Mahesh

Learning these TS Inter 1st Year Maths 1B Formulas Chapter 2 Transformation of Axes will help students to solve mathematical problems quickly.

TS Inter 1st Year Maths 1B Transformation of Axes Formulas

→ The transformation obtained, by shifting the origin to a given different point in the plane without changing the directions of coordinate axes therein is called a Translation of axes.
If the origin is shifted to (h, k) by translation of axes, then

• The coordinates of a point P(x, y) are transformed as P(x – h, y – k) and
• The equation f(x, y) = 0 of the curve is transformed as f(X + h. Y + k) = 0

→ The transformation obtained, by rotating both the coordinate axes in the plane by an equal angle, without changing the position of the origin is called a Rotation of axes.
x = X cos θ – Y sin θ, X = x cos θ – y sin θ
y = X sin θ + Y cos θ, Y = – x sin θ + y cos θ

→ To make the first degree terms absent, origin should be shifted to \(\left(\frac{h f-b g}{a b-h^2}, \frac{g h-a f}{a b-h^2}\right)\)

→ To make xy term to be absent, axes should be rotated through an angle 0 given by tan 2θ = \(\frac{2 h}{a-b}\)
⇒ θ = \(\frac{1}{2}\)tan^-1\(\left(\frac{2 h}{a-b}\right)\)

Learning these TS Inter 1st Year Maths 1B Formulas Chapter 1 Locus will help students to solve mathematical problems quickly.

TS Inter 1st Year Maths 1B Locus Formulas

→ Consider a pair of mutually perpendicular lines of reference X’ X, Y’ Y in a plane. These are called the coordinate axes and their point of intersection is called the origin denoted by ‘O’.

→ Consider a point P in the plane. Let x, denotes the perpendicular distance of P from Y – axis and yx denotes the perpendicular distance of P from X – axis. Then P is represented as ordered pair in the following quadrants.
1st Quadrant → P (x₁, y₁)
2ndQuadrant → P(-x₁, y₁)
3rd Quadrant → P (- x₁, – y₁)
4th Quadrant → P(x₁, -y₁)
The first element is called the x – coordinate (abscissa) and the second element is called the y- coordinate (ordinate).

• The distance between the points P(x₁, y₁) and Q(x₂, y₂) in the plane denoted by
  PQ = \(\sqrt{\left(x_1-x_2\right)^2+\left(y_1-y_2\right)^2}=\sqrt{(\text { difference of } x \text {-coordinates })^2+(\text { difference of } y \text {-coordinates })^2}\)
• The distance of P(x₁, y₁) from the origin (0, 0) is OP = \(\sqrt{\mathrm{x}_1^2+\mathrm{y}_1^2}\)
• The distance between [joints A (x₁, 0) and B (x₂,0) is AB = \(\sqrt{\left(x_1-x_2\right)^2+(0-0)^2}\) = (x₁ – x₂).
• The distance between points A (0, y₁) and B (0, y₂) is y₁ – y₂.

→ Section Formulae:

• The coordinates of the point ‘P’ which divides the line segment joining points A(x₁, y₁) and B(x₂, y₂) internally in the ratio m₁ : m₂ is \(\left(\frac{m_1 x_2+m_2 x_1}{m_1+m_2}, \frac{m_1 y_2+m_2 y_1}{m_1+m_2}\right)\)
• The coordinates of the point P’ which divides the line segment joining points A(x₂, y₂) and B(x₂, y₂)Externally in the ratio m₁ : m₂ is \(\left(\frac{m_1 x_2-m_2 x_1}{m_1-m_2}, \frac{m_1 y_2-m_2 y_1}{m_1-m_2}\right)\)
• Coordinates of midpoint of the line segment \(\overline{\mathrm{AB}}\) is \(\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)\)
• The points which divide a line segment in the ratio 1 : 2 or 2 : 1 are called the points of trisection.

→ (i) The area of the triangle with vertices A(x₁, y₁), B(x₂, y₂) and C(x₃, y₃) is given by
Δ = \(\frac{1}{2}\)|x₁(y₂ – y₃) + x₂(y₃ – y₁)+ x₃(y₁ – y₂) = \(\frac{1}{2}\)Σx₁(y₂ – y₃)
(ii) The area of the triangle OAB with vertices O (0, 0), A(x₁, y₁) and B(x₂, y₂) is given by
Δ = \(\frac{1}{2}\)|x₁y₂ – x₂y₁|

→ The area of the quadrilateral with vertices A(x₁, y₁), B(x₂, y₂) C(x₃, y₃) and D(x₄, y₄) is
= \(\frac{1}{2}\)|x₁(y₂ – y₄) + x₂(y₃ – y₁) + x₃(y₄ – y₂) + x₄(y₁ – y₃) = \(\frac{1}{2}\)Σx₁(y₂ – y₄)

→ In a triangle the line segment joining a vertex to the midpoint of opposite side is called the median. Point of intersection of the 3 medians of the triangle is called the centroid of the triangle denoted by G. This point G divides every median internally in the ratio 2 : 1.

→ The coordinates of centroid of the triangle having vertices A (x₁, y₁), B ( x₂, y₂) and C (x₃, y₃) is G = \(\)

→ The bisectors of internal angles of a triangle are concurrent and the point of concurrence is called in center of the triangle denoted by 1.
This is equidistant from three sides and this distance is called the in radius denoted by r. The circle drawn with I as centre and r as radius touches all the three sides internally and this circle is called the in- circle.

→ If A (x₁, y₁), B ( x₂, y₂) and C (x₃, y₃) are the vertices and a,b, and c are respectively the sides BC, CA and AB of triangle ABC, then the coordinates of the in center are
I = \(\left(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}\right)\)

→ In a triangle, one internal angular bisector and two externa! angular bisectors are concurrent j and the point of concurrence is called the Ex-centre of the triangie denoted by I₁.
This is equidistant from one side and extensions of the other sides. This distance is called i the Ex-radius of the lriangie denoted by r₁. The circle drawn with I₁ as centre and r₁ as radius touches these sides. This circle is called the Ex-circle of the triangle opposite to the vertex A. Similarly we have two more centres I₂ and I₃ with radii r₂ r₃. Coordinates of ex-centres of the triangle are given by
I₁ = \(\left(\frac{-a x_1+b x_2+c x_3}{-a+b+c}, \frac{-a y_1+b y_2+c y_3}{-a+b+c}\right)\)
I₂ = \(\left(\frac{a x_1-b x_2+c x_3}{a-b+c}, \frac{a y_1-b y_2+c y_3}{a-b+c}\right)\)
I₃ = \(\left(\frac{a x_1+b x_2-c x_3}{a+b-c}, \frac{a y_1+b y_2-c y_3}{a+b-c}\right)\)
where A (x₁, y₁), B ( x₂, y₂), C (x₃, y₃) are vertices of the triangle.

→ A set of geometric conditions is said to be consistent if there exists atleast one point satisfying the set of conditions. As an example, if A = (1, 0) and B = (3, 0) then PA + PB = 2 represents the sum of the distances of a point P from A and B is equal to 2, is a consistent condition whereas the distances of a point Q from A and B. ie., QA + QB = 1 is not consistent (∵ AB = 2).

→ Locus is the set of points (and only those points) that satisfy the given consistent geometric conditions. Hence

• Every point satisfying the given condition (s) is a point on the locus.
• Every point on the locus satisfies the given conditions).

→ Equation of locus of a point is an algebraic equation in ‘x’ and y’ satisfied by the points (x, y) on the locus alone. To get the full description of the locus, the exact part of the curve, the points of which satisfy the given geometric description need to be specified.

Learning these TS Inter 1st Year Maths 1A Formulas Chapter 5 Products of Vectors will help students to solve mathematical problems quickly.

TS Inter 1st Year Maths 1A Products of Vectors Formulas

→ The dot or scalar product of two vectors which are non zero denoted by a̅. b̅ and defineci by a̅. b̅ = |a̅| |b̅| cos θ where θ is the angle between a̅ and b̅ which is geometrically equal to product of magnitude of one of the vectors and the projection of the other on the first vector.

→ Dot product is a scalar, if a̅ = 0 or b̅ = 0 then wre define a̅ . b̅ = 0; If we write (a, b) = 9 then a̅ . b̅ = |a̅| |b̅| cos θ if a ≠ 0. b ≠ 0. a̅ .b̅ ⇔ a̅ and b̅ are perpendicular.

• Projection of b̅ on a̅ = \(\frac{|\bar{a} \cdot \bar{b}|}{|\bar{a}|}\)
• Orthogonal projection b̅ on a̅ = \(\frac{(\bar{a} \cdot \bar{b}) \bar{a}}{|\bar{a}|^2}\); a̅ ≠ 0
  (or) \(\left(\frac{\overline{\mathrm{a}} \cdot \overline{\mathrm{b}}}{|\overline{\mathrm{a}}|^2}\right)\)a̅ and its magnitude = \(\frac{|\bar{a} \cdot \bar{b}|}{|\bar{a}|}\)
• Component vector of b̅ along a̅ (or) parallel to a̅ is \(\left(\frac{\bar{a} \cdot \bar{b}}{|\bar{a}|^2}\right)\)a̅.
• Component vector of b̅ along a̅ (or) parallel to a̅ is \(\left(\frac{\bar{a} \cdot \bar{b}}{|\bar{b}|^2}\right)\) b̅ and component vector of a̅ perpendicular to b̅ is b̅ = a̅ – \(\frac{(\bar{a} \cdot \bar{b}) \bar{b}}{|\bar{b}|^2}\)

→ If i̅, j̅, k̅ are orthogonal unit vectors then i̅. j̅ = j̅.k̅ = k̅.i̅ = 0 and i̅ . i̅ = j̅ . j̅ = k̅ . k̅ = 1

→ If a̅, b̅, c̅ are any three vectors then

• (a̅ + b̅)² = |a̅|² + |b̅|² + 2(a̅ . b̅)
• (a̅ – b̅)² = |a̅|² – |b̅|² + 2(a̅ . b̅)
• (a̅ + b̅)² + (a̅ – b̅)² = 2(|a̅|² + |b̅|²)

→ If a̅ = a₁i̅ + a₂j̅ + a₃k̅ and b̅ = b₁i̅ + b₂j̅ + b₃k̅ then

• a̅.b̅ = a₁b₁ + a₂b₂ + a₃b₃
  a̅ is perpendicular to b̅ ⇔ a₁b₁ + a₂b₂ + a₃b₃ = 0
• cos θ = \(\frac{\bar{a} \cdot \bar{b}}{|\bar{a}||\bar{b}|}=\frac{a_1 b_1+a_2 b_2+a_3 b_3}{\sqrt{\Sigma a_1^2} \sqrt{\Sigma b_1^2}}\)
• a̅ is parallel to b̅ ⇔ \(\frac{a_1}{b_1}=\frac{a_2}{b_2}=\frac{a_3}{b_3}\)
• a̅.a̅ ≥ 0; |a̅.b̅| ≤ |a̅||b̅|
  |a̅ + b̅| ≤ |a̅| + |b̅|; |a̅ – b̅| ≤ |a̅| + |b̅|

→ a̅ × b̅ = |a̅||b̅| sin θ n̂ is the vector product of two vectors a̅ and b̅ and n̂ is a unit vector perpendicular to the plane containing a̅ and b̅.
sin θ = \(\frac{|\bar{a} \times \bar{b}|}{|\bar{a}||\bar{b}|}\); n = \(\frac{\bar{a} \times \bar{b}}{|\bar{a} \times \bar{b}|}\)
Also a̅ × b̅ ≠ b̅ × a̅ and a̅ × b̅ = -(b̅ × a̅)

→ (i) a̅ × a̅ = 0̅ , a̅, b̅ are parallel ⇒ a̅ × b̅ = 0

• i̅ × i̅ = j̅ × j̅ = k̅ × k̅ = 0̅
• i̅ × j̅ = k̅; j̅ × k̅ = i̅ , k̅ × i̅ = j̅

(ii) If a̅ = a₁i̅ + a₂j̅ + a₃k̅, b̅ = b₁i̅ + b₂j̅ + b₃k̅ ⇒ a̅ × b̅ = \(\left|\begin{array}{ccc}
\overline{\mathrm{i}} & \overline{\mathrm{j}} & \overline{\mathrm{k}} \\
\mathrm{a}_1 & \mathrm{a}_2 & \mathrm{a}_3 \\
\mathrm{~b}_1 & \mathrm{~b}_2 & \mathrm{~b}_3
\end{array}\right|\)

→ (i) Vector area of parallelogram with adjacent sides a̅, b̅ = |a̅ x b̅|
(ii) Vector area of parallelogram with diagonals \(\overline{\mathrm{d}}_1, \overline{\mathrm{d}}_2=\frac{1}{2}\left|\overline{\mathrm{d}}_1 \times \overline{\mathrm{d}}_2\right|\)
(iii) Area of the quadrilateral with diagonals \(\overline{\mathrm{AC}}, \overline{\mathrm{BD}}=\frac{1}{2}|\overline{\mathrm{AC}} \times \overline{\mathrm{BD}}|\)
(iv) Area of ΔABC = \(\frac{1}{2}|\overline{\mathrm{AB}} \times \overline{\mathrm{AC}}|\)

→ The vector equation of a plane in the normal form is r̅.n̅ = p, where n̅ is a unit normal vector from the origin to the plane and p is the perpendicular distance from the origin to the plane.

→ The vector equation of a plane passing through the point A(a̅) and perpendicular to n̅ is (r̅ – a̅)n̅ = 0 or r̅.n̅ = a̅. n̅

→ The angle 0 between the planes r̅,n̅₁ = p₁ and r̅ . n̅₂ = p₂ is 0 = cos^-1\(\frac{\bar{n}_1 \cdot \bar{n}_2}{\left|\bar{n}_1\right|\left|\bar{n}_2\right|}\)

→ The scalar triple product (STP) of the vectors a,b, c is (a̅ × b̅) . c̅ or a̅ . (b̅ × c̅) and is denoted by [a̅ b̅ c̅].

→ The magnitude |[a̅ b̅ c̅]| gives the volume of the parallelopiped with a̅, b̅, c̅ as its coterminus edges.

→ If a̅ = a₁i̅ + a₂ j̅ + a₃k̅, b̅ = b₁ i̅ + b₂ j̅ + b₃k̅, c̅ = c₁ i̅ + c₂ j̅ + c₃k̅ then
[a̅ b̅ c̅] = \(\left|\begin{array}{lll}
a_1 & a_2 & a_3 \\
b_1 & b_2 & b_3 \\
c_1 & c_2 & c_3
\end{array}\right|\)

• Volume of the tetrahedron with a, b,c as its coterminus edges is V = \(\frac{1}{6}\)| [a̅ b̅ c̅]|
• Volume of tetrahedron ABCD is V = \(\frac{1}{6}|[\overline{\mathrm{AB}} \overline{\mathrm{AC}} \overline{\mathrm{AD}}]|\)

→ Three vectors a, b,c are coplanar ⇔ [a̅ b̅ c̅] = 0 and a, b,c are non – coplanar ⇔ [a̅ b̅ c̅] ≠ 0

→ The shortest distance between the skew lines r̅ = a̅ + tb̅ and r̅ = c̅ + sd̅ where s, t are scalars is \(\)

→ Vector triple product of three vectors a̅, b̅, c̅ is a vector defined by
(a̅ × b̅) × c̅ = (a̅. c̅)b̅ – (c̅ . b̅)a̅ (or) a̅ × (b̅ × c̅) = (a̅. c̅)b̅ – (a̅ . b̅)c̅

→ The vector equation of a plane passing through the point A(a̅) and parallel to two non – collinear vectors b̅ and c̅ is [r̅ b̅ c̅] = [a̅ b̅ c̅]

→ The vector equation of a plane passing through A(a̅), B(b̅) and parallel to the vector is [r̅ b̅ c̅] + [r̅ c̅ a̅] = [a̅ b̅ c̅]

→ The vector equation of a plane passing through three non colilnear points A(a̅), B(b̅) and C(c̅) is [r̅ b̅ c̅] + [r̅ c̅ a̅] + [r̅ a̅ b̅] = [a̅ b̅ c̅]

→ The vector equation of the plane containing the line r̅ = a̅ + tb̅; t ∈ R and perpendicular to the plane r̅.c̅ = q is [r̅ b̅ c̅] = [a̅ b̅ c̅]

• If a̅, b̅ are two non zero and non parallel vectors then |a̅ × b̅|² = a²b² – (a̅ . b̅) = \(\left|\begin{array}{cc}
  \bar{a} \cdot \bar{a} & \bar{a} \cdot \bar{b} \\
  \bar{a} \cdot \bar{b} & \bar{b} \cdot \bar{b}
  \end{array}\right|\)
• For any vector a̅, (a̅ × i̅) + (a̅ × j̅) + (a̅ × k̅) = 2|a̅|.
• If a̅, b̅, c̅ are the position vectors of the points A, B, C then the perpendicular distance from C to the line AB is = \(\frac{|\overline{\mathrm{AC}} \times \overline{\mathrm{AB}}|}{|\overline{\mathrm{AB}}|}=\frac{|(\overline{\mathrm{b}} \times \overline{\mathrm{c}})+(\overline{\mathrm{c}} \times \overline{\mathrm{a}})+(\overline{\mathrm{a}} \times \overline{\mathrm{b}})|}{|\overline{\mathrm{b}}-\overline{\mathrm{a}}|}\).

Learning these TS Inter 1st Year Maths 1A Formulas Chapter 4 Addition of Vectors will help students to solve mathematical problems quickly.

TS Inter 1st Year Maths 1A Addition of Vectors Formulas

→ Scalar : A physical quantity having magnitude is called a scalar.
E.g. : Length, mass, area, volume, temperature, speed etc.

→ Vector : A physical quantity having both magnitude and direction is called a vector.
Ex : Displacement, velocity, acceleration, force, angular momentum.

→ Modulus of a vector : If a vector \(\overline{\mathrm{AB}}\) is denoted by a̅ then |a̅|. denote the length oi the vector of a̅ also |a̅| is called the magnitude or modulus of a vector a .

→ Collinear or parallel vectors : Vectors along the same line or along the parallel line are called collinear vectors. In figure \(\overline{\mathrm{AB}}, \overline{\mathrm{BC}}, \overline{\mathrm{CA}}\) are collinear vectors. Two vectors a̅ b̅ are parallel or collinear iff a̅ = tb̅ . t ∈ R.

→ Like vectors: Collinear or parallel vectors having the same direction are called like vectors.

→ Unlike vectors:
Collinear or parallel vectors having opposite direction are called unlike vectors.

→ Unit vector:
A vector whose modulus is unity is called a unit vector. The unit vector in the direction of vector a̅ is denoted by a̅̂. Thus modulus of |a̅̂| = 1.

• Unit vector in the direction of a̅ is \(\frac{\overline{\mathrm{a}}}{|\overline{\mathrm{a}}|}\).
• Unit vector in the opposite direction of a̅ is \(\frac{-a}{|\bar{a}|}\).

→ Position vector : If a point ‘O’ is fixed as origin in the plane and ‘A’ is any point then \(\overline{\mathrm{OA}}\) is called the position vector of A’ with respect to ‘O’.

→ Triangle law of addition of vectors: In a triangle OAB, let \(\overline{\mathrm{OA}}\) = a̅, \(\overline{\mathrm{AB}}\) = b̅ then the resultant vector \(\overline{\mathrm{OB}}\) is defined as \(\overline{\mathrm{OB}}=\overline{\mathrm{OA}}+\overline{\mathrm{AB}}\) = a̅ + b̅.
This is known as triangle law of addition of vectors.

→ Section formula:

• Let A and B be two points with position vectors a̅ and b̅ respectively. Let ‘C’ be a point dividing AB internally in the ratio m : n. The position of ‘C’ is \(\overline{O C}=\frac{m b+n a}{m+n}\)
• Let A and B be two points with position vectors a̅ and b̅. Let be a point dividing the line segment AB externally in the ratio in : n then the position vector of C is given by \(\overline{\mathrm{OC}}=\frac{\mathrm{mb}-n \bar{a}}{m-n}\)

→ The position vector of the midpoint of the line segment joining two vectors with position vector is \(\frac{\bar{a}+\bar{b}}{2}\).

→ Coplanar vectors: Two or more vectors are said to be coplanar if they lie on the same plane.

• The vectors a̅, b̅, c̅ are said to be coplanar iff [a̅ b̅ c̅] = 0.
• Four points A, B. C. D are said to be coplanar iff \(\left[\begin{array}{lll}
  \overline{\mathrm{AB}} & \overline{\mathrm{AC}} & \overline{\mathrm{AD}}
  \end{array}\right]\) = 0.
• Three vectors a̅, b̅, c̅ are said to be linearly dependent iff [a̅ b̅ c̅] = 0.
• Three vectors a̅, b, c̅ are said to be linearly independent iff [a̅ b̅ c̅] = 0.

→ Vector equations of a straight line :

• The vector equation of the straight line passing through the point A (a̅) and parallel to the vector b̅ is r̅ = a̅ + tb̅ , t ∈ R.
• The vector of the line passing through origin ‘O’ and parallel to the vector b̅ is r̅ = tb̅, t ∈ R.
  Cartesian form : Cartesian equation for the line equation passing through A (x₁, y₁, z₁) and parallel to the vector b̅ = li + mj + nk is \(\frac{x-x_1}{l}=\frac{y-y_1}{m}=\frac{z-z_1}{n}\)
• The vector equat ion of the line passing through the points A (a) and B(b) is r = (1 – t) a̅ + tb̅. t ∈ R.
  Cartesian form : Cartesian equation for the line through A(x₁, y₁, z₁) and B(x₂, y₂, z₂) is \(\frac{\mathrm{x}-\mathrm{x}_1}{\mathrm{x}_2-\mathrm{x}_1}=\frac{\mathrm{y}-\mathrm{y}_1}{\mathrm{y}_2-\mathrm{y}_1}=\frac{\mathrm{z}-\mathrm{z}_1}{\mathrm{z}_2-\mathrm{z}_1}\)

→ Vector equations of a plane :

• The vector equation of the plane passing through the points A(a̅) and parallel to the vectors b̅ & c̅ is r̅ = a̅ + tb̅ + sc̅ : t. s ∈ R.
• The equation of the plane passing through the points A(a̅).B(b̅) and parallel to the vector c is r̅ = (1 – t) a̅ + tb + sc̅ ; t, s ∈ R.
• The equation of the plane passing through three non-collinear points A(a̅), B(b̅) and C(c̅) is r̅ = (1 – t – s)a̅ + tb̅ + sc̅ ; t, s ∈ R.

→ Linear combinations : Let \(\overline{a_1}, \overline{a_2}, \ldots \ldots, \overline{a_n}\) be n vectors and l₁, l₂, …………. l_n be n scalars.
Then \(l_1 \overline{\mathrm{a}_1}+l_2 \overline{\mathrm{a}_2}, \ldots \ldots \ldots+l_{\mathrm{n}} \overline{\mathrm{a}_{\mathrm{n}}}\) is called a linear combination of \(\overline{\mathrm{a}_1}, \overline{\mathrm{a}_2}, \ldots \overline{\mathrm{a}_{\mathrm{n}}}\).

Learning these TS Inter 1st Year Maths 1A Formulas Chapter 6 Trigonometric Ratios up to Transformations will help students to solve mathematical problems quickly.

TS Inter 1st Year Maths 1A Trigonometric Ratios up to Transformations Formulas

→ sin θ, cos θ, tan θ, cot θ, cosec θ and sec θ are called trigonometric functions. The reciprocals of sin θ, cos θ, tan θ are cosec θ, sec θ and cot θ respectively.

→ The main identities are sin²θ + cos²θ =1, sec²θ – tan²θ = 1 and cosec²θ – cot²θ = 1.

→ The bounds for sin θ, cos θ and sec θ are |sin θ| ≤ 1, |cosec θ| ≤ 1 and |sec θ| ≥ 1.

→ The three main tables are given by
Table 1 :

→ Using “All Silver Tea Cups”, the following tables may be comitted to memory.
Table 2:

Table 3:

→ sin 0° = 0 = cos 90°

• sin 15° = \(\left(\frac{\sqrt{3}-1}{2 \sqrt{2}}\right)\) = cos 75°, sin 18° = \(\left(\frac{\sqrt{5}-1}{4}\right)\) = cos 72°
• sin 36° = \(\left(\frac{\sqrt{10}-2 \sqrt{5}}{4}\right)\) = cos 54°, sin 54° = \(\left(\frac{\sqrt{5}+1}{4}\right)\) = cos 36°
• sin 72° = \(\left(\frac{\sqrt{10+2 \sqrt{5}}}{4}\right)\) = cos 18°
• sin 75° = \(\left(\frac{\sqrt{3}+1}{2 \sqrt{2}}\right)\) = cos 15°
• tan 15° = 2 – √3 , tan 75° = 2 + √3

→ Any non constant function f: R → R is said to be Periodic” if there exists a real number p (* 0) such that f (x + p) = f(x) for each x e R. The least positive value of p with this period is called the “Period of f”.

→ (a) If f (x) is a periodic function with period p then f(ax +b) is also a periodic function with period \(\left(\frac{p}{|a|}\right)\)
(b) If y = f(x), y = g(x) are periodic functions with l, m as the periods respectively then h(x) = a f(x) + b g(x) where a, b R is a periodic function and LCM of {l, m} if exist is a period of h.
(c) The period of sin x, cosec x, cos x and sec x is 2π.
(d) The period of tan x and cot x is π.

→ (a) Range of a sin x + b cos x is \(\left[-\sqrt{a^2+b^2}, \sqrt{a^2+b^2}\right]\)
(b) Range of a sin x + b cos x + c is \(\left[c-\sqrt{a^2+b^2}, c+\sqrt{a^2+b^2}\right]\)

→ (a) sin(A ± B) = sin A cos B ± cos A sin B
(b) cos(A ± B) = cos A cos B ∓ sin A sin B
(c) tan (A ± B) = \(\frac{\tan A \pm \tan B}{1 \mp \tan A \tan B}\)
(d) cot (A ± B) = \(\frac{\cot A \cot B \pm 1}{\cot B \mp \cot A}\)
(e) sin (A + B) sin (A – B) = sin²A – sin²B = cos²B – cos²A
(f) cos (A + B) cos (A – B) = cos2A – sin’B = cos2B – sin2A

→ (a) sin (A + B + C) = sin A cos B cos C + cos A sin B cos C + cos A cos B sin C – sin A sin B sin C
(b) cos (A + B + C) = cos A cos B cos C + cos A sin B sin C – sin A cos B sin C – sin A sin B cos C
(c) tan (A + B + C) = \(\frac{\sum \tan A-\pi \tan A}{1-\sum \tan A \tan B}\)

→ (a) sin 2A = 2 sinA cos A, sinA = 2 sin\(\frac{\mathrm{A}}{2}\) cos \(\frac{\mathrm{A}}{2}\)
(b) cos 2A = cos²A – sin²A = 1 – 2 sin²A = 2 cos²A – 1
cos A = cos² \(\frac{\mathrm{A}}{2}\) – sin² \(\frac{\mathrm{A}}{2}\) = 1 – 2 sin²\(\frac{\mathrm{A}}{2}\) = 2 cos² \(\frac{\mathrm{A}}{2}\) – 1
(c) tan 2A = \(\frac{2 \tan A}{1-\tan ^2 A}\), tan A = \(\frac{2 \tan \frac{A}{2}}{1-\tan ^2 \frac{A}{2}}\) (\(\frac{A}{2}\), A are not odd multiples of \(\frac{\pi}{2}\))
(d) cot 2A = \(\frac{\cot ^2 A-1}{2 \cot A}\), cot A = \(\frac{\cot ^2 \frac{A}{2}-1}{2 \cot \frac{A}{2}}\) (A is not an integral multiple of π)
(e) sin 2A = \(\frac{2 \tan A}{1+\tan ^2 A}\), sin A = \(\frac{2 \tan \frac{A}{2}-1}{1+\tan ^2 \frac{A}{2}}\) (\(\frac{A}{2}\) is not an integral multiple of \(\frac{\pi}{2}\))
(f) cos 2A = \(\frac{1-\tan ^2 \mathrm{~A}}{1+\tan ^2 \mathrm{~A}}\), cos A = \(\frac{1-\tan ^2 \frac{A}{2}}{1+\tan ^2 \frac{A}{2}}\)

→ (a) sin 3A = 3 sin A – 4 sin³A
(b) cos 3A = 4cos³A – 3 cos A
(c) tan 3A = \(\frac{3 \tan A-\tan ^3 A}{1-3 \tan ^2 A}\)
(d) cot 3A = \(\frac{3 \cot A-\cot ^3 A}{1-3 \cot ^2 A}\)

→ Sums into formulae:
(a) sin (A + B) + sin (A – B) = 2 sin A cos B
(b) sin(A + B) – sin (A – B) = 2 cos A sin B
(c) cos (A + B) – cos (A – B) = 2 cos A cos B
(d) cos (A – B) – cos (A + B) = 2 sin A sin B

→ (a) sin C + sin D = 2 sin \(\left(\frac{C+D}{2}\right)\) cos \(\left(\frac{C-D}{2}\right)\)
(b) sin C – sin D = 2 cos\(\left(\frac{C+D}{2}\right)\) sin \(\left(\frac{C-D}{2}\right)\)
(c) cos C + cos D = 2 cos\(\left(\frac{C+D}{2}\right)\) cos \(\left(\frac{C-D}{2}\right)\)
(d) cos C – cos D = 2 sin \(\left(\frac{C+D}{2}\right)\) sin \(\left(\frac{D-C}{2}\right)\)

→ (a) sin A = ±\(\sqrt{\frac{1-\cos 2 A}{2}}\)
(b) cos A = ±\(\sqrt{\frac{1+\cos 2 A}{2}}\)
(c) tan A = ±\(\pm \sqrt{\frac{1-\cos 2 A}{1+\cos 2 A}}\), if A is not an odd multiple of \(\frac{\pi}{2}\)

→ (a) sin \(\frac{A}{2}=\pm \sqrt{\frac{1-\cos A}{2}}\)
(b) cos \(\frac{A}{2}=\pm \sqrt{\frac{1+\cos A}{2}}\)
(c) tan\(\frac{A}{2}=\pm \sqrt{\frac{1-\cos A}{1+\cos A}}\), if A is not an odd mutiple of π.

→ (a) sin x and cos x are continuous functions on IR.
(b) tan x is discontinuous at x = (2n – 1) \(\frac{\pi}{2}\), n ∈ Z
(c) sec x is discontinuous at x = (2x + 1) \(\frac{\pi}{2}\), n ∈ Z
(d) cosec x is discontinuous at x = nπ, n ∈ Z

Learning these TS Inter 1st Year Maths 1A Formulas Chapter 7 Trigonometric Equations will help students to solve mathematical problems quickly.

TS Inter 1st Year Maths 1A Trigonometric Equations Formulas

→ Trigonometric equation :
An equation involving trigonometric functions is called a trigonometric equation.
Ex – 1 : a cos² θ – b sin θ + c = 0
Ex – 2 : a cos θ + b sin θ + c = 0
Ex – 3 : a tan θ + b sec θ + c = 0

→ General solution (or) solution set:
The set of all values of θ which satisfy a trigonometric equation f(θ) = 0 is called general solution (or) solution set of f(θ) = 0.

→ Principle value:
1. There exists a unique value of θ in \(\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]\) satisfying sin θ = k, k ∈ R. |k| ≤ 1. This value of θ is called principle value of θ (or) principle solution of sin θ = k.
Ex :

• Principle solution of sin θ = \(\frac{1}{2}\) is \(\frac{\pi}{6}\).
• Principle solution of sin θ = \(\frac{-1}{\sqrt{2}}\) is \(\frac{-\pi}{4}\).

2. There exists a unique value of θ in [0, n] satisfying cos θ = k. k ∈ R. |k| < 1. This value of θ is called principle value of θ (or) principle solution of cos θ = k.
Ex :

• Principle solution of cos θ = \(\frac{1}{\sqrt{2}}\) is \(\frac{\pi}{4}\).
• Principle solution of cos θ = \(\frac{-1}{2}\) is \(\frac{2 \pi}{3}\).

3. There exists a unique value of θ in \(\left(\frac{-\pi}{2}, \frac{\pi}{2}\right)\) satisfying tan θ = k, k ∈ R. This value of θ is called principle value of θ (or) principle solution of tan θ = k.
Ex :

• Principle solution of tan θ = √3 is \(\frac{\pi}{3}\).
• Principle solution of tan θ = \(\frac{-1}{\sqrt{3}}\) is \(\frac{-\pi}{6}\)

→ General solutions of trigonometric equations:

Learning these TS Inter 1st Year Maths 1A Formulas Chapter 8 Inverse Trigonometric Functions will help students to solve mathematical problems quickly.

TS Inter 1st Year Maths 1A Inverse Trigonometric Functions Formulas

→ sin^-1(sin θ) = θ, if θ ∈ \(\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]\)

→ sin (sin^-1x) = x, if x ∈ [-1, 1]

→ cos^-1(cos θ) = θ, if θ ∈ [0, π]

→ cos (cos^-1x) = x, if x ∈ [-1, 1]

→ tan^-1(tan θ) = θ, if θ ∈ \(\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]\)

→ tan(tan^-1x) = x, if x ∈ R

→ cot^-1(cot θ) = θ, if θ ∈ (0, π)

→ cot(cot^-1x) = x, if x ∈ R

→ sin^-1 (- x) = – sin^-1 x if x ∈ [-1, 1]

→ cos^-1(- x) = π – cos^-1 x if x ∈ [-1, 1]

→ tan^-1 (- x) = – tan^-1 x, if x ∈ R

→ cot^-1 (- x) = π – cot^-1 x if x ∈ R

→ sin^-1 x – cos 1 x = \(\frac{\pi}{2}\) , if x ∈ [-1, 1]

→ tan^-1x + cot^-1x = \(\frac{\pi}{2}\), for any x ∈ R

→ sec^-1x + cosec^-1x = \(\frac{\pi}{2}\) , if x ∈ (-∞, -1] ∪ [1, ∞)

→ sin^-1x = cosec^-1\(\left(\frac{1}{x}\right)\) for x ∈ [-1, 0) ∪[1, ∞)

→ cos^-1x = sec^-1\(\left(\frac{1}{x}\right)\) for x ∈ [-1, 0) ∪(0, 1]

→ cot^-1x = tan^-1\(\frac{1}{x}\) if x > 0

→ cot^-1x = + tan^-1\(\frac{1}{x}\) , if x < 0

→ sin^-1x + sin^-1y = sin^-1 (x\(\sqrt{1-y^2}\) + y\(\sqrt{1-x^2}\)) if x, y ∈ [0, 1] and x² + y² < 1

→ sin^-1x + sin^-1 y = π – sin^-1 (x\(\sqrt{1-y^2}\) + y\(\sqrt{1-x^2}\)) if x, y ∈ [0, 1] and x² + y² > 1

→ sin^-1x – sin^-1y = sin^-1(x\(\sqrt{1-y^2}\) – y\(\sqrt{1-x^2}\)) if x. y ∈ [0, 1]

→ cos^-1x + cos^-1y = cos^-1(xy – \(\sqrt{1-x^2} \sqrt{1-y^2}\)) if x, y ∈ [0, 1 ]

→ cos^-1x – cos^-1y = cos^-1(xy + \(\sqrt{1-x^2} \sqrt{1-y^2}\)) if 0 ≤ x ≤ y ≤ 1

→ tan^-1x – tan^-1y = tan^-1\(\left(\frac{x-y}{1+x y}\right)\) if x > 0, y >0 (or) x < 0, y < 0

→ 2sin^-1x = sin^-1 (2x\(\sqrt{1-\mathrm{x}^2}\))

→ 2cos^-1x = cos^-1(2x² – 1)

→ 2tan^-1x = tan^-1\(\left(\frac{2 x}{1-x^2}\right)\)

→ 3sin^-1x = sin^-1(3x -4x³)

→ 3cos^-1x = cos^-1(4x³ – 3x)

→ 3tan^-1x = tan^-1\(\left(\frac{3 x-x^3}{1-3 x^2}\right)\)

Learning these TS Inter 1st Year Maths 1A Formulas Chapter 9 Hyperbolic Functions will help students to solve mathematical problems quickly.

TS Inter 1st Year Maths 1A Hyperbolic Functions Formulas

→ Hyperbolic Functions:

→ Inverse Hyperbolic Functions:

→ Hyperbolic Identities:

• cosh²x – slnh²x = 1
• sech²x – tanh²x = 1
• coth²x – cosech²x = 1
• sinh(2x) = 2sinhx coshx
• cosh(2x) = cosh²x + sinh²x = 1 + 2sinh²x = 2cosh²x – 1

→ sinh (- x) = – sin hx

→ cosh (- x) = cosh x

→ tanh(-x)= -tanhx

→ coth(-x) = -coth x

→ cosech(-x) = -cosech x

→ sech(-x) = sech x

→ sinh (x + y) = shih x . cosh y + cosh x sin h y

→ cosh (x + y) = cosh x. cosh y + sin h x sinb y

→ sinh(x – y) sinhx.cosh y – cosh x sinh y

→ cosh(x – y)=coshx.coshy – sinhxsinhy

→ tanh (x + y) = \(\frac{\tanh x+\tanh y}{1+\tanh x \tanh y}\)

→ tanh (x – y) = \(\frac{\tanh x-\tanh y}{1-\tanh x \tanh y}\)

→ sinh3x = 3sinhx + 4sinh³x

→ cosh3x = 4cosh³x – 3coshx

→ tanh3x = \(\frac{3 \tanh x+\tanh ^3 x}{1+3 \tanh ^2 x}\)

Learning these TS Inter 1st Year Maths 1A Formulas Chapter 10 Properties of Triangles will help students to solve mathematical problems quickly.

TS Inter 1st Year Maths 1A Properties of Triangles Formulas

→ In any ΔABC, \(\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}\) = 2R where a, b, c are the lengths of sides BC, CA and AB of a triangle ABC. A, B, C are the angles at the vertices of ΔABC, and R is the circum radius of the ΔABC. This is called the SINE RULE.

→ In any ΔABC,

• a² = b² + c² – 2bc cos A
• b² = c² + a² – 2ca cos B
• c² = a² + b² – 2ab cos C is the COSINE RULE.

→ The angles A, B, C can be found by the formulae

• cos A = \(\frac{b^2+c^2-a^2}{2 b c}\)
• cos B = \(\frac{c^2+a^2-b^2}{2 c a}\)
• cos C = \(\frac{a^2+b^2-c^2}{2 a b}\)

→ In any ΔABC,

• a = b cos C + c cos C
• b = a cos C + c cos A
• c = a cos B + b cos A (projection formulae)

→ tan\(\left(\frac{B-C}{2}\right)=\frac{b-c}{b+c}\) cot\(\frac{A}{2}\) (or) tan\(\left(\frac{c-A}{2}\right)=\frac{c-a}{c+a}\)cot\(\frac{B}{2}\) (or)
tan\(\left(\frac{A-B}{2}\right)=\frac{a-b}{a+b}\) cot \(\frac{C}{2}\) is the Napier’s analogy

→ If a + b + c = 2s which is the perimeter of ΔABC then
sin\(\frac{A}{2}=\sqrt{\frac{(s-b)(s-c)}{b c}}\), sin\(\frac{B}{2}=\sqrt{\frac{(s-a)(s-c)}{a c}}\) and sin\(\frac{c}{2}=\sqrt{\frac{(s-a)(s-b)}{a b}}\)
cos\(\frac{A}{2}=\sqrt{\frac{s(s-a)}{b c}}\), cos\(\frac{\mathrm{B}}{2}=\sqrt{\frac{s(s-b)}{a c}}\), cos \(\frac{c}{2}=\sqrt{\frac{s(s-c)}{a b}}\) and
tan\(\frac{A}{2}=\sqrt{\frac{(s-b)(s-c)}{s(s-a)}}\), tan\(\frac{B}{2}=\sqrt{\frac{(s-a)(s-c)}{s(s-b)}}\), tan\(\frac{c}{2}=\sqrt{\frac{(s-a)(s-b)}{s(s-c)}}\)

→ Area of ΔABC Δ = \(\frac{1}{2}\)bc sin A = \(\frac{1}{2}\)ca sin B = \(\frac{1}{2}\)ab sin C
= \(\sqrt{s(s-a)(s-b)(s-c)}=\frac{a b c}{4 R}\) = 2R sin A sin B sin C

→ If ‘r’ is the radius of incircle of ΔABC; r₁, r₂, r₃ are the radii of excircle then
r = \(\frac{\Delta}{s}\), r₁ = \(\frac{\Delta}{s-a}\), r₂ = \(\frac{\Delta}{s-b}\) and r₃ = \(\frac{\Delta}{s-c}\)

→ Also r = 4R sin\(\frac{A}{2}\)sin\(\frac{B}{2}\)sin\(\frac{C}{2}\)

• r₁ = 4R sin\(\frac{A}{2}\)cos\(\frac{B}{2}\)cos\(\frac{C}{2}\)
• r₂ = 4R sin\(\frac{B}{2}\) cos\(\frac{C}{2}\) cos\(\frac{A}{2}\)
• r₃ = 4R sin\(\frac{C}{2}\) cos\(\frac{A}{2}\) cos\(\frac{B}{2}\)

→ In any ΔABC, \(\frac{a+b}{c}=\frac{\cos \left(\frac{A-B}{2}\right)}{\sin \frac{C}{2}}\)
\(\frac{b+c}{a}=\frac{\cos \left(\frac{B-C}{2}\right)}{\sin \frac{A}{2}}\)
and \(\frac{c+a}{b}=\frac{\cos \left(\frac{C-A}{2}\right)}{\sin \frac{B}{2}}\)

Learning these TS Inter 1st Year Maths 1A Formulas Chapter 3 Matrices will help students to solve mathematical problems quickly.

TS Inter 1st Year Maths 1A Matrices Formulas

→ Matrix: If the real or complex numbers are arranged in the form of a rectangular or square array consisting the complex numbers in horizontal and vertical lines, then that arrangement is called a matrix.
Ex: A = \(\left[\begin{array}{ccc}
1 & 2 & 4 \\
3 & 0 & -6
\end{array}\right]\), B = \(\left[\begin{array}{cc}
1 & 2 \\
4 & -3
\end{array}\right]\)

→ Order of a matrix: A matrix ‘A1 is said to be of type (or) order (or) size m × n (read as m by n), if the matrix A’ has m rows and n’ columns.

→ Square matrix: A matrix ‘A’ is said to be a square matrix if the number of rows in A is equal to the number of columns in A.
Ex: \(\left[\begin{array}{cc}
1 & -1 \\
0 & 4
\end{array}\right]\)_2×2
\(\left[\begin{array}{ccc}
2 & 0 & 1 \\
4 & -1 & 2 \\
7 & 6 & 9
\end{array}\right]\)_2×2

→ Trace of a matrix : If ‘A’ is a square matrix then the sum of elements in the principle diagonal of A’ is called trace of A’. It is denoted by tra A .
Ex: A = \(\left[\begin{array}{ccc}
2 & 0 & 1 \\
4 & -1 & 2 \\
7 & 6 & 9
\end{array}\right]\)
The elements of the principle diagonal = 2, – 1, 9
Tra (A) = 2 + (- 1) + 9 = 10.

→ Null matrix : A matrix ‘A’ is said to be a zero matrix or null matrix of every element of A is equal to zero.
Ex: O₂ = \(\left[\begin{array}{ll}
0 & 0 \\
0 & 0
\end{array}\right]\)
O_3×2 = \(\left[\begin{array}{ll}
0 & 0 \\
0 & 0 \\
0 & 0
\end{array}\right]\)

→ Upper triangular matrix: A square matrix A = [a_ij]_n×n is said to be an upper triangular matrix if a_ij, = 0, whenever i > i.
Ex: \(\left[\begin{array}{ccc}
2 & -1 & 5 \\
0 & 3 & 6 \\
0 & 0 & 1
\end{array}\right]_{3 \times 3}\)

→ Lower triangular matrix :
A square matrix A = [a_ij]_n×n is said to he a lower triangular matrix if a_ij, = 0 whenever i < j .
Ex: \(\left[\begin{array}{ccc}
2 & 0 & 0 \\
1 & 3 & 0 \\
5 & 4 & 6
\end{array}\right]_{3 \times 3}\)

→ Triangular matrix : A square matrix. A is said to be a triangular matrix ifA is either an upper triangular matrix or a lower triangular matrix.
Ex: \(\left[\begin{array}{ccc}
-1 & 0 & 0 \\
0 & 3 & 0 \\
7 & 5 & 2
\end{array}\right]_{3 \times 3}\)

→ Diagonal matrIx : A square matrix, A is said to be a diagonal matrix if A is both upper triangnlar and lower triangular matrix. (or) A square matrix in which every element is equal to zero except those of principle diagonal of the matrix Is a diagonal matrix.
Ex: \(\left[\begin{array}{ccc}
2 & 0 & 0 \\
0 & 3 & 0 \\
0 & 0 & -1
\end{array}\right]_{3 \times 3}\)

→ Scalar matrix : A diagonal matrix, A is said to be a scalar matrix if all elements in the principle diagonal are equal.
Ex: \(\left[\begin{array}{ll}
2 & 0 \\
0 & 2
\end{array}\right]_{2 \times 2}\)
\(\left[\begin{array}{lll}
3 & 0 & 0 \\
0 & 3 & 0 \\
0 & 0 & 3
\end{array}\right]_{3 \times 3}\)

→ Unit matrix : A diagonal matrix is said to be a unit matrix if every element in the principle diagonal is equal to unity. It is denoted by 1.
Ex: I₂ = \(\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]\)
I₃ = \(\left[\begin{array}{lll}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right]\)

→ Transpose of a matrix: The matrix obtained by changing the rows of a given matrix, A into columns is called transpose of ‘A’. It is denoted by A^T or A’.
Ex: A = \(\left[\begin{array}{ccc}
2 & 3 & -1 \\
1 & 2 & 3
\end{array}\right]\) then A^T = \(\left[\begin{array}{cc}
2 & 1 \\
3 & 2 \\
-1 & 3
\end{array}\right]\)

→ Symmetric matrix : A square matrix, A is said to be a symmetric matrix, if A^T = A.
Ex: If A = \(\left[\begin{array}{lll}
2 & 3 & 1 \\
3 & 4 & 5 \\
1 & 5 & 7
\end{array}\right]\), then A^T = \(\left[\begin{array}{lll}
2 & 3 & 1 \\
3 & 4 & 5 \\
1 & 5 & 7
\end{array}\right]^{\mathrm{T}}=\left[\begin{array}{lll}
2 & 3 & 1 \\
3 & 4 & 5 \\
1 & 5 & 7
\end{array}\right]\) = A
A is a symmetric matrix.

→ Skew symmetric matrix : A square matrix A’ is said to be a skew symmetric matrix, if A^T = – A.
Ex: A = \(\left[\begin{array}{ccc}
0 & 1 & -2 \\
-1 & 0 & 3 \\
2 & -3 & 0
\end{array}\right]\), then AT = \(\left[\begin{array}{ccc}
0 & 1 & -2 \\
-1 & 0 & 3 \\
2 & -3 & 0
\end{array}\right]^{\mathrm{T}}=\left[\begin{array}{ccc}
0 & -1 & 2 \\
1 & 0 & -3 \\
-2 & 3 & 0
\end{array}\right]=\left[\begin{array}{ccc}
0 & 1 & -2 \\
-1 & 0 & 3 \\
2 & -3 & 0
\end{array}\right]\) = -A
A is a skew symmetric matrix.

→ Adjoint of a matrix : The transpose of the matrix obtained by replacing the elements of a square matrix. A by the corresponding cofactors is called the adjoint matrix of A. It is denoted by Adi A or adj A.

→ Inverse of a square matrix : A sqiictre matrix A is said to be an invertible matrix, if there exists a square matrix, B such that AB = BA = I. The matrix B is called inverse of A’.
If A is a non-singular matrix, then A is invertible and A^-1 = \(\frac{{adj} A}{{det} A}\)

→ Sub matrix : A matrix obtained by deleting some rows or columns or both of a matrix is called a sub matrix of the given matrix.
Ex: If A = \(\left[\begin{array}{cc}
1 & 2 \\
-1 & 2
\end{array}\right] \cdot\left[\begin{array}{lll}
1 & 2 & 3 \\
2 & 3 & 1
\end{array}\right] \cdot\left[\begin{array}{ll}
2 & 3 \\
3 & 1 \\
2 & 0
\end{array}\right]\) then some matrices of A are \(\left[\begin{array}{cc}
1 & 2 \\
-1 & 2
\end{array}\right] \cdot\left[\begin{array}{lll}
1 & 2 & 3 \\
2 & 3 & 1
\end{array}\right] \cdot\left[\begin{array}{ll}
2 & 3 \\
3 & 1 \\
2 & 0
\end{array}\right]\), [0]

→ Rank of a matrix : Let ‘A’ be a non zero matrix. The rank of A is defined as the maximum of the orders of the non singular square sub-matrices of A. The rank of a null matrix is defined ^ as zero. The rank of a matrix A is denoted by rank [A].

→ Rank of 3 × 3 matrix : Suppose A is a non-zero 3 × 3 matrix, then

• If A is a non – singular then its rank is 3.
• If A is a singular matrix and if at least one of its 2 × 2 sub matrix is non-singular, then the rank of A is 2.
• If A is a singular matrix and every 2×2, sub matrix is also singular, then the rank of ’A’ is 1.

→ Properties of matrices :

• If A and B are two matrices of same type, then A + B = B + A.
• If A, B and C are three matrices of same type then (A + B) + C = A + (B + C).
• If confirmability is assured for the matrices A, B and C then A(BC) = (AB)C.
• If confirmability is assured for the matrices A, B and C then
  (a) A(B + C) = AB + AC
  (b) (B + C) A = BA + CA.
• If A is any matrix then (A^T)^T = A.
• If A and B are two matrices of same type then (A + B)^T = A^T + B^T.
• If A and B are two matrices for which confirmability for multiplication is assured then (AB)^T = B^T. A^T.
• If I is the identity matrix of order n then for every square matrix A of order n. AI = IA = A.
• If A is an invertible matrix then A^-1 is also invertible and (A^-1)^-1 = A.
• If A and B are two invertible matrices of same type then AB is also invertible and (AB)^-1 = B^-1. A^-1.
• If A is an invertible matrix then A is also invertible and (A^T)^-1 = (A^-1)^T

→ Methods of solving linear equations :
1. Cramer’s rule : Let a₁x + b₁y + c₁z = d₁
a₂x + b₂y + c₂z = d₂
a₃x + b₃y + c₃z = d₃
be a system of linear equations

2. Matrix inversion method : If ‘A’ is a non-singular matrix then the solution of AX’ = B is X = A^-1 B.

3. Gauss Jordan method : Let a₁x + b₁y + c₁z = d₁
a₂x + b₂y + c₂z = d₂
a₃x + b₃y + c₃z = d₃
be a system of linear equations. If the augmented matrix \(\left[\begin{array}{llll}
\mathrm{a}_1 & \mathrm{~b}_1 & \mathrm{c}_1 & \mathrm{~d}_1 \\
\mathrm{a}_2 & \mathrm{~b}_2 & \mathrm{c}_2 & \mathrm{~d}_2 \\
\mathrm{a}_3 & \mathrm{~b}_3 & \mathrm{c}_3 & \mathrm{~d}_3
\end{array}\right]\) can be reduced to the form \(\left[\begin{array}{llll}
1 & 0 & 0 & \alpha \\
0 & 1 & 0 & \beta \\
0 & 0 & 1 & \gamma
\end{array}\right]\) by using elementary row transformations then x = α, y = β, z = γ, i.e., unique solution is the solution.

i) In the above matrix \(\left[\begin{array}{llll}
1 & 0 & 0 & \alpha \\
0 & 1 & 0 & \beta \\
0 & 0 & 1 & \gamma
\end{array}\right]\) is called final matrix of the system of equations.

ii) In the final matrix, if ‘O’ is obtained in place of 1 and in the same rows last element α or β or γ is
a) ‘O’ then the system has infinite number of solutions.
b) non-zero then the system of equations has no solution.

→ The system of non-homogeneous equations AX = D has

• a unique solution if rank [A] = rank [AD] = 3
• Infinitely many solutions if rank [A] = rank [AD] < 3
• No solution if rank [A] * rank [AD],

→ The system of homogeneous equations AX = O has

• The trivial solution only if Rank [A] = 3 = The number of unknowns (variables)
• An infinite number of solutions (non-trivial solution), if Rank of A less than the number of unknowns (variables) < 3.