Algebra, Trigonometry, and Geometry Revision Study Guide

Fundamental Definitions and Types of Matrices

  • Order of the Matrix: If a matrix has mm rows and nn columns, it is of order m×nm \times n.     * Example: Matrix A=(2amp;3amp;1 0amp;4amp;5)A = \begin{pmatrix} 2 & 3 & -1 \ 0 & -4 & 5 \end{pmatrix} is of order 2×32 \times 3. The element a11=2a_{11} = 2, a12=3a_{12} = 3, and a23=5a_{23} = 5.

  • The Row Matrix: A matrix consisting of exactly one row and any number of columns. Example: (1,2,3)(1, 2, 3).

  • The Column Matrix: A matrix consisting of exactly one column and any number of rows. Examples: (1 0)\begin{pmatrix} 1 \ 0 \end{pmatrix} and (3 1 5)\begin{pmatrix} 3 \ -1 \ 5 \end{pmatrix}.

  • The Square Matrix: A matrix where the number of rows equals the number of columns. Example: (3amp;2 1amp;0)\begin{pmatrix} 3 & 2 \ -1 & 0 \end{pmatrix}.

  • The Diagonal Matrix: A square matrix where all elements are zero except for the main diagonal, where at least one element is not zero. Examples: (3amp;0 0amp;1)\begin{pmatrix} 3 & 0 \ 0 & 1 \end{pmatrix} and (1amp;0amp;0 0amp;0amp;0 0amp;0amp;5)\begin{pmatrix} 1 & 0 & 0 \ 0 & 0 & 0 \ 0 & 0 & -5 \end{pmatrix}.

  • The Zero Matrix (OO): A matrix in which all elements are zero. Examples: (0amp;0amp;0)\begin{pmatrix} 0 & 0 & 0 \end{pmatrix} or (0amp;0 0amp;0)\begin{pmatrix} 0 & 0 \ 0 & 0 \end{pmatrix}.

  • The Unit Matrix (II): A diagonal matrix where all elements of the main diagonal are equal to one. Examples: (1amp;0 0amp;1)\begin{pmatrix} 1 & 0 \ 0 & 1 \end{pmatrix} and (1amp;0amp;0 0amp;1amp;0 0amp;0amp;1)\begin{pmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{pmatrix}.

Matrix Operations and Transposition

  • Matrix Transpose: The transpose of matrix AA, denoted by AtA^t, is created by replacing rows with columns and columns with rows.     * Example: If A=(5amp;2amp;1 3amp;4amp;6)A = \begin{pmatrix} 5 & -2 & 1 \ 3 & 4 & 6 \end{pmatrix}, then At=(5amp;3 2amp;4 1amp;6)A^t = \begin{pmatrix} 5 & 3 \ -2 & 4 \ 1 & 6 \end{pmatrix}.

  • Symmetric Matrix: A square matrix AA is symmetric if and only if A=AtA = A^t.     * Example: A=(1amp;2amp;5 2amp;3amp;6 5amp;6amp;4)A = \begin{pmatrix} 1 & -2 & 5 \ -2 & 3 & 6 \ 5 & 6 & 4 \end{pmatrix} is symmetric because At=AA^t = A.

  • Skew-Symmetric Matrix: A square matrix AA is skew-symmetric if and only if A=AtA = -A^t.     * Example: A=(0amp;3amp;2 3amp;0amp;4 2amp;4amp;0)A = \begin{pmatrix} 0 & -3 & 2 \ 3 & 0 & 4 \ -2 & -4 & 0 \end{pmatrix} is skew-symmetric because At=(0amp;3amp;2 3amp;0amp;4 2amp;4amp;0)=AA^t = \begin{pmatrix} 0 & 3 & -2 \ -3 & 0 & -4 \ 2 & 4 & 0 \end{pmatrix} = -A.

  • The Equality of Two Matrices: Two matrices AA and BB are equal if they have the same order and their corresponding elements are equal (aij=bija_{ij} = b_{ij}).     * Example: Given (xamp;2amp;3 yxamp;1amp;2)=(3amp;2amp;3 5amp;1amp;2)\begin{pmatrix} x & 2 & -3 \ y - x & 1 & -2 \end{pmatrix} = \begin{pmatrix} 3 & 2 & -3 \ 5 & 1 & -2 \end{pmatrix}, then x=3x = 3 and yx=5y - x = 5. Solving for yy gives y=8y = 8.

  • Scalar Multiplication: To multiply a real number by a matrix, multiply that number by every element in the matrix.     * Example: If A=(4amp;1amp;5 2amp;3amp;1)A = \begin{pmatrix} -4 & 1 & 5 \ 2 & -3 & -1 \end{pmatrix}, then 2A=(8amp;2amp;10 4amp;6amp;2)2A = \begin{pmatrix} -8 & 2 & 10 \ 4 & -6 & -2 \end{pmatrix}.

  • Adding and Subtracting Matrices: Matrices must be of the same order m×nm \times n.     * A+BA + B: A matrix where every element is the sum of corresponding elements in AA and BB.     * AB=A+(B)A - B = A + (-B): A matrix where every element is the sum of corresponding elements in AA and B-B.     * Example: A=(3amp;2amp;2 2amp;5amp;4)A = \begin{pmatrix} 3 & -2 & 2 \ -2 & 5 & -4 \end{pmatrix}, B=(0amp;3amp;4 1amp;2amp;2)B = \begin{pmatrix} 0 & -3 & 4 \ -1 & 2 & -2 \end{pmatrix}. AB=(3amp;1amp;2 1amp;3amp;2)A - B = \begin{pmatrix} 3 & 1 & -2 \ -1 & 3 & -2 \end{pmatrix}.

  • Multiplying Matrices: For matrices AA (order m×lm \times l) and BB (order r×nr \times n), product ABAB is possible if and only if l=rl = r (columns of AA = rows of BB). The resulting matrix is of order m×nm \times n.     * Example: A=(2amp;1amp;0 3amp;1amp;2)A = \begin{pmatrix} 2 & -1 & 0 \ 3 & 1 & -2 \end{pmatrix} (2×32 \times 3) and B=(3amp;2 1amp;0 4amp;5)B = \begin{pmatrix} 3 & 2 \ -1 & 0 \ 4 & 5 \end{pmatrix} (3×23 \times 2). The product ABAB is order 2×22 \times 2.     * Calculation: AB=((2)(3)+(1)(1)+(0)(4)amp;(2)(2)+(1)(0)+(0)(5) (3)(3)+(1)(1)+(2)(4)amp;(3)(2)+(1)(0)+(2)(5))=(7amp;4 0amp;4)AB = \begin{pmatrix} (2)(3)+(-1)(-1)+(0)(4) & (2)(2)+(-1)(0)+(0)(5) \ (3)(3)+(1)(-1)+(-2)(4) & (3)(2)+(1)(0)+(-2)(5) \end{pmatrix} = \begin{pmatrix} 7 & 4 \ 0 & -4 \end{pmatrix}.

  • Matrix Properties Remarks:     * (A+B)t=At+Bt(A + B)^t = A^t + B^t     * (AB)t=BtAt(AB)^t = B^t A^t

Determinants and Inverses

  • 2×22 \times 2 Determinants: If A=(a11amp;a12 a21amp;a22)A = \begin{pmatrix} a_{11} & a_{12} \ a_{21} & a_{22} \end{pmatrix}, then A=a11a22a21a12|A| = a_{11}a_{22} - a_{21}a_{12}.     * Example: (4amp;7 2amp;6)=(4×6)(2×7)=24+14=38|\begin{pmatrix} 4 & -7 \ 2 & 6 \end{pmatrix}| = (4 \times 6) - (2 \times -7) = 24 + 14 = 38.

  • 3×33 \times 3 Determinants: Expand using minor determinants along any row/column following the sign pattern (+amp;amp;+ amp;+amp; +amp;amp;+)\begin{pmatrix} + & - & + \ - & + & - \ + & - & + \end{pmatrix}.     * Formula: A=a11a22amp;a23 a32amp;a33a12a21amp;a23 a31amp;a33+a13a21amp;a22 a31amp;a32|A| = a_{11} \begin{vmatrix} a_{22} & a_{23} \ a_{32} & a_{33} \end{vmatrix} - a_{12} \begin{vmatrix} a_{21} & a_{23} \ a_{31} & a_{33} \end{vmatrix} + a_{13} \begin{vmatrix} a_{21} & a_{22} \ a_{31} & a_{32} \end{vmatrix}.     * Triangular Matrix Property: Determinant equals the product of main diagonal elements: |\begin{pmatrix} a_{11} & a_{12} & a_{13} \ 0 & a_{22} & a_{23} \ 0 & 0 & a_{33} \end{vmatrix}| = a_{11}a_{22}a_{33}.

  • Area of Triangle XYZXYZ: If X(a,b)X(a, b), Y(c,d)Y(c, d), and Z(e,f)Z(e, f), the area is A|A| where A=12aamp;bamp;1 camp;damp;1 eamp;famp;1A = \frac{1}{2} \begin{vmatrix} a & b & 1 \ c & d & 1 \ e & f & 1 \end{vmatrix}. If A=0|A| = 0, points are collinear.

  • Multiplicative Inverse (A1A^{-1}): Exists if A=Δ0|A| = \Delta \neq 0. For 2×22 \times 2 matrix A=(aamp;b camp;d)A = \begin{pmatrix} a & b \ c & d \end{pmatrix}, A1=1Δ(damp;b camp;a)A^{-1} = \frac{1}{\Delta} \begin{pmatrix} d & -b \ -c & a \end{pmatrix}.     * Example: A=(2amp;2 3amp;4)A = \begin{pmatrix} -2 & 2 \ 3 & -4 \end{pmatrix}. Δ=(2×4)(3×2)=2\Delta = (-2 \times -4) - (3 \times 2) = 2. A1=12(4amp;2 3amp;2)=(2amp;1 32amp;1)A^{-1} = \frac{1}{2} \begin{pmatrix} -4 & -2 \ -3 & -2 \end{pmatrix} = \begin{pmatrix} -2 & -1 \ -\frac{3}{2} & -1 \end{pmatrix}.

Systems of Equations

  • Simultaneous Equations (2×22 \times 2): Solving a1x+b1y=c1a_1x + b_1y = c_1 and a2x+b2y=c2a_2x + b_2y = c_2.     * Cramer's Rule: Δ=a1amp;b1 a2amp;b2\Delta = \begin{vmatrix} a_1 & b_1 \ a_2 & b_2 \end{vmatrix}, Δx=c1amp;b1 c2amp;b2\Delta_x = \begin{vmatrix} c_1 & b_1 \ c_2 & b_2 \end{vmatrix}, Δy=a1amp;c1 a2amp;c2\Delta_y = \begin{vmatrix} a_1 & c_1 \ a_2 & c_2 \end{vmatrix}. Then x=ΔxΔx = \frac{\Delta_x}{\Delta} and y=ΔyΔy = \frac{\Delta_y}{\Delta}.     * Matrix Inverse Method: Write as AX=CAX = C where A=(a1amp;b1 a2amp;b2)A = \begin{pmatrix} a_1 & b_1 \ a_2 & b_2 \end{pmatrix}, X=(x y)X = \begin{pmatrix} x \ y \end{pmatrix}, and C=(c1 c2)C = \begin{pmatrix} c_1 \ c_2 \end{pmatrix}. Then X=A1CX = A^{-1}C.

  • Simultaneous Equations (3×33 \times 3): For equations with variables x,y,zx, y, z, calculate Δ\Delta, Δx\Delta_x, Δy\Delta_y, and Δz\Delta_z using determinants. x=ΔxΔx = \frac{\Delta_x}{\Delta}, y=ΔyΔy = \frac{\Delta_y}{\Delta}, z=ΔzΔz = \frac{\Delta_z}{\Delta}.

Linear Programming and Optimization

  • Boundary Lines:     * Inequalities with \geq or \leq are represented by solid lines (boundary is part of the solution).     * Inequalities with > or < are represented by dashed lines (boundary is not part of the solution).

  • Graphical Solution Set: Determine the region for each individual inequality. The overlap of all regions is the common solution set.

  • Objective Function Optimization: To maximize or minimize P=ax+byP = ax + by, evaluate the function at the corner points (vertices) of the shaded solution region.     * Example: Region defined by x0,y0,x+2y8,3x+2y12x \geq 0, y \geq 0, x + 2y \leq 8, 3x + 2y \leq 12. Objective function P=50x+75yP = 50x + 75y. Evaluated at: A(4,0)P=200A(4, 0) \rightarrow P = 200, B(2,3)P=325B(2, 3) \rightarrow P = 325, C(0,4)P=300C(0, 4) \rightarrow P = 300. Maximum value is 325 at (2,3)(2, 3).

Trigonometric Measures and Conversion

  • Radian and Degree Relation: θradπ=x180\frac{\theta_{rad}}{\pi} = \frac{x^\circ}{180^\circ}.     * Converter: x=θrad×180πx^\circ = \theta_{rad} \times \frac{180^\circ}{\pi}.     * Converter: θrad=x×π180\theta_{rad} = x^\circ \times \frac{\pi}{180^\circ}.

  • Arc Length (ll): If rr is radius and θrad\theta_{rad} is the central angle, l=θradrl = \theta_{rad}r.

  • Trigonometric Ratios on Unit Circle: For point (x,y)(x, y), sin(θ)=y\sin(\theta) = y, cos(θ)=x\cos(\theta) = x, tan(θ)=yx\tan(\theta) = \frac{y}{x}.

  • Relation between θ\theta and θ-\theta:     * sin(θ)=sin(θ)\sin(-\theta) = -\sin(\theta), csc(θ)=csc(θ)\csc(-\theta) = -\csc(\theta).     * cos(θ)=cos(θ)\cos(-\theta) = \cos(\theta), sec(θ)=sec(θ)\sec(-\theta) = \sec(\theta).     * tan(θ)=tan(θ)\tan(-\theta) = -\tan(\theta), cot(θ)=cot(θ)\cot(-\theta) = -\cot(\theta).

Trigonometric Identities and Equations

  • Pythagorean Identities:     * sin2(θ)+cos2(θ)=1sin2(θ)=1cos2(θ)\sin^2(\theta) + \cos^2(\theta) = 1 \Rightarrow \sin^2(\theta) = 1 - \cos^2(\theta), cos2(θ)=1sin2(θ)\cos^2(\theta) = 1 - \sin^2(\theta).     * 1+tan2(θ)=sec2(θ)sec2(θ)tan2(θ)=11 + \tan^2(\theta) = \sec^2(\theta) \Rightarrow \sec^2(\theta) - \tan^2(\theta) = 1.     * cot2(θ)+1=csc2(θ)csc2(θ)cot2(θ)=1\cot^2(\theta) + 1 = \csc^2(\theta) \Rightarrow \csc^2(\theta) - \cot^2(\theta) = 1.

  • General Solutions for Trigonometric Equations (β\beta is smallest positive angle, nZn \in \mathbb{Z}):     * cos(θ)=aθ=±β+2πn\cos(\theta) = a \Rightarrow \theta = \pm\beta + 2\pi n.     * sin(θ)=aθ=β+2πn\sin(\theta) = a \Rightarrow \theta = \beta + 2\pi n or θ=(πβ)+2πn\theta = (\pi - \beta) + 2\pi n.     * tan(θ)=aθ=β+πn\tan(\theta) = a \Rightarrow \theta = \beta + \pi n.

  • Quadrantal Angle General Solutions:     * sin(θ)=0θ=πn\sin(\theta) = 0 \Rightarrow \theta = \pi n     * sin(θ)=1θ=π2+2πn\sin(\theta) = 1 \Rightarrow \theta = \frac{\pi}{2} + 2\pi n     * cos(θ)=0θ=π2+πn\cos(\theta) = 0 \Rightarrow \theta = \frac{\pi}{2} + \pi n     * cos(θ)=1θ=2πn\cos(\theta) = 1 \Rightarrow \theta = 2\pi n

  • Angles in Quadrants:     * 1st Quadrant: θ=β\theta = \beta     * 2nd Quadrant: θ=180β\theta = 180^\circ - \beta     * 3rd Quadrant: θ=180+β\theta = 180^\circ + \beta     * 4th Quadrant: θ=360β\theta = 360^\circ - \beta

Geometric Applications of Trigonometry

  • Solving Right-Angled Triangles: Use Pythagoras (AC2=AB2+BC2AC^2 = AB^2 + BC^2) and ratios (SOH CAH TOA).

  • Angles of Elevation and Depression:     * Elevation: Point of target is above horizontal line.     * Depression: Point of target is below horizontal line.     * Elevation angle of BB from AA equals depression angle of AA from BB.

  • Triangles and Polygons Areas:     * General Triangle: 12base×height\frac{1}{2} \text{base} \times \text{height} or 12absin(θ)\frac{1}{2} ab \sin(\theta).     * Equilateral Triangle: 34s2\frac{\sqrt{3}}{4}s^2.     * Regular Hexagon: 332s2\frac{3\sqrt{3}}{2}s^2.     * Quadrilateral: 12d1d2sin(θ)\frac{1}{2} d_1 d_2 \sin(\theta).     * Regular Polygon (nn sides, side xx): Area A=14nx2cot(πn)A = \frac{1}{4} nx^2 \cot(\frac{\pi}{n}).

  • Circular Sector and Segment:     * Sector Area: 12lr=12r2θrad\frac{1}{2} lr = \frac{1}{2} r^2 \theta_{rad}.     * Sector Perimeter: l+2rl + 2r.     * Segment Area: 12r2[θradsin(θ)]\frac{1}{2} r^2 [\theta_{rad} - \sin(\theta)].     * Chord Length: 2rsin(θ2)2r \sin(\frac{\theta}{2}).

Vector Geometry

  • Definitions: A directed line segment has starting/ending points and direction (AB=BA\vec{AB} = -\vec{BA}).

  • Equivalence: AB=CD\vec{AB} = \vec{CD} if they have the same norm (length) and same direction.

  • Norm: For point A(x,y)A(x, y), A=x2+y2\lVert \vec{A} \rVert = \sqrt{x^2 + y^2}.

  • Polar Form: Expressed as (A,θ)(\lVert \vec{A} \rVert, \theta).     * Components: x=Acos(θ)x = \lVert \vec{A} \rVert \cos(\theta), y=Asin(θ)y = \lVert \vec{A} \rVert \sin(\theta).

  • Unit Vector: A vector with norm 1. Fundamental unit vectors: i=(1,0)\mathbf{i} = (1, 0) and j=(0,1)\mathbf{j} = (0, 1).

  • Vector Operations:     * Geometric: AB+BC=AC\vec{AB} + \vec{BC} = \vec{AC}.     * Parallelogram rule: AB+AC=AD\vec{AB} + \vec{AC} = \vec{AD} where ADAD is the diagonal.     * Algebraic: A+B=(x1+x2,y1+y2)\vec{A} + \vec{B} = (x_1+x_2, y_1+y_2).

  • Parallel and Perpendicular Vectors:     * Parallel (A//B\vec{A} // \vec{B}): Slopes are equal (y1x1=y2x2\frac{y_1}{x_1} = \frac{y_2}{x_2}).     * Perpendicular (AB\vec{A} \perp \vec{B}): Product of slopes is 1-1 (x1x2+y1y2=0x_1x_2 + y_1y_2 = 0).

Lines and Physical Applications

  • Resultant Force: F=F1+F2+\vec{F} = \vec{F_1} + \vec{F_2} + \dots

  • Relative Velocity: Velocity of BB relative to AA is VBA=VBVA\vec{V_{BA}} = \vec{V_B} - \vec{V_A}.

  • Division of Line Segment: Point C(x,y)C(x, y) divides segment between (x1,y1)(x_1, y_1) and (x2,y2)(x_2, y_2) with ratio m2:m1m_2:m_1.     * Formula: x=m1x1+m2x2m1+m2x = \frac{m_1x_1 + m_2x_2}{m_1 + m_2}, y=m1y1+m2y2m1+m2y = \frac{m_1y_1 + m_2y_2}{m_1 + m_2}.     * Internally: m_2/m_1 > 0. Externally: m_2/m_1 < 0.

  • Slope of Straight Line:     * From points: m=y2y1x2x1m = \frac{y_2-y_1}{x_2-x_1}.     * From angle: m=tan(θ)m = \tan(\theta).     * From equation ax+by+c=0ax + by + c = 0: m=abm = -\frac{a}{b}.

  • Different Forms of Line Equation:     * Vector Form: r=A+ku\vec{r} = \vec{A} + k\vec{u}.     * Cartesian Form: yy1=m(xx1)y - y_1 = m(x - x_1).     * Intercept Form: xa+yb=1\frac{x}{a} + \frac{y}{b} = 1.

  • Angle between Lines: tan(θ)=m1m21+m1m2\tan(\theta) = |\frac{m_1 - m_2}{1 + m_1m_2}| for θ[0,π2]\theta \in [0, \frac{\pi}{2}].

  • Perpendicular Length from Point (x1,y1)(x_1, y_1) to line ax+by+c=0ax + by + c = 0:     * L=ax1+by1+ca2+b2L = \frac{|ax_1 + by_1 + c|}{\sqrt{a^2 + b^2}}.

Questions & Discussion

  • Q: Given B=(2amp;4 3amp;1)B = \begin{pmatrix} 2 &amp; 4 \ 3 &amp; 1 \end{pmatrix} and AB=IAB = I, find matrix AA.     * A: A=B1A = B^{-1}. Construct Δ=(2×1)(3×4)=10\Delta = (2 \times 1) - (3 \times 4) = -10. A=110(1amp;4 3amp;2)=(110amp;25 310amp;15)A = -\frac{1}{10} \begin{pmatrix} 1 &amp; -4 \ -3 &amp; 2 \end{pmatrix} = \begin{pmatrix} -\frac{1}{10} &amp; \frac{2}{5} \ \frac{3}{10} &amp; -\frac{1}{5} \end{pmatrix}.

  • Q: Does (xamp;4 9amp;x)\begin{pmatrix} x &amp; 4 \ 9 &amp; x \end{pmatrix} have no inverse?     * A: If Δ=x236=0\Delta = x^2 - 36 = 0, then x=±6x = \pm 6.

  • Q: Determine if points (3,5)(3, 5) and (1,5)(1, 5) satisfy x+y8x + y \leq 8.     * A: 3+5=883 + 5 = 8 \leq 8 (True) and 1+5=681 + 5 = 6 \leq 8 (True).

  • Q: Find the area of the triangle with vertices (1,2)(1, -2), (3,2)(3, 2), (5,3)(5, 3).     * A: Using determinant formula: 121(23)(2)(35)+1(910)=12141=3\frac{1}{2} |1(2-3) - (-2)(3-5) + 1(9-10)| = \frac{1}{2} |-1 - 4 - 1| = 3. Result: 3 square units.

  • Q: A person observes a plane at 1000m height with elevation angle 4040^\circ. Find the observer-plane distance.     * A: sin(40)=1000xx=1000sin(40)1556 meters\sin(40^\circ) = \frac{1000}{x} \Rightarrow x = \frac{1000}{\sin(40^\circ)} \approx 1556 \text{ meters}.

  • Q: Convert vector A=(3,3)\vec{A} = (3, -\sqrt{3}) to polar form.     * A: A=9+3=23\lVert \vec{A} \rVert = \sqrt{9 + 3} = 2\sqrt{3}. tan(θ)=33θ=330\tan(\theta) = -\frac{\sqrt{3}}{3} \Rightarrow \theta = 330^\circ. Form: (23,330)(2\sqrt{3}, 330^\circ).