 2t = 0$. At $t=1/2$: $-4+1+1=-2\ne 0$. Try calculus or specific substitution: at $x=1 \to f = 1+1+1 = 3$. AM-GM of 3 terms: min = 3 when all equal → $x=1$ ✓.

✓ $f(1) = 1+1+1 = $ 3 (minimum). When all three AM-GM terms are equal: $x^{2}=1/x^{2}=1/x^{4} \to x=1$.

Find the term independent of $x$ in the expansion of $\left(x^2 - \dfrac{2}{x}\right)^{9}$, and evaluate it.

General term: $T_{r+1} = \binom{9}{r}(x^2)^{9-r}\left(-\dfrac{2}{x}\right)^{r} = \binom{9}{r}(-2)^{r} x^{18-2r-r} = \binom{9}{r}(-2)^{r} x^{18-3r}.$
Independent of $x$: power $= 0 \Rightarrow 18 - 3r = 0 \Rightarrow r = 6.$
$T_{7} = \binom{9}{6}(-2)^{6} = 84 \times 64 = 5376.$

✓ The term independent of $x$ is $T_{7} = 5376$.

Find the sum to $n$ terms of the series $1 + 3x + 5x^2 + 7x^3 + \cdots$ (with $|x| < 1$, then the infinite sum).

This is an AGP: arithmetic part $1,3,5,\dots$ ($a=1,\ d=2$); geometric part $1,x,x^2,\dots$ ($r=x$). Let $S = \sum_{k=0}^{n-1}(2k+1)x^{k}.$
Multiply by $x$: $xS = \sum (2k+1)x^{k+1}$. Subtract: $(1-x)S = 1 + 2x + 2x^2 + \cdots + 2x^{n-1} - (2n-1)x^{n}.$
The middle geometric run sums to $2x\dfrac{1-x^{n-1}}{1-x}$, giving $(1-x)S = 1 + \dfrac{2x(1-x^{n-1})}{1-x} - (2n-1)x^{n}.$

✓ As $n\to\infty$ with $|x|<1$: $(1-x)S = 1 + \dfrac{2x}{1-x} = \dfrac{1+x}{1-x}$, so $S_{\infty} = \dfrac{1+x}{(1-x)^2}.$

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Sequences & Binomial — Visual Concepts

Click any diagram to zoom

Pascal's Triangle — Binomial Coefficients

Pascal's triangle. Row n contains the binomial coefficients C(n,0), C(n,1), …, C(n,n). Each entry equals the sum of the two entries above it. Row sum = 2ⁿ (put x=1 in binomial theorem). Diagonal sums give Fibonacci numbers.

AM-GM-HM Inequality — Geometric Interpretation

For positive numbers a and b, HM ≤ GM ≤ AM, with equality only when a = b. Illustrated with a=4, b=9: HM=72/13≈5.54 < GM=6 < AM=6.5. The AM is the midpoint of a and b on the number line; GM is the "geometric mean" (where it lies between them depends on ratio b/a).

⚡

Sequences & Binomial — Quick Quiz

Question 1

The sum of the infinite GP $1 + 1/3 + 1/9 + 1/27 + \ldots$ is:

$S_\infty = a/(1-r) = 1/(1-1/3) = 1/(2/3) = 3/2$. Here $a=1, r=1/3, |r|<1$ so infinite sum exists. Answer: C ✓

Question 2

The middle term in the expansion of $(x + 1/x)^{10}$ is:

$n=10$ (even) → single middle term is $T_{n/2+1} = T_6$. $T_6 = C(10,5)\cdot x^{10-5}\cdot(1/x)^5 = C(10,5)\cdot x^5\cdot x^{-5} = C(10,5)\cdot x^0 = 252$. Answer: B ✓. Note the middle term is also independent of x here.

Question 3

The value of $1^3 + 2^3 + 3^3 + \ldots + 10^3$ is:

$\Sigma k^3 = [n(n+1)/2]^2 = [10\times 11/2]^2 = [55]^2 = 3025$. Check: $\Sigma k = n(n+1)/2 = 55$. So $\Sigma k^3 = 55^2 = 3025$. Note: $385 = \Sigma k^2 = 10\times 11\times 21/6$. Answer: A ✓

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Permutations, Combinations & Probability

Counting · P(n,r) · C(n,r) · Conditional Probability · Bayes' Theorem · Distributions

Fundamental Counting Principle

Multiplication rule: If task 1 has m ways AND task 2 has n ways (independently), total $= m\times n$ ways.
Addition rule: If task 1 has m ways OR task 2 has n ways (mutually exclusive), total $= m+n$ ways.
Permutation $P(n,r)$: Ordered selection of r items from n distinct items $= n!/(n-r)!$. (Order matters.)
Combination $C(n,r)$: Unordered selection of r items from n $= n!/[r!(n-r)!]$. (Order doesn't matter.)
Relation: $P(n,r) = r! \times C(n,r)$.

Permutations — Special Cases

Scenario	Count
Arrange n distinct objects	$n!$
n objects with p alike, q alike, r alike ($p+q+r=n$)	$n!/(p!q!r!)$
Circular arrangement of n distinct objects	$(n-1)!$
Necklace / beads on ring (flipping counts as same)	$(n-1)!/2$
r objects from n, with repetition allowed	$n^r$
Words from n letters (repetition allowed)	$n^n$ (no), $n!$ (no repeat)

Combinations — Special Results

$C(n,0) = C(n,n) = 1$; $C(n,1) = n$; $C(n,r) = C(n,n-r)$.
Distributing n identical objects into r distinct boxes (each box $\ge 0$): $C(n+r-1, r-1)$.
Distributing n identical objects into r distinct boxes (each box $\ge 1$): $C(n-1, r-1)$.
Number of subsets of a set with n elements: $2^n$ (include empty set).
Number of non-empty proper subsets: $2^n - 2$.
Dearrangement (no fixed point): $D_n = n!\left[1 - 1/1! + 1/2! - 1/3! + \ldots + (-1)^n/n!\right]$.

Classical Probability

$P(A) = n(A)/n(S)$, where $n(S)$ = total equally likely outcomes (sample space).
$0 \le P(A) \le 1$; $P(A) + P(A') = 1$ (complement).
Addition rule: $P(A\cup B) = P(A) + P(B) - P(A\cap B)$. If mutually exclusive: $P(A\cup B) = P(A)+P(B)$.
Conditional probability: $P(A|B) = P(A\cap B)/P(B), P(B)\ne 0$.
Multiplication rule: $P(A\cap B) = P(A)\cdot P(B|A) = P(B)\cdot P(A|B)$.
Independent events: $P(A\cap B) = P(A)\cdot P(B)$. Then $P(A|B) = P(A)$ (occurrence of B doesn't affect A).

Total Probability and Bayes' Theorem

Law of Total Probability: If $A_1, \ldots, A_n$ are mutually exclusive and exhaustive, then $P(B) = \sum_i P(A_i)\cdot P(B|A_i)$.
Bayes' Theorem: $P(A_i|B) = P(A_i)\cdot P(B|A_i) / \sum_j P(A_j)\cdot P(B|A_j)$. Posterior = (prior × likelihood) / evidence.
Binomial Distribution: $P(X=r) = C(n,r)\cdot p^r\cdot q^{n-r}$ where $p+q=1, r=0,1,\ldots,n$. Mean $= np$, Variance $= npq$.

💡 JEE Tip: For circular permutations: always fix one object and arrange the rest $(n-1)!$. For necklace/bracelet: divide by 2 further (clockwise = counter-clockwise). For "at least one" type probability: $P(\text{at least one}) = 1 - P(\text{none})$. This complement approach is almost always faster than direct counting.

⚠ Common Error: Independent ≠ Mutually Exclusive. Mutually exclusive events cannot both occur → if $P(A)>0$ and $P(B)>0$, they CANNOT be independent AND mutually exclusive. Independent: $P(A\cap B)=P(A)P(B)$. Mutually exclusive: $P(A\cap B)=0$.

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P&C & Probability — Formula Sheet

Counting Formulas

$P(n,r) = n!/(n-r)!$ | $C(n,r) = n!/[r!(n-r)!]$ | $P(n,r) = r!\cdot C(n,r)$

P: ordered (permutation); C: unordered (combination). $C(n,r) = C(n,n-r)$ by symmetry.

Distributing n identical into r distinct boxes ($\ge 0$): $C(n+r-1, r-1)$

Stars and bars formula. If each box must have $\ge 1$: $C(n-1, r-1)$.

Probability Laws

$P(A\cup B) = P(A)+P(B)-P(A\cap B)$ | $P(A|B) = P(A\cap B)/P(B)$

Addition rule + conditional probability definition. For independent: $P(A\cap B)=P(A)\cdot P(B)$.

Bayes': $P(A_i|B) = P(A_i)\cdot P(B|A_i) / \sum_j P(A_j)\cdot P(B|A_j)$

Requires $\{A_i\}$ to be mutually exclusive and exhaustive. Denominator $= P(B)$ by law of total probability.

Binomial: $P(X=r) = C(n,r)\cdot p^r\cdot q^{n-r}$ | Mean $=np$ | Var $=npq$

n independent trials, $p=P(\text{success}), q=1-p$. $X \sim B(n,p)$.

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P&C & Probability — Solved Examples

In how many ways can 8 people be seated around a circular table? If 2 specific people must sit together, how many ways?

Circular arrangement of n distinct: $(n-1)! = 7! = 5040$ ways.
2 specific people together: treat them as one unit → 7 units in circle → $(7-1)! = 6! = 720$ ways to arrange units. Within the pair, they can swap: ×2 = 1440 ways.

✓ Total with pair together: $6!\times 2 = 720\times 2 =$ 1440. Total without restriction: $7! =$ 5040.

A disease affects 1% of the population. A test for it is 95% accurate (95% true positive rate; 5% false positive rate). If a person tests positive, what is the probability they actually have the disease?

Let D = has disease, T = tests positive. $P(D) = 0.01, P(D') = 0.99, P(T|D) = 0.95, P(T|D') = 0.05$.
$$ P(T) = P(T|D)\cdot P(D) + P(T|D')\cdot P(D') = 0.95\times 0.01 + 0.05\times 0.99 = 0.0095 + 0.0495 = 0.059. $$
Bayes': $P(D|T) = P(D)\cdot P(T|D)/P(T) = (0.01\times 0.95)/0.059 = 0.0095/0.059 \approx$ 0.161 = 16.1%.

✓ Despite a 95% accurate test, only a 16% chance of having the disease when testing positive — because the disease is rare (base rate 1%). This is the base rate fallacy and a classic JEE/real-world Bayes application.

How many 4-letter words (with or without meaning) can be formed from the letters of MATHEMATICS such that: (a) no restrictions, (b) all letters distinct?

MATHEMATICS: M(2), A(2), T(2), H, E, I, C, S → 11 letters total with 3 pairs of duplicates. Distinct letters: M, A, T, H, E, I, C, S = 8 types.
All 4 letters distinct: choose 4 from 8 types → $C(8,4)$, then arrange → $C(8,4)\times 4! = 70\times 24 =$ 1680.
No restriction — consider cases: (i) all distinct: 1680. (ii) exactly one pair (e.g., MM, AA, or TT): $C(3,1)\times C(7,2)\times 4!/2! = 3\times 21\times 12 = 756$. (iii) two pairs: $C(3,2)\times 4!/(2!2!) = 3\times 6 = 18$.

✓ Total = 1680 + 756 + 18 = 2454. Key: systematically split into cases by the number of repeated pairs.

A disease affects 1% of a population. A test is 99% sensitive (correctly flags the sick) and 95% specific (correctly clears the healthy). If a person tests positive, what is the probability they actually have the disease?

Let $D$ = has disease, $+$ = tests positive. $P(D)=0.01,\ P(+\mid D)=0.99,\ P(+\mid D^c)=1-0.95=0.05.$
Total probability of a positive: $P(+) = P(+\mid D)P(D) + P(+\mid D^c)P(D^c) = 0.99(0.01) + 0.05(0.99) = 0.0099 + 0.0495 = 0.0594.$
Bayes: $P(D\mid +) = \dfrac{P(+\mid D)P(D)}{P(+)} = \dfrac{0.0099}{0.0594} \approx 0.1667.$

✓ $P(D\mid +) \approx \tfrac16 \approx 16.7\%$ — counter-intuitively low, because the disease is rare (base-rate effect).

A fair die is thrown 4 times. Find the probability of getting (a) exactly two sixes, (b) at least one six.

Each throw: $p = P(\text{six}) = \tfrac16,\ q = \tfrac56,\ n = 4$. Use $P(X=r) = \binom{n}{r}p^r q^{n-r}.$
(a) $P(X=2) = \binom{4}{2}\left(\tfrac16\right)^2\left(\tfrac56\right)^2 = 6\cdot\dfrac{1}{36}\cdot\dfrac{25}{36} = \dfrac{150}{1296} = \dfrac{25}{216}.$
(b) $P(X\ge 1) = 1 - P(X=0) = 1 - \left(\tfrac56\right)^4 = 1 - \dfrac{625}{1296} = \dfrac{671}{1296}.$

✓ (a) $\dfrac{25}{216}$; (b) $\dfrac{671}{1296} \approx 0.518.$

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P&C & Probability — Visual Concepts

Click any diagram to zoom

Bayes' Theorem — Tree Diagram (Disease Test)

Bayes' tree: $P(\text{Disease}|\text{Positive test}) = 0.0095/(0.0095+0.0495) = 0.0095/0.059 \approx 16.1\%$. The base rate (1% prevalence) dominates — even with a 95% accurate test, most positives are false positives when the disease is rare. Always draw the tree in Bayes' problems to avoid errors.

⚡

P&C & Probability — Quick Quiz

Question 1

The number of ways to arrange n people around a circular table is:

In circular arrangement, one position is fixed (rotations of same arrangement are identical). So n people → fix 1 → arrange remaining $(n-1) \to (n-1)!$ ways. Note: $(n-1)!/2$ is for necklaces/bracelets where reflections are also identical. Answer: B ✓

Question 2

If $P(A) = 0.4, P(B) = 0.3$, and A and B are independent, then $P(A\cup B) = ?$

Independent: $P(A\cap B)=P(A)\cdot P(B)=0.4\times 0.3=0.12$. $P(A\cup B)=P(A)+P(B)-P(A\cap B)=0.4+0.3-0.12=0.58$. Answer: D ✓. Note: 0.7 would be the answer if they were mutually exclusive (which is different from independent).

Question 3

The number of ways to distribute 7 identical balls into 3 distinct boxes (each box can be empty) is:

Stars and bars: n identical into r distinct boxes (empty allowed) $= C(n+r-1, r-1) = C(7+3-1, 3-1) = C(9,2) = 36$. Note: $7^3=343$ would be for 7 DISTINCT balls into 3 boxes. Answer: C ✓

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Matrices & Determinants

Matrix types · Operations · Determinants · Inverse · Cramer's Rule · System consistency

Types of Matrices

Type	Definition	Example / Property
Square	$n\times n$ matrix (rows = cols)	Determinant is defined for square matrices
Diagonal	All off-diagonal = 0; $a_{ij}=0$ for $i\ne j$	det = product of diagonal entries
Identity ($I$)	Diagonal with all 1s	$AI = IA = A$; $\det(I) = 1$
Symmetric	$A = A^T$ ($a_{ij} = a_{ji}$)	$A+A^T$ is symmetric for any A
Skew-symmetric	$A = -A^T$ ($a_{ij} = -a_{ji}$)	Diagonal = 0; $A-A^T$ is skew-symmetric; det = 0 for odd order
Orthogonal	$AA^T = I$ (i.e., $A^T = A^{-1}$)	$\det(A) = \pm 1$; preserves lengths
Idempotent	$A^2 = A$	Projection matrices; eigenvalues are 0 or 1
Nilpotent	$A^k = 0$ for some k	All eigenvalues = 0; $\det(A)=0$

Matrix Operations

Multiplication rule: $A_{m\times n} \times B_{n\times p} = AB_{m\times p}$. Row of A × Column of B. NOT commutative: $AB \ne BA$ in general.
Transpose properties: $(A+B)^T = A^T+B^T$; $(AB)^T = B^T A^T$ (order reverses); $(A^T)^T = A$; $(kA)^T = kA^T$.
Every matrix = symmetric + skew-symmetric: $A = \tfrac{1}{2}(A+A^T) + \tfrac{1}{2}(A-A^T)$.
Trace: $\operatorname{tr}(A)$ = sum of diagonal elements = sum of eigenvalues.

Determinants

$2\times 2$: $\det\begin{vmatrix} a & b \\ c & d \end{vmatrix} = ad - bc$.
$3\times 3$ (cofactor expansion along row 1): $a_{11}C_{11} + a_{12}C_{12} + a_{13}C_{13}$, where $C_{ij} = (-1)^{i+j}\cdot M_{ij}$ ($M_{ij}$ = minor).
Properties:
— Interchange two rows/cols: det changes sign.
— Two identical rows/cols: $\det = 0$.
— Multiply one row by $k$: det multiplied by $k$.
— Add $k\times$(row i) to row j: det unchanged.
— $\det(AB) = \det(A)\cdot \det(B)$.
— $\det(A^T) = \det(A)$; $\det(A^{-1}) = 1/\det(A)$; $\det(kA) = k^n\cdot \det(A)$ for $n\times n$ matrix.
Skew-symmetric matrix of odd order: $\det = 0$.

Inverse and Adjoint

Cofactor matrix: $C_{ij} = (-1)^{i+j}\cdot(\text{minor } M_{ij})$.
Adjoint: $\operatorname{adj}(A)$ = transpose of cofactor matrix = $C^T$.
Inverse: $A^{-1} = \operatorname{adj}(A)/\det(A)$, provided $\det(A) \ne 0$. ($AA^{-1} = I$.)
Identity: $A\cdot\operatorname{adj}(A) = \operatorname{adj}(A)\cdot A = \det(A)\cdot I$.
$2\times 2$ inverse shortcut: For $A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$, $A^{-1} = (1/\det)\cdot\begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$.
Cayley-Hamilton theorem: Every matrix satisfies its own characteristic equation. If $\lambda^2-5\lambda+6=0$ is char. poly., then $A^2-5A+6I=0$.

System of Linear Equations AX = B

Unique solution: $\det(A) \ne 0 \to X = A^{-1}B$.
No solution (inconsistent): $\det(A) = 0$ and $(\operatorname{adj} A)B \ne 0$.
Infinite solutions: $\det(A) = 0$ and $(\operatorname{adj} A)B = 0$.
Cramer's rule: $x_1 = D_1/D, x_2 = D_2/D, x_3 = D_3/D$ where $D = \det(A), D_i = \det$ with the i-th column replaced by B.
Homogeneous system $AX = 0$: always has trivial solution $(0,0,0)$. Non-trivial solution exists iff $\det(A) = 0$.

💡 JEE Tip: For a $3\times 3$ matrix A with elements, $\det(A)$ can often be reduced by row operations ($R_1 \to R_1-R_2$, etc.) to create zeros, then expand along a row/col with many zeros. Also: if $\det(A)=0$ for a homogeneous system, the system has infinitely many non-trivial solutions — this is the key condition in JEE questions asking "for what k does the system have non-trivial solution?"

⚠ Common Error: $\det(kA) = k^n\cdot\det(A)$ for an $n\times n$ matrix — NOT $k\cdot\det(A)$. So $\det(2A)$ for $3\times 3 = 2^3\cdot\det(A) = 8\cdot\det(A)$. Many students write $\det(2A) = 2\cdot\det(A)$ which is wrong. Also: $(AB)^T = B^T A^T$ (reverse order — same as for inverse: $(AB)^{-1} = B^{-1}A^{-1}$).

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Matrices & Determinants — Formula Sheet

Inverse and Adjoint

$A^{-1} = \operatorname{adj}(A) / \det(A)$ | $A\cdot\operatorname{adj}(A) = \det(A)\cdot I$

$\operatorname{adj}(A)$ = transpose of cofactor matrix. Exists only if $\det(A)\ne 0$ (A is non-singular).

$2\times 2$: $A^{-1} = (1/\det)\cdot\begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$ where $A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$, $\det = ad-bc$

Swap diagonal, negate off-diagonal, divide by determinant. Fastest method for $2\times 2$.

Determinant Properties

$\det(AB) = \det(A)\cdot\det(B)$ | $\det(A^T) = \det(A)$ | $\det(kA) = k^n\cdot\det(A)$

For $n\times n$ matrix. $\det(A^{-1}) = 1/\det(A)$. $\det(A^n) = [\det(A)]^n$. Skew-symmetric odd order: $\det = 0$.

System Consistency (AX=B)

$\det(A)\ne 0 \to$ unique solution $X=A^{-1}B$ (consistent)

Cramer's rule: $x_i = D_i/D$ where $D=\det(A), D_i=\det$ with i-th column replaced by B.

$\det(A)=0$ and $(\operatorname{adj} A)B=0 \to \infty$ solutions | $(\operatorname{adj} A)B\ne 0 \to$ no solution

For homogeneous ($B=0$): $\det(A)=0$ iff non-trivial solution exists.

🧮

Matrices & Determinants — Solved Examples

If $A = \begin{bmatrix} 2 & 1 \\ 5 & 3 \end{bmatrix}$, find $A^{-1}$ and verify $AA^{-1} = I$.

$\det(A) = 2\times 3 - 1\times 5 = 6-5 = 1$.
$A^{-1} = (1/\det)\cdot\begin{bmatrix} d & -b \\ -c & a \end{bmatrix} = \begin{bmatrix} 3 & -1 \\ -5 & 2 \end{bmatrix}$.
Verify: $AA^{-1} = \begin{bmatrix} 2 & 1 \\ 5 & 3 \end{bmatrix}\begin{bmatrix} 3 & -1 \\ -5 & 2 \end{bmatrix} = \begin{bmatrix} 6-5 & -2+2 \\ 15-15 & -5+6 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = I$ ✓

Solve using Cramer's rule: $2x+y = 5,\ x+3y = 10$.

$A = \begin{bmatrix} 2 & 1 \\ 1 & 3 \end{bmatrix}, B = \begin{bmatrix} 5 \\ 10 \end{bmatrix}$. $D = \det(A) = 2\times 3-1\times 1 = 5$.
$D_1 = \det\begin{vmatrix} 5 & 1 \\ 10 & 3 \end{vmatrix} = 5\times 3-1\times 10 = 15-10 = 5$. $x = D_1/D = 5/5 = 1$.
$D_2 = \det\begin{vmatrix} 2 & 5 \\ 1 & 10 \end{vmatrix} = 2\times 10-5\times 1 = 20-5 = 15$. $y = D_2/D = 15/5 = 3$.

✓ $x=1, y=3$. Verify: $2(1)+3=5$ ✓; $1+3(3)=10$ ✓. Since $D\ne 0$, unique solution exists.

For $A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}$, find $A^{-1}$ using the Cayley-Hamilton theorem. Also compute $A^2$.

Characteristic equation: $\det(A-\lambda I) = 0 \to (1-\lambda)(4-\lambda)-6 = 0 \to \lambda^2-5\lambda-2 = 0$.
Cayley-Hamilton: $A^2-5A-2I = 0$. So $A^2 = 5A+2I$.
$A^2 = 5\begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} + 2\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 5 & 10 \\ 15 & 20 \end{bmatrix} + \begin{bmatrix} 2 & 0 \\ 0 & 2 \end{bmatrix} = \begin{bmatrix} 7 & 10 \\ 15 & 22 \end{bmatrix}$.
For $A^{-1}$: from $A^2=5A+2I$, multiply both sides by $A^{-1}$: $A = 5I + 2A^{-1} \to 2A^{-1} = A-5I \to A^{-1} = (A-5I)/2$.

✓ $A^{-1} = \left(\begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}-\begin{bmatrix} 5 & 0 \\ 0 & 5 \end{bmatrix}\right)/2 = \begin{bmatrix} -2 & 1 \\ 3/2 & -1/2 \end{bmatrix}$. Verify using $A^{-1} = \operatorname{adj}(A)/\det(A)$: $\det=-2,\ \operatorname{adj}=\begin{bmatrix} 4 & -2 \\ -3 & 1 \end{bmatrix}$, $A^{-1}=\begin{bmatrix} -2 & 1 \\ 3/2 & -1/2 \end{bmatrix}$ ✓.

Evaluate $\Delta = \begin{vmatrix} 1 & a & a^2 \\ 1 & b & b^2 \\ 1 & c & c^2 \end{vmatrix}$ (the Vandermonde determinant).

Apply $R_2 \to R_2 - R_1$ and $R_3 \to R_3 - R_1$: $\Delta = \begin{vmatrix} 1 & a & a^2 \\ 0 & b-a & b^2-a^2 \\ 0 & c-a & c^2-a^2 \end{vmatrix}.$
Expand along column 1: $\Delta = (b-a)(c^2-a^2) - (c-a)(b^2-a^2).$ Factor: $(b-a)(c-a)\big[(c+a) - (b+a)\big] = (b-a)(c-a)(c-b).$
So $\Delta = (a-b)(b-c)(c-a)$ (rearranged sign). It vanishes whenever any two of $a,b,c$ are equal — the factor theorem for determinants.

✓ $\Delta = (a-b)(b-c)(c-a).$

For what value of $\lambda$ does the system $x+y+z=6,\ x+2y+3z=14,\ x+2y+\lambda z = \mu$ have (a) a unique solution, (b) infinitely many solutions, (c) no solution?

Coefficient determinant $D = \begin{vmatrix} 1 & 1 & 1 \\ 1 & 2 & 3 \\ 1 & 2 & \lambda \end{vmatrix} = 1(2\lambda-6) - 1(\lambda-3) + 1(2-2) = 2\lambda-6-\lambda+3 = \lambda-3.$
(a) Unique solution: $D \ne 0 \Rightarrow \lambda \ne 3$ (any $\mu$).
When $\lambda = 3$, rows 2 and 3 have identical LHS coefficients $(1,2,3)$; the equations are consistent only if their RHS agree: $\mu = 14$.

✓ (a) $\lambda \ne 3$: unique. (b) $\lambda = 3,\ \mu = 14$: infinitely many. (c) $\lambda = 3,\ \mu \ne 14$: no solution.

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Matrices & Determinants — Visual Concepts

Click any diagram to zoom

System Consistency — Decision Tree for AX = B

Decision tree for $AX=B$. If $\det(A)\ne 0$: unique solution $X=A^{-1}B$ (green). If $\det(A)=0$: check $(\operatorname{adj} A)B$ — if zero, infinitely many solutions (blue); if non-zero, no solution (red/inconsistent). For homogeneous $AX=0$: always consistent (zero solution), non-trivial solutions exist iff $\det(A)=0$.

⚡

Matrices & Determinants — Quick Quiz

Question 1

If A is a $3\times 3$ matrix with $\det(A) = 5$, then $\det(2A) = ?$

For an $n\times n$ matrix: $\det(kA) = k^n\cdot\det(A)$. Here $n=3, k=2$: $\det(2A) = 2^3\cdot\det(A) = 8\times 5 = 40$. A common mistake is writing $\det(2A)=2\cdot\det(A)=10$, which ignores the n-th power. Answer: C ✓

Question 2

For the system $AX=0$ (homogeneous), a non-trivial solution (other than $X=0$) exists if and only if:

For homogeneous $AX=0$: if $\det(A)\ne 0$, the only solution is $X=A^{-1}\cdot 0=0$ (trivial). For non-trivial solutions, we need $\det(A)=0$ (singular matrix). This is the key JEE condition: "find k so that the system has non-trivial solution" → set $\det(A)=0$ and solve for k. Answer: A ✓

Question 3

If A is a skew-symmetric matrix of order 3, then $\det(A) = ?$

Skew-symmetric matrix: $A = -A^T$. Taking determinants: $\det(A) = \det(-A^T) = (-1)^3\cdot\det(A^T) = -\det(A)$ (since $n=3$ is odd). So $\det(A) = -\det(A) \to 2\det(A)=0 \to \det(A) = 0$. For EVEN order, det can be non-zero. Answer: B ✓

D	Nature of roots	Comment
\(D > 0\)	Two distinct real roots	Rational if D is a perfect square and \(a, b, c \in \mathbb{Q}\)
\(D = 0\)	Two equal real roots (repeated)	\(x = -b/(2a)\)
\(D < 0\)	Two complex conjugate roots	\(\alpha = p+qi,\ \beta = p-qi\); a, b, c real

Condition	Locus
\(\|z - z_{0}\| = r\)	Circle centre \(z_0\), radius \(r\)
\(\|z - z_{1}\| = \|z - z_{2}\|\)	Perpendicular bisector of segment \(z_1 z_2\)
\(\|z - z_{1}\| + \|z - z_{2}\| = 2a\ (2a > \|z_{1}-z_{2}\|)\)	Ellipse with foci \(z_1, z_2\)
\(\arg(z - z_{0}) = \theta\)	Ray from \(z_0\) at angle θ
\(\operatorname{Im}(z) = 0\)	Real axis (x-axis)
\(\operatorname{Re}(z) = 0\)	Imaginary axis (y-axis)
\(\|z\| = 1\)	Unit circle

Series	Sum
\(\Sigma k = 1+2+\ldots+n\)	\(n(n+1)/2\)
\(\Sigma k^{2} = 1^{2}+2^{2}+\ldots+n^{2}\)	\(n(n+1)(2n+1)/6\)
\(\Sigma k^{3} = 1^{3}+2^{3}+\ldots+n^{3}\)	\([n(n+1)/2]^{2} = (\Sigma k)^{2}\)
\(\Sigma (2k-1) = 1+3+5+\ldots+(2n-1)\)	\(n^2\) (sum of first n odd numbers)

Concept	Definition	Key point
Relation R on A	Subset of A×A	Total relations = \(2^{n^2}\) for \|A\|=n
Reflexive	(a,a) ∈ R ∀a	Diagonal must be included
Symmetric	(a,b)∈R ⟹ (b,a)∈R	R = Rᵀ (symmetric matrix analog)
Transitive	(a,b),(b,c)∈R ⟹ (a,c)∈R	Most commonly tested property
Equivalence	Reflexive + Symmetric + Transitive	Partitions set into equivalence classes
One-one (injective)	\(f(x_{1})=f(x_{2}) ⟹ x_{1}=x_{2}\)	Horizontal line cuts graph at most once
Onto (surjective)	Range = Codomain	Every y in codomain has a pre-image
Bijective	One-one + Onto	Invertible; \|Domain\|=\|Codomain\|

Scenario	Count
Arrange n distinct objects	\(n!\)
n objects with p alike, q alike, r alike (\(p+q+r=n\))	\(n!/(p!q!r!)\)
Circular arrangement of n distinct objects	\((n-1)!\)
Necklace / beads on ring (flipping counts as same)	\((n-1)!/2\)
r objects from n, with repetition allowed	\(n^r\)
Words from n letters (repetition allowed)	\(n^n\) (no), \(n!\) (no repeat)

Type	Definition	Example / Property
Square	\(n\times n\) matrix (rows = cols)	Determinant is defined for square matrices
Diagonal	All off-diagonal = 0; \(a_{ij}=0\) for \(i\ne j\)	det = product of diagonal entries
Identity (\(I\))	Diagonal with all 1s	\(AI = IA = A\); \(\det(I) = 1\)
Symmetric	\(A = A^T\) (\(a_{ij} = a_{ji}\))	\(A+A^T\) is symmetric for any A
Skew-symmetric	\(A = -A^T\) (\(a_{ij} = -a_{ji}\))	Diagonal = 0; \(A-A^T\) is skew-symmetric; det = 0 for odd order
Orthogonal	\(AA^T = I\) (i.e., \(A^T = A^{-1}\))	\(\det(A) = \pm 1\); preserves lengths
Idempotent	\(A^2 = A\)	Projection matrices; eigenvalues are 0 or 1
Nilpotent	\(A^k = 0\) for some k	All eigenvalues = 0; \(\det(A)=0\)